p$}\cr \Lambda^n(I)&=\Z_p\oplus \Z_p[G]^{({p-1\choose n}-1)/p} \hbox{ if $n$ is even and $0\le n

0$, and $F$ is the sum of all pieces of degree $(m,n)$ of $\m$ for $m<0$. The $\Z[1/2]$ form on $V$ induces a self dual $\Z[1/2]$-form and hence a $\Z_p$-form on $\m$. From now on we will write $\m[1/2]$ for this $\Z[1/2]$-form. In the preprint version of this paper I implicitly assumed that the degree $(m,n)$ piece of $\m$ was self dual and isomorphic as a module over the monster to $V_{mn}$. At the last moment I realized that this is not at all clear. The problem is that although the no-ghost theorem gives an isomorphism of both spaces tensored with $\Q$, and both spaces have $\Z[1/2]$-forms, there is no obvious reason why this isomorphism should map one $\Z[1/2]$ form to the other. (In fact for the fake monster Lie algebra the corresponding isomorphism does not preserve the integral forms.) Fortunately in this paper we only need the following weaker statement: \proclaim Assumption. If $m

0,n\in \Z}r^mq^n [V_{mn}])$$ lies in $R(1)$. Proof. The Laplace operator $\Omega=dd^*+d^*d$ acts as multiplication by $(m-1)n$ on the degree $(m,n)$ piece of $\Lambda^*(E[1/2])$, so the result follows from lemma 2.9 if we let $A_i$ be the degree $(m,n)$ piece of $\Lambda^i(E[1/2])$. This proves lemma 3.2. \proclaim Lemma 3.3. If $A$ is a $\Z_p[G]$-module with $\Tr(g|A)=c^+$, $f([A])=c^-$, (and $m

0,n\in \Z}(1-r^mq^n)^{(c^-_g(mn)+c^+_g(mn))/2} (1+r^mq^n)^{(c^+_g(mn)-c^-_g(mn))/2}$$ vanishes. In other words this power series lies in $R(1)$. Proof. We apply the homomorphism $f$ to the expression in lemma 3.2 and use lemma 3.3. We find that the expression in this proposition is equal to an element of $R(1)$ times a unit in $R(2)$, and is therefore still an element of $R(1)$. Therefore its coefficients of $r^mq^n$ vanish unless $p|mn$. This proves proposition 3.4. Proposition 3.4 can be generalized in the obvious way to any generalized Kac-Moody algebra which has a self dual $\Z_p$-form, a Weyl vector, and an integral root lattice. We can also ask whether or not the coefficients of $r^mq^n$ in proposition 3.4 vanish when $p|mn$. Some numerical calculations suggest that they usually do not. \proclaim 4.~The modular moonshine conjectures for $p\ge 13$. In this section we prove the following theorem, which completes the proof of the modular moonshine conjectures of [R section 6] (apart from a small technicality in the case $p=2$.) \proclaim Theorem 4.1. If $g\in M$ is an element of prime order $p\ge 17$ or an element of type $13A$ then $\H^1(g,V[1/2])=0$. This implies that $^gV=\H^0(g,V[1/2])=\H^*(g,V[1/2])$ is a vertex algebra whose modular character is given by Hauptmoduls, and whose homogeneous components have the characters of [R definition 2]. The proof of this theorem will occupy the rest of this section. We have to show that the numbers $c^-_g(n)$ of proposition 3.4 are equal to the numbers $c^+_g(n)$, because the difference is twice the dimension of $\H^1(g,V[1/2]_n)$. We start by summarizing what we know about these numbers. \proclaim Lemma 4.2. \item {1.} The numbers $c^-_g(n)$ and $c^+_g(n)$ satisfy the relations given in proposition 3.4. \item {2.} The numbers $c^-_g(n)$ are integers, with $c^-_g(n)\equiv c^+_g(n)\bmod 2$. \item {3.} $c^-_g(n)\ge |c^+_g(n)|$ and $(p-2)c^+_g(n)+pc^-_g(n)\le 2c^+_1(n)$, (where $c^+_1(n)$ is the coefficient of $q^n$ in the elliptic modular function). \item {4.} The numbers $c^+_g(n)$ are the coefficients of the Hauptmodul of the element $g\in M$. Proof. These properties follow easily from proposition 3.4 together with the fact that $\Z_p\otimes V_n$ has dimension $c^+_1(n)$ and is the sum of $(c^-_g(n)+c^+_g(n))/2$ copies of $\Z_p$, $(c^-_g(n)-c^+_g(n))/2$ copies of $I$, and some copies of $\Z_p[G]$. This proves lemma 4.2. We will prove theorem 4.1 by showing that if $g$ satisfies the conditions of theorem 4.1 then the conditions 1 to 4 in lemma 4.2 imply that $c^-_g(n)=c^+_g(n)$. We find a finite set of possible solutions of conditions 1, 2, and 4 of lemma 4.2, and find that for each $p$ only one of these satisfies the condition 3. (There are often several solutions not satisfying condition 3.) The proof is just a long messy calculation and the reader should not waste time looking at it. We first put the relations into a more convenient form. \proclaim Proposition 4.3. There are integers $c_{m,n}$ defined for $m,n>0$ (and $m

0,n>0}c_{m,n}r^mq^n\right) \equiv
\sum_{m>0,n>0}\sum_{d|(m,n)} {c^{(-)^d}_g(mn/d^2)\over d}r^mq^n\bmod r^p$$
\item{2.} If $(p,mn)=1$ then $c_{m+1,n}=c_{m,n+1}$.
Proof. This follows from proposition 3.4 if we multiply both sides by
$(r-q)/rq$ and then take the logarithm of both sides. This proves
proposition 4.3.
\proclaim Lemma 4.4. If $p>13$ then the values
of $c^-_g(n)$ for $n\le 21$, $c^-_g(2n)$ for $2n\le 32$, and
$c^-_g(36)$ and $c^-_g(45)$ are given by polynomials in the numbers
$c^-_g(i)$ for $i=1,2,4,5$ with coefficients in $\Z_3$.
Proof. We can evaluate the elements $c(n), n\le 21$ and $c(2n),n\le
16$ using the argument for cases 1 and 2 of lemma 4.7 below. We can
evaluate $c(36)$ by looking at the coefficient of $r^4q^9$ in
proposition 4.3 and using the fact that we know $c_{m,n}$ for $m\le
3$, $n\le 8$. Similarly we can evaluate $c(45)$ by looking at the
coefficient of $r^5q^9$. Notice that the term
$(\sum_{m,n}c_{m,n}r^mq^n)^3/3$ has 3's only in the denominators of
coefficients of $r^mq^n$ when $3|m$ and $3|n$, so we do not get
problems from this for the coefficients of $r^4q^9$ and $r^5q^9$ (but
we do get problems from this if we try to work out $c^-_g(27) $ using
the same method). This proves lemma 4.4.
If $p=13$ we run into trouble when we try to determine $c^-_g(26)$;
this is why the argument in lemma 4.4 does not work in this case. Also
the argument breaks down if we try to work out $c(27)$, because when
we look at the coefficient of $r^3q^9$ we get an extra term
$c^-_g(3)/3$ which does not have coefficients in $\Z_3$. This is why
we use the coefficients $c^-_g(1)$, $c^-_g(2)$, $c^-_g(4)$, and
$c^-_g(5)$ rather than $c^-_g(1)$, $c^-_g(2)$, $c^-_g(3)$, and
$c^-_g(5)$.
Lemma 4.4 is the reason why our arguments do not work for $p<13$
(and do not work so smoothly for $p=13$). As $p$ gets smaller
we have fewer relations to work with and it gets more difficult
to determine all coefficients in terms of the $c^-_g(m)$'s for small $m$.
It is probably possible to extend lemma 4.4 to cover some smaller primes
with a lot of effort. Fortunately it is not necessary
to do the cases $p<13$ because these cases have already been
done by explicit calculations in [B-R].
\proclaim Lemma 4.5. If $c^-_g(1)$, $c^-_g(2)$, $c^-_g(4)$, and $c^-_g(5)$
are known mod $3^n$ for some $n\ge 1$, and $p> 13$, and
$(c^-_g(1),3)=1$ if $p=13$, then these values of $c^-_g(m)$ are
determined mod $3^{n+1}$.
Proof. By lemma 4.4 we see that we can determine the values mod $3^n$
of $c^-_g(n)$ for $n\le 16$, $c^-_g(2n)$ for $2n\le 32$, and
$c^-_g(36)$ and $c^-_g(45)$.
But now if we look at the coefficient of $r^3q^{3m}$ of 3.4 with
$m=1,2,4,$ or $5$, we see that $c^-_g(9m)+c^+_g(2m)/2+c^-_g(m)/3=
c^-_g(m)^3/3 + $ (some polynomial in known $c^-_g(i)$'s with
coefficients that are $3$-adic integers). But $c^-_g(m)^3/3\bmod 3^n$
depends only on $c^-_g(m)\bmod 3^n$ if $n\ge 1$, so we can determine
$c^-_g(m)/3\bmod 3^n$ and hence $c^-_g(m)\bmod 3^{n+1}$. This proves
lemma 4.5.
This lemma is the reason that we use 3-adic rather than 2-adic
approximation. If we try to prove the lemma above for $2^{n+1}$
instead of $3^{n+1}$, all we find is that we can determine the
$c^+_g(m)$'s mod $2^{n+1}$, which is useless because we already know
these numbers.
\proclaim Proposition 4.6.
If the numbers $c^-_g(n)$ satisfy the conditions 1, 2, and 4 of lemma
4.2 then the numbers $c^-_g(1)$, $c^-_g(2)$, $c^-_g(4)$, and
$c^-_g(5)$ are congruent mod $3^{29}$ to the coefficients of $q$,
$q^2$, $q^4$, and $q^5$ of one of the following power series. (The
second column is a genus zero group whose Hauptmodul appears to be the
function with coefficients $c^-_g(n)$. This has not been checked
rigorously as it is not necessary for the proof of theorem 4.1.)
\vfill\line{}
$$\matrix{
p=71&\Gamma_0(71)+&
\quad& q^{-1} &+q &+q^2 &+q^3 &+q^4 &+2q^5 &+\cdots\cr
p=59&\Gamma_0(59)+&
\quad& q^{-1} &+q &+q^2 &+2q^3 &+2q^4 &+3q^5 &+\cdots\cr
p=47&\Gamma_0(47)+&
\quad& q^{-1} &+q &+2q^2 &+3q^3 &+3q^4 &+5q^5 &+\cdots\cr
p=47&\Gamma_0(94)+&
\quad& q^{-1} &+q & &+q^3 &+q^4 &+q^5 &+\cdots\cr
p=41&\Gamma_0(41)+&
\quad& q^{-1} &+2q &+2q^2 &+3q^3 &+4q^4 &+7q^5 &+\cdots\cr
p=41&\Gamma_0(82|2)+&
\quad& q^{-1} & & &+q^3 & &+q^5 &+\cdots\cr
p=31&\Gamma_0(31)+&
\quad& q^{-1} &+3q &+3q^2 &+6q^3 &+9q^4 &+13q^5 &+\cdots\cr
p=31&\Gamma_0(62)+&
\quad& q^{-1} &+q &+q^2 &+2q^3 &+q^4 &+3q^5 &+\cdots\cr
p=29&\Gamma_0(29)+&
\quad& q^{-1} &+3q &+4q^2 &+7q^3 &+10q^4 &+17q^5 &+\cdots\cr
p=29&\Gamma_0(58|2)+&
\quad& q^{-1} &+q & &+q^3 & &+q^5 &+\cdots\cr
p=23&\Gamma_0(23)+&
\quad& q^{-1} &+4q &+7q^2 &+13q^3 &+19q^4 &+33q^5 &+\cdots\cr
p=23&\Gamma_0(46)+23&
\quad& q^{-1} & &-q^2 &+q^3 &-q^4 &+q^5 &+\cdots\cr
p=23&\Gamma_0(46)+&
\quad& q^{-1} &+2q &+q^2 &+3q^3 &+3q^4 &+5q^5 &+\cdots\cr
p=19&\Gamma_0(19)+&
\quad& q^{-1} &+6q &+10q^2 &+21q^3 &+36q^4 &+61q^5 &+\cdots\cr
p=19&\Gamma_0(38)+&
\quad& q^{-1} &+2q &+2q^2 &+5q^3 &+4q^4 &+9q^5 &+\cdots\cr
p=19&\Gamma_0(38|2)+&
\quad& q^{-1} &+2q & &+q^3 & &+3q^5 &+\cdots\cr
p=17&\Gamma_0(17)+&
\quad& q^{-1} &+7q &+14q^2 &+29q^3 &+50q^4&+92q^5&+\cdots\cr
p=17&\Gamma_0(34)+&
\quad& q^{-1} &+3q &+2q^2 &+5q^3 &+6q^4 &+12q^5 &+\cdots\cr
p=17&\Gamma_0(34|2)+&
\quad& q^{-1} &+q & &+3q^3 & &+4q^5 &+\cdots\cr
p=13&\Gamma_0(13)+&
\quad& q^{-1} &+12q &+28q^2 &+66q^3 &+132q^4&+258q^5&+\cdots\cr
p=13&\Gamma_0(26)+&
\quad& q^{-1} &+4q &+4q^2 &+10q^3 &+12q^4 &+26q^5 &+\cdots\cr
p=13&\Gamma_0(26|2)+&
\quad& q^{-1} &+2q & &+4q^3 & &+6q^5 &+\cdots\cr
}$$
It is also possible (but unlikely) that there are other solutions for
$p=13$ for which $c^-_g(1)$ does not satisfy the inequalities
$0\le c^-_g(1)< 3^{10}$.
Proof. For each prime $p$ we use a computer to test all 81
possibilities for $c^-_g(1)$, $c^-_g(2)$, $c^-_g(4)$, and
$c^-_g(5)\bmod 3$. If $p>13$ then we can calculate the $p$-adic
expansion of all the coefficients $c^-_g(n)$ recursively using
lemma 4.5 above, and we reach a contradiction by looking at
coefficients of proposition 4.3 except in the cases above. For $p=13$
this does not quite work as we have not shown the values mod $3^n$
determine those mod $3^{n+1}$, so we can adopt the crude procedure of
just testing all $3^4$ possibilities for the $c^-_g(i)$'s mod
$3^{n+1}$ for each solution mod $3^n$ we have found, and checking to
see which of them leads to contradictions. There were some cases for
$p=13$ where this did not lead to a contradiction but did at least
lead to the conclusion that $c^-_g(1)<0$ or $c^-_g(1)\ge 3^{10}$. (If
$c^-_g(1)$ is not divisible by 3 then even when $p=13$ the numbers
$c^-_g(i)\bmod 3^n$ determine the numbers $c^-_g(n)\bmod 3^{n+1}$.
When $c^-_g(1)$ was divisible by 3, there were several cases where the
coefficient $c^-_g(5)$ was not determined mod $3^{n+1}$ by the
identities used by the computer program. However in all the cases
looked at by the computer with $3|c^-_g(1)$, the values of $c^-_g(1)$,
$c^-_g(2)$, $c^-_g(4)$, and $3c^-_g(5)\bmod 3^n$ uniquely determined
their values mod $3^{n+1}$.) This proves proposition 4.6, at least if
one believes the computer calculations. (Anyone who does not like
computer calculations in a proof is welcome to redo the calculations
by hand.)
\proclaim Lemma 4.7. If the coefficients
$c^-_g(1)$, $c^-_g(2)$, $c^-_g(4)$, and $c^-_g(5)$ are equal to the
coefficients $c^+_g(1)$, $c^+_g(2)$, $c^+_g(4)$, and $c^+_g(5)$ for
$g$ an element of prime order $p>13$ or an element of type $13A$ then
all the coefficients are determined by the relations of proposition
3.4.
Proof. This proof is just a long case by case check using induction.
We will repeatedly use the fact that the coefficients of $r^mq^n$ of
both sides of proposition 4.3 are equal.
Case 1. $c^-_g(2n)$ for $2n\not\equiv 0\bmod p$. By looking at the
coefficient of $r^2q^n$ and using the fact that $c_{2,n}=c_{1,n+1}$
(as $(p,n)=1$) we see that $c^-_g(2n)$ can be written as a polynomial
in $c(n+1)$ and $c(i)$ for $1\le i m/p$
and the free exterior algebra generated by super elements of each
degree $n$ with $0