$(2+)$-Replication and the Baby Monster

$\boldsymbol{(2+)}$-Replication and the Baby Monster††This paper is a
contribution to the Special Issue on Moonshine and String Theory. The full
collection is available at https://www.emis.de/journals/SIGMA/moonshine.html

Chris CUMMINS † and Rodrigo MATIAS ‡

C. Cummins and R. Matias

† Department of Mathematics and Statistics, Concordia University,
† 1455 de Maisonneuve Blvd Ouest, Montréal, H3G 1M8, Québec, Canada
<EMAIL_ADDRESS>
‡ Departamento de Matemática, Faculdade de Ciências e Tecnologia,
‡ Universidade de Coimbra, Portugal<EMAIL_ADDRESS>
Received October 04, 2017, in final form May 31, 2018; Published online June
16, 2018

The definitions of replicable and completely replicable functions are
intimately related to the Hecke operators for the modular group. We define the
notions of “$(2+)$-replicable” and “completely $(2+)$-replicable” functions by
considering the Hecke operators for $\Gamma_{0}(2)^{+}$. We prove that the
McKay–Thompson series for $2\cdot\mathbb{B}$, as computed by Höhn, are
completely $(2+)$-replicable.

moonshine; baby monster; replication

11F22; 11F25

##  1 Introduction

The monstrous moonshine conjectures of Conway and Norton [8], led, via rapid
developments in VOA theory [2, 16] and generalized Kac–Moody algebras [3], to
Borcherds’ proof of the conjectures [4]. In this paper our aim is to
generalize one aspect of this proof to the case of the baby monster,
$\mathbb{B}$, namely the connection between an appropriate generalization of
“complete replication” of modular functions and the power map structure of
$2\cdot\mathbb{B}$.

To explain this connection, recall the monstrous case. For this, Norton [22],
introduced the idea of replicable and completely replicable functions. A
formal $q$-series $f=q^{-1}+c_{1}q+c_{2}q^{2}+\cdots$ with rational
coefficients is said to be replicable if and only if there exist formal
$q$-series $f^{(a)}=q^{-1}+c_{1}^{(a)}q+c_{2}^{(a)}q^{2}+\cdots$,
$a=1,2,3,\dots$ such that

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{a\tau+b}{d}\right)=P_{n,f}(f(q)),$ (1.1)

where $P_{n,f}(X)$ is the $n$-th Faber polynomial of $f$. This is the unique
polynomial such that $P_{n,f}(f(q))-\frac{1}{q^{n}}$ has only positive powers
of $q$.

The series $f^{(a)}$ are called the replicates of $f$. Norton conjectured that
$f$ is replicable if and only if it either has the form $q^{-1}+cq$ or is the
$q$ expansion of a Haupmodul of certain congruence subgroups. A completely
replicable $f$ is a replicable function for which $f^{(a)}$ is also replicable
for all positive integers $a$. Koike showed that a certain set of modular
functions, which includes those occurring in monstrous moonshine, are
completely replicable and that their replicates agree with those predicted by
the monstrous moonshine conjectures.

The Hauptmoduls for genus zero groups with rational integer coefficients are
known to be replicable functions [11]. The genus zero congruence subgroups of
${\rm PSL}(2,\mathbb{R})$ have been classified in [9] and the completely
replicable functions (with rational integer coefficients) were computed in
[1].

The connection with the monster, $\mathbb{M}$, is that the trace functions of
the action of $\mathbb{M}$ on $V^{\natural}$ are completely replicable as a
consequence of the twisted denominator formula for the monster Lie algebra [4,
8, 15].

Thus we have the surprising fact that the relatively simple characterization
of completely replicable functions in a sense captures the power map structure
of the monster.

These facts have been generalized. In particular Norton [23] introduced
generalized moonshine by associating functions to pairs of commuting elements
in $\mathbb{M}$. There has been considerable progress in understanding this
phenomenon and Carnahan has announced a proof of the generalized moonshine
conjectures [5, 6, 7].

Here we take a complementary approach which aims to understand and extend
complete replicability to the case of the baby monster. The idea is that the
replication equations (1.1) were discovered by Conway and Norton by modifying
the Hecke operators for ${\rm PSL}(2,\mathbb{Z})$ to take into account the
conjectured trace functions on classes of $\mathbb{M}$ other than the class of
the identity. Instead we start here with the Hecke operators for
$\Gamma_{0}(2)^{+}$ and then look for a natural way to introduce replication
and complete replication in this context (see [8] for notation). The
motivation is that $2\cdot\mathbb{B}$ is the centralizer in $\mathbb{M}$ of an
element from the conjugacy class labeled $2A$ in the atlas and in monstrous
moonshine the class $2A$ is associated with the Hauptmodul for
$\Gamma_{0}(2)^{+}$.

This is done in Section 2 and we find a different form of replicability and
prove that it reflects the power map structure in $2\cdot\mathbb{B}$. We call
it $(2{+})$-replicability as it is motivated by the Hecke operators of
$\Gamma_{0}(2)^{+}$ (which is denoted by $(2{+})$in Conway and Norton’s
notation). Although the functional equations we use are implicit in the work
of Carnahan [5, 6, 7] and Borcherds [4, Section 10], we find that they have a
natural interpretation as $(2{+})$-replication identities.

Our main motivation for this approach is that Höhn [18] has given a proof of
the baby moonshine conjectures which closely parallels Borcherds’ monstrous
proof, but with a method which uses replicability rather than complete
replicability. Introducing complete $(2{+})$-replicability allows us to modify
part of Höhn’s proof so as to clone Borcherds’ use of complete replicability
in his monstrous proof. To do this, in Section 3 we show that a large class of
Hauptmoduls are completely $(2{+})$-replicable by working out candidate
replicates. This class includes all but 13 of the Hauptmoduls which occur in
moonshine for the baby monster. In Section 4 we prove that a completely
$(2{+})$-replicable function is completely determined by the first $5$
coefficients of the function and its replicates. In the final Section 5 we
prove that the baby monster McKay–Thompson series are completely
$(2{+})$-replicable with $(2{+})$-replicability respecting the power map
structure in $2\cdot\mathbb{B}$. This is sufficient to establish that
McKay–Thompson series are Hauptmoduls once their first 5 coefficients are
calculated, except for the 13 exceptions noted above. For these 13 cases our
proof is similar that of Höhn, but simplified somewhat since we have
additional recurrence relations satisfied by the McKay–Thompson series. This
establishes, incidentally, that these exceptions are also $(2{+})$-replicable
– however we have not found a way to include them in the cases covered in
Section 3.

More generally, it is our belief that the notions of $(2{+})$-replicability
and complete $(2{+})$-replicability will lead to a better understanding of the
connections between moonshine and Hecke operators.

##  2 Hecke operators and replication

Some references for the background material needed for this section are:
Chapter 5 in [12] for Hecke operators and [8] for the basics of moonshine and
the concept of replicability.

Let $G$ be a discrete subgroup of ${\rm PSL}(2,\mathbb{R})$ which is
commensurable with ${\rm PSL}(2,\mathbb{Z})$. We consider $\mathcal{F}(G)$ the
set of functions meromorphic on the upper half plane and at cusps that are
invariant under the action of $G$. For each element $\alpha\in{\rm
PSL}(2,\mathbb{R})$ in the commensurator of $G$ we define a Hecke operator
$T_{\alpha}\colon\mathcal{F}(G)\longrightarrow\mathcal{F}(G)$ in the following
way. Consider a decomposition $G\alpha G=\bigcup\limits_{j=1}^{n}G\gamma_{j}$
as a disjoint union, guaranteed to be finite as $\alpha$ is in the
commensurator of $G$. Define

$\displaystyle T_{\alpha}f(z)=\sum_{j=1}^{n}f(\gamma_{j}z).$

An important example is given by the Hecke operators of $G=\Gamma={\rm
PSL}(2,\mathbb{Z})$ (see, for example, [25, pp. 60–63]. Let
$\Delta=\\{\alpha\in M_{2}(\mathbb{Z})\,|\,\det(\alpha)>0\\}$ and let
${\tilde{\mathbb{T}}}_{m}$ be the sum over all double cosets
$\Gamma\alpha\Gamma$ with $\alpha\in\Delta$ and $\det(\alpha)=m$. Also define
${\mathbb{T}}_{a,d}$ to be the double coset
$\Gamma\left[\begin{smallmatrix}a&0\\\ 0&d\end{smallmatrix}\right]\Gamma$ and
set ${\mathbb{T}}_{m}={\mathbb{T}}_{1,m}$. From the theory of elementary
divisors we then have formula
${\mathbb{\tilde{T}}}_{m}=\sum{\mathbb{T}}_{a,d}$ where the sum is over all
positive $a$ and $d$ such that $ad=m$ such that $a$ divides $d$. As explained
above, the Hecke action of these (distinct) double cosets on modular functions
is obtained by expressing the double cosets as a union of left cosets (by a
slight abuse of notation we use the same symbol for both a sum of double
cosets and the corresponding operator). For example, if $f$ is a modular
function for $\Gamma$ then

$\displaystyle\tilde{\mathbb{T}}_{4}f(\tau)={\mathbb{T}}_{1,4}f(\tau)+{\mathbb{T}}_{2,2}f(\tau)=f(4\tau)+f\left(\frac{2\tau}{2}\right)+f\left(\frac{2\tau+1}{2}\right)$
$\displaystyle\hphantom{\tilde{\mathbb{T}}_{4}f(\tau)=}{}+f\left(\frac{\tau}{4}\right)+f\left(\frac{\tau+1}{4}\right)+f\left(\frac{\tau+2}{4}\right)+f\left(\frac{\tau+3}{4}\right).$
(2.1)

The relationship to the Conway–Norton replication formulas of “classical”
moonshine is as follows. When $f=f_{g}$ is a Thompson–McKay series for an
element $g$ in the monster, the Conway–Norton replication formula
corresponding to (2.1) is

$\displaystyle
f^{(4)}(4\tau)+f^{(2)}\left(\frac{2\tau}{2}\right)+f^{(2)}\left(\frac{2\tau+1}{2}\right)+f\left(\frac{\tau}{4}\right)$
$\displaystyle\qquad{}+f\left(\frac{\tau+1}{4}\right)+f\left(\frac{\tau+2}{4}\right)+f\left(\frac{\tau+3}{4}\right)=P_{4,f}(f),$
(2.2)

where $P_{4,f}(f)$ is the 4th. Faber polynomial of $f$ and the functions
$f^{(4)}=f_{g^{4}}$ and $f^{(2)}=f_{g^{2}}$ are the Thompson–McKay series for
the monstrous elements $g^{4}$ and $g^{2}$ respectively. Thus the left side of
this replication formula is a modification of the Hecke action of
${\tilde{\mathbb{T}}}_{4}$. So in this sense the Conway–Norton replication
identities say that the result of a modified Hecke operator acting on a
Thompson–McKay series is equal to a Faber polynomial in that series.

We will show that a similar phenomenon occurs if we replace the monster by
$2\cdot\mathbb{B}$ and ${\rm PSL}(2,\mathbb{Z})$ by $\Gamma_{0}(2)^{+}$. For
the $\Gamma_{0}(2)^{+}$ case we set

$\displaystyle\Delta_{2}=\left\\{\begin{pmatrix}a&b\\\
2c&d\end{pmatrix}|\,a,b,c,d\in\mathbb{Z},\,ad-2bc>0\right\\}$

and let

$\displaystyle{\bf\tilde{T}}_{m}=\sum_{\begin{subarray}{c}\alpha\in\Delta_{2}\\\
\det(\alpha)=m\end{subarray}}\Gamma_{0}(2)^{+}\alpha\Gamma_{0}(2)^{+}.$ (2.3)

Also define ${\bf T}_{a,d}$ to be the double coset
$\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\right]\Gamma_{0}(2)^{+}$ and ${\bf T}_{m}={\bf T}_{m,1}$.
As we will see ${\bf\tilde{T}}_{m}=\sum{\bf T}_{a,d}$ where in this case the
sum is over all positive integers $a$ and $d$ such that $a$ divides $d$ and
either $ad=m$ or $ad=m/2$ (the second option being absent if $m$ is odd).

As for the modular group we have an action on functions. For example

$\displaystyle{\bf\tilde{T}}_{2}f(\tau)={\bf T}_{1,2}f(\tau)+{\bf
T}_{1,1}f(\tau)$
$\displaystyle\hphantom{{\bf\tilde{T}}_{2}f(\tau)}{}=f(2\tau)+f(\tau)+f\left(\tau+\frac{1}{2}\right)+f\left(\frac{\tau}{2}\right)+f\left(\frac{\tau+1}{2}\right),$
(2.4)

where we have used Proposition 2.1 below. As we will see, there is a
“replication formula” for $2\cdot\mathbb{B}$ which involves a modification of
(2.4) in the same way that (2.2) involves a modification of (2.1). It is

$\displaystyle
f^{[2]}(2\tau)+f^{[\sqrt{2}]}(\tau)+f^{[\sqrt{2}]}\left(\tau+\frac{1}{2}\right)+f\left(\frac{\tau}{2}\right)+f\left(\frac{\tau+1}{2}\right)=P_{2,f}(f).$
(2.5)

The novelty here is not in the existence of these identities, as mentioned
above they are implicit in previous work. Nor is it the existence of the
$(2{+})$-replicates, which, as we shall see, are trace functions on
appropriate VOA-modules. Rather we believe that the key points are firstly
that a form of complete replication is restored by introducing “half step”
replicates as in (2.5). So, for example,
$\big{(}f^{[\sqrt{2}]}\big{)}^{[\sqrt{2}]}=f^{[2]}$ (which explains our chosen
normalization for the exponents of $(2{+})$-replicates). The absence of
complete replicability is a complicating factor for groups other than the
monster and introducing complete $(2{+})$-replicability means that we can give
a modified version of Höhn’s $2\cdot\mathbb{B}$ proof which is closer to
Borcherds’ monstrous proof. The second key point is that this approach
emphasizes a close connection between replication identities and Hecke
algebras of groups other than ${\rm PSL}(2,\mathbb{Z})$. A connection which,
we believe, may prove fruitful.

The purpose of the next section is to define $(2{+})$-replicability based on
the Hecke operators for $\Gamma_{0}(2)^{+}$. We then define complete
$(2{+})$-replicability and we will see that the set of completely
$(2{+})$-replicable functions includes the McKay–Thompson series for the baby
monster group.

###  2.1 Hecke operators for $\boldsymbol{\Gamma_{0}(2)^{+}}$

In this subsection we find expressions for the Hecke operators of
$\Gamma_{0}(2)^{+}$. These will be used in Section 2.2 to motivate the
definition of $(2{+})$-replication.

Let $m$ be a positive integer. We define the following sets

$\displaystyle M_{1}^{m}=\left\\{\begin{bmatrix}x&y\\\
0&z\end{bmatrix}|\,xz=m,\,0\leq y<z,\,\gcd(x,y,z)=1,\,x\ \hbox{odd}\right\\},$
$\displaystyle S_{1}^{m}=\left\\{\begin{bmatrix}x&y\\\
0&z\end{bmatrix}|\,xz=m,\,0\leq y<z,\,\gcd(x,y,z)=1,\,z\ \hbox{odd}\right\\},$
$\displaystyle S_{2}^{m}=\left\\{2^{-1/2}\begin{bmatrix}x&y\\\
0&z\end{bmatrix}|\,xz=2m,\,0\leq y<z,\,\gcd(x,y,z)=1,\,x,z\
\hbox{even}\right\\},$ $\displaystyle M_{2}^{m}=S_{1}^{m}\cup S_{2}^{m},\qquad
M^{m}=M_{1}^{m}\cup M_{2}^{m}.$ (2.6)

The main result of this subsection is the following

######  Proposition 2.1.

Let $m$ be an even positive integer then we have the following

  1. $1)$

$\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]\Gamma_{0}(2)=\bigcup_{\gamma\in
M^{m}_{1}}\Gamma_{0}(2)^{+}\gamma$,

  2. $2)$

$\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}m&0\\\
0&1\end{smallmatrix}\right]\Gamma_{0}(2)=\bigcup_{\gamma\in
M^{m}_{2}}\Gamma_{0}(2)^{+}\gamma$,

  3. $3)$

$\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}m&0\\\
0&1\end{smallmatrix}\right]\Gamma_{0}(2)^{+}=\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]\Gamma_{0}(2)^{+}=\bigcup_{\gamma\in
M^{m}}\Gamma_{0}(2)^{+}\gamma$.

The proof will proceed by several lemmas. Until Proposition 2.8, $m$ will
always be a positive even integer.

######  Lemma 2.2.

The cosets $\Gamma_{0}(2)^{+}\gamma$ for $\gamma$ in $M^{m}$ are distinct.

###### Proof.

For any $\gamma$ and $\gamma^{\prime}$ in $M^{m}$ we have that
$r=\gamma^{\prime}\gamma^{-1}$ fixes $\infty$. So if $r$ is in
$\Gamma_{0}(2)^{+}$ it must be of the form $\pm\left[\begin{smallmatrix}1&k\\\
0&1\end{smallmatrix}\right]$. A short calculation now shows that
$\gamma=\gamma^{\prime}$. ∎

In what follows, for a positive integer $n$ we define
$\psi(n)=n\prod\limits_{\begin{subarray}{c}p|n\\\ p\
\text{prime}\end{subarray}}\big{(}1-\frac{1}{p}\big{)}$.

######  Lemma 2.3.

Let $m=2^{a}b$ with $a>0$ and $b$ odd. Then we have

$\displaystyle\big{|}M_{1}^{m}\big{|}=2^{a}\psi(b)=\hbox{\rm
Index}\left(\Gamma_{0}(2),\Gamma_{0}(2)\cap\begin{bmatrix}1&0\\\
0&m\end{bmatrix}^{-1}\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\right).$

###### Proof.

First observe that each element of $M^{m}_{1}$ has the form
$\left[\begin{smallmatrix}d&y\\\ 0&2^{a}(b/d)\end{smallmatrix}\right]$ where
$d$ is a divisor of $b$ (and hence odd) and $0\leq y<2^{a}(b/d)$. The number
of such matrices is $2^{a}$ times the number of matrices of the form
$\left[\begin{smallmatrix}d&y\\\ 0&(b/d)\end{smallmatrix}\right]$, but where
$0\leq y<(b/d)$. By the standard theory of Hecke operators for the modular
groups, the number of these matrices is $\psi(b)$, the index of
$\Gamma_{0}(b)$ in the modular group, and so the first equation follows.

For the second equation, if $w$ is in $\Gamma_{0}(2)^{+}$ we have that
$\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]^{-1}w\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]$ is in $\Gamma_{0}(2)$ if and only if $w$ is in
$\Gamma_{0}(2m)$. Since conjugation preserves areas of fundamental domains, it
preserves indexes and so the required index is that of $\Gamma_{0}(2m)$ in
$\Gamma_{0}(2)$ which is $\psi(2m)/3=\psi(2^{a+1}b)/3=2^{a}\psi(b)$ as
required. ∎

######  Lemma 2.4.

The cosets $\Gamma_{0}(2)^{+}\gamma$ for $\gamma$ in $M_{m}^{1}$ are closed
under right multiplication by elements of $\Gamma_{0}(2)$.

###### Proof.

The group $\Gamma_{0}(2)$ is generated by $\left[\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}1&0\\\
2&1\end{smallmatrix}\right]$. Closure under right multiplication by the former
is easy to verify.

For the latter we have

$\displaystyle\begin{bmatrix}x&y\\\ 0&z\end{bmatrix}\begin{bmatrix}1&0\\\
2&1\end{bmatrix}=\begin{bmatrix}x+2y&y\\\ 2z&z\end{bmatrix}.$

If $g=\gcd(x+2y,2z)$ then $g$ is odd since $x$ is odd. Let $a$ and $b$ be such
that $a2z+b(x+2y)=g$. Then $r=\left[\begin{smallmatrix}b&a\\\
-(2z/g)&(x+2y)/g\end{smallmatrix}\right]$ is in $\Gamma_{0}(2)$ and we have

$\displaystyle r\begin{bmatrix}x+2y&y\\\
2z&z\end{bmatrix}=\begin{bmatrix}g&(g-bx)/2\\\ 0&m/g\end{bmatrix}.$

Up to left multiplication by a translation this last element is in
$M_{1}^{m}$, as required. ∎

######  Lemma 2.5.

Part $1$ of Proposition 2.1 holds:

$\displaystyle\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)=\bigcup_{\gamma\in
M^{m}_{1}}\Gamma_{0}(2)^{+}\gamma.$

###### Proof.

By Lemma 2.4 and the fact that $\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]$ is in $M_{1}^{m}$ we can conclude that
$\bigcup\limits_{\gamma\in M^{m}_{1}}\Gamma_{0}(2)^{+}\gamma$ is a union of
$\Gamma_{0}(2)^{+}-\Gamma_{0}(2)$ double cosets which includes the double
coset $\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]\Gamma_{0}(2)$. According to [25, Proposition 3.1],
if $G_{1}$ and $G_{2}$ are commensurable subgroups of a group $G$ then for
each $g$ in $G$ the double coset $G_{1}gG_{2}$ is a union of disjoint single
cosets $\bigcup\limits_{i=1}^{d}G_{1}g_{i}$ where $d$ is the index of
$G_{2}\cap g^{-1}G_{1}g$ in $G_{2}$. Using this result and Lemma 2.3, the
double coset $\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}1&0\\\
0&m\end{smallmatrix}\right]\Gamma_{0}(2)$ is a union of $|M_{1}^{m}|$ distinct
left $\Gamma_{0}(2)^{+}$ cosets. By Lemma 2.2 the left cosets defined by
elements of $M_{1}^{m}$ are distinct and so we have equality. ∎

######  Lemma 2.6.

$|M_{1}^{m}|=|M_{2}^{m}|$.

###### Proof.

With $m=2^{a}b$ as above, we have to show $|M_{2}^{m}|=2^{a}\psi(b)$. Recall
that $M_{2}^{m}=S_{1}^{m}\cup S_{2}^{m}$ with $S_{1}^{m}$ and $S_{2}^{m}$
defined as in (2.6). We will find the sizes of these two sets.

A typical element of $S_{1}^{m}$ has the form
$\left[\begin{smallmatrix}2^{a}(b/s)&y\\\ 0&s\end{smallmatrix}\right]$ where
$s$ is a divisor of $b$ and hence odd. The elements of this matrix are coprime
if and only if the elements of the matrix $\left[\begin{smallmatrix}(b/s)&y\\\
0&s\end{smallmatrix}\right]$ are coprime and there is a one to one
correspondence between these two sets of matrices. But the number of matrices
of the latter form is $\psi(b)$ and so this is the size of $S_{1}^{m}$.

To count the elements of $S_{2}^{m}$ consider a typical element, which has the
form $2^{-1/2}\left[\begin{smallmatrix}(2m/z)&y\\\
0&z\end{smallmatrix}\right]$ with $z$ an even divisor of $2m$ such that $2m/z$
is even and $y$ is such that $0\leq y<z$ and the entries are coprime.

If we ignore the condition that $z$ and $2m/z$ are even, there are $\psi(2m)$
such matrices. But this over-counts by matrices of the form
$2^{-1/2}\left[\begin{smallmatrix}b/s&y\\\ 0&2^{a+1}s\end{smallmatrix}\right]$
of which there are $2^{a+1}\psi(b)$ (as in the proof of Lemma 2.3), and also
matrices of the form $2^{-1/2}\left[\begin{smallmatrix}2^{a+1}b/s&y\\\
0&s\end{smallmatrix}\right]$ of which there are $\psi(b)$ (as in case of
$S_{1}^{M}$ above). This gives
$|S_{2}^{m}|=\psi(2^{a+1}b)-2^{a+1}\psi(b)-\psi(b)=3\times
2^{a}\psi(b)-2^{a+1}\psi(b)-\psi(b)=2^{a}\psi(b)-\psi(b)$. So finally
$|M_{2}^{m}|=2^{a}\psi(b)$ as required. ∎

Let $w_{2}=2^{-1/2}\left[\begin{smallmatrix}0&-1\\\
2&0\end{smallmatrix}\right]$.

######  Lemma 2.7.

$\displaystyle\bigcup_{\gamma\in
M_{1}^{m}}\Gamma_{0}(2)^{+}\gamma=\bigcup_{\gamma\in
M_{2}^{m}}\Gamma_{0}(2)^{+}\gamma w_{2}.$

###### Proof.

By Lemmas 2.2 and 2.6 it suffices to show that each $\gamma w_{2}$ with
$\gamma$ in $M_{2}^{m}$ is equal, up to left multiplication by elements of
$\Gamma_{0}(2)^{+}$, to an element of $M_{1}^{m}$.

Case I: Suppose $\gamma=\left[\begin{smallmatrix}x&y\\\
0&z\end{smallmatrix}\right]$ is in $S_{1}^{m}$. Then $w_{2}^{-1}\gamma
w_{2}=\left[\begin{smallmatrix}z&0\\\ -2y&x\end{smallmatrix}\right]$. Let
$g=\gcd(z,2y)$ which is odd since $z$ is odd, and let $a$ and $b$ be such that
$az-b2y=g$. Then $r=\left[\begin{smallmatrix}a&b\\\
2y/g&z/g\end{smallmatrix}\right]$ is in $\Gamma_{0}(2)$ and
$r\left[\begin{smallmatrix}z&0\\\
-2y&x\end{smallmatrix}\right]=\left[\begin{smallmatrix}g&bx\\\
0&m/g\end{smallmatrix}\right]$ which, up to a left translation, is in
$M_{1}^{m}$.

Case II: Suppose $\gamma=2^{-1/2}\left[\begin{smallmatrix}x&y\\\
0&z\end{smallmatrix}\right]$ is in $S_{2}^{m}$. Then $\gamma
w_{2}=\left[\begin{smallmatrix}y&-x/2\\\ z&0\end{smallmatrix}\right]$. Let
$g=\gcd(z,y)$ which is odd since $y$ is odd, and let $a$ and $b$ be such that
$az+by=g$ with $a$ even (which is possible since $y/g$ is odd). Then
$r=\left[\begin{smallmatrix}b&a\\\ -z/g&y/g\end{smallmatrix}\right]$ is in
$\Gamma_{0}(2)$ and $r\left[\begin{smallmatrix}y&-x/2\\\
z&0\end{smallmatrix}\right]=\left[\begin{smallmatrix}g&-bx/2\\\
0&m/g\end{smallmatrix}\right]$ which again, up to a left translation, is in
$M_{1}^{m}$. ∎

###### Proof of Proposition 2.1.

Part 1 of the Proposition was proved in Lemma 2.5.

Part 2 follows from

$\displaystyle\bigcup_{\gamma\in
M_{2}^{m}}\Gamma_{0}(2)^{+}\gamma=\bigcup_{\gamma\in
M_{1}^{m}}\Gamma_{0}(2)^{+}\gamma w_{2}=\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)w_{2}$ $\displaystyle\hphantom{\bigcup_{\gamma\in
M_{2}^{m}}\Gamma_{0}(2)^{+}\gamma}{}=\Gamma_{0}(2)^{+}w_{2}^{-1}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}w_{2}\Gamma_{0}(2)=\Gamma_{0}(2)^{+}\begin{bmatrix}m&0\\\
0&1\end{bmatrix}\Gamma_{0}(2),$

where the first equality follows from Lemma 2.7 and the fact that
$w_{2}^{2}=-1_{2}$.

Finally, part 3 follows from

$\displaystyle\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)^{+}=\left(\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)\right)\bigcup\left(\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)w_{2}\right)$
$\displaystyle\hphantom{\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)^{+}}{}=\left(\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)\right)\bigcup\left(\Gamma_{0}(2)^{+}\begin{bmatrix}m&0\\\
0&1\end{bmatrix}\Gamma_{0}(2)\right)$
$\displaystyle\hphantom{\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)^{+}}{}=\bigg{(}\bigcup_{\gamma\in
M_{1}^{m}}\Gamma_{0}(2)^{+}\gamma\bigg{)}\bigcup\bigg{(}\bigcup_{\gamma\in
M_{2}^{m}}\Gamma_{0}(2)^{+}\gamma\bigg{)}=\bigcup_{\gamma\in
M^{m}}\Gamma_{0}(2)^{+}\gamma,$

where we have used parts 1 and 2. ∎

For the case where $m$ is odd we have from the definition (2.6) above that

$\displaystyle M^{m}=\left\\{\begin{bmatrix}x&y\\\
0&z\end{bmatrix}\,|\,xz=m,\,0\leq y<z,\,\gcd(x,y,z)=1\right\\}.$

We then have the following:

######  Proposition 2.8.

Let $m$ be an odd positive integer then

$\displaystyle\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)=\Gamma_{0}(2)^{+}\begin{bmatrix}m&0\\\
0&1\end{bmatrix}\Gamma_{0}(2)=\Gamma_{0}(2)^{+}\begin{bmatrix}m&0\\\
0&1\end{bmatrix}\Gamma_{0}(2)^{+}$
$\displaystyle\hphantom{\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)}{}=\Gamma_{0}(2)^{+}\begin{bmatrix}1&0\\\
0&m\end{bmatrix}\Gamma_{0}(2)^{+}=\bigcup_{\gamma\in
M^{m}}\Gamma_{0}(2)^{+}\gamma.$

###### Proof.

The proof is similar, but simpler, than the proof of Proposition 2.1 and we
omit it. ∎

We now wish to introduce Hecke operators for $\Gamma_{0}(2)^{+}$. In the
introduction to this section, we defined Hecke operators by starting with
double cosets and then writing these double cosets as single cosets. We will
find it convenient here to reverse this ordering and start by defining
operators ${\bf\tilde{T}}_{m}$ and ${\bf T}_{m}$ by their actions and then
deducing their characterizations as double cosets given in (2.3). Moreover, to
further simplify the presentation we only consider actions on functions (i.e.,
weight zero modular forms) so that the action of ${\bf T}_{a,d}$ is equal to
that of ${\bf T}_{a/d,1}$ which, as mentioned earlier, we write as ${\bf
T}_{a/d}$. Thus we make the following definitions:

######  Definition 2.9.

Let $m$ be a positive integer

  1. 1)

${\bf{T}}_{m}f(\tau)=\displaystyle\sum\limits_{\gamma\in M^{m}}f(\gamma(z))$,

  2. 2)

${\bf\tilde{T}}_{m}f(\tau)=\begin{cases}\displaystyle\sum\limits_{\begin{subarray}{c}xz=m\\\
0\leq y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right),&m\ \hbox{\rm odd},\\\
\displaystyle\sum_{\begin{subarray}{c}xz=m\\\ 0\leq
y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right)+\sum_{\begin{subarray}{c}xz=2m\\\
x,z\ {\rm even}\\\ 0\leq y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right),&m\
\hbox{\rm even}.\end{cases}$

By Proposition 2.1 this definition of the action ${\bf T}_{m}$ is the same as
the action of the double coset
$\Gamma_{0}(2)^{+}\left[\begin{smallmatrix}m&0\\\
0&1\end{smallmatrix}\right]\Gamma_{0}(2)^{+}$ as required. Moreover, if we
observe that every element of $\Delta_{2}$ is, up to left multiplication by an
element of $\Gamma_{0}(2)^{+}$, equivalent to one of the matrices which occur
on the right side of part 2 of Definition 2.9 and that these matrices
represent distinct left cosets of $\Gamma_{0}(2)^{+}$, then it is
straightforward to verify that this definition of $\tilde{T}_{m}$ is
equivalent to that give in (2.3).

The relationship between ${\bf T}_{m}$ and ${\bf\tilde{T}}_{m}$ is given by
the following:

######  Proposition 2.10.

Let $m=2^{\alpha}\beta$ be a positive integer where $\beta$ is odd, then

$\displaystyle{\bf\tilde{T}}_{m}=\sum_{\begin{subarray}{c}r^{2}|m\\\ r\ {\rm
odd}\end{subarray}}\sum_{i=0}^{\alpha}{\bf T}_{m/(r^{2}2^{i})}.$

###### Proof.

Define first the following operators

$\displaystyle{\bf R}_{m}f(\tau)=\sum_{\begin{subarray}{c}xz=m\\\ 0\leq
y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right),\qquad$ $\displaystyle{\bf
D}_{m}f(\tau)=\sum_{\begin{subarray}{c}xz=m\\\ x\ {\rm odd}\\\ 0\leq
y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right),$ $\displaystyle{\bf
U}_{m}f(\tau)=\sum_{\begin{subarray}{c}xz=m\\\ z\ {\rm odd}\\\ 0\leq
y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right),\qquad$ $\displaystyle{\bf
O}_{m}f(\tau)=\sum_{\begin{subarray}{c}xz=m\\\ x,z\ {\rm even},\ y\ {\rm
odd}\\\ 0\leq y<z\end{subarray}}f\left(\frac{x\tau+y}{z}\right).$

Then we have

$\displaystyle{\bf\tilde{T}}_{m}=\begin{cases}{\bf R}_{m}+{\bf R}_{2m}-{\bf
U}_{2m}-{\bf D}_{2m},&m\ \hbox{\rm even},\\\ {\bf R}_{m},&m\ \hbox{\rm
odd},\end{cases}$ (2.7)

since when $m$ is even the second term of ${\bf\tilde{T}}_{m}$ excludes the
cases when $x$ is odd and when $z$ is odd which are disjoint.

We also have

$\displaystyle{\bf O}_{4m}={\bf R}_{4m}-{\bf R}_{m}-{\bf U}_{4m}-{\bf
D}_{4m},$ (2.8)

since the cases when $x$, $y$ and $z$ are all even, when $x$ is odd and when
$z$ is odd must be excluded from the sum in ${\bf R}_{4m}$ and these are
disjoint.

If $m$ is odd then

$\displaystyle{\bf O}_{2m}=0={\bf R}_{2m}-{\bf U}_{2m}-{\bf D}_{2m},$ (2.9)

since in this case either $x$ or $z$ is odd (but not both).

For ${\bf T}_{m}$ we have the following

$\displaystyle\sum_{\begin{subarray}{c}r^{2}|m\\\ r\ {\rm
odd}\end{subarray}}{\bf T}_{m/r^{2}}=\begin{cases}{\bf O}_{2m}+{\bf
U}_{m}+{\bf D}_{m},&m\ {\rm even},\\\ {\bf R}_{m},&m\ {\rm odd}.\end{cases}$

This follows from the terms in $M^{m}$ in each case and the fact that the
condition that the $gcd$ of the entries is 1 for each element of $M^{m}$. In
particular, this establishes the required result if $m$ is odd.

We also have

$\displaystyle{\bf\tilde{T}}_{2m}-{\bf\tilde{T}}_{m}={\bf O}_{4m}+{\bf
U}_{2m}+{\bf D}_{2m}.$ (2.10)

This follows from (2.7) as follows. If $m$ is even then

$\displaystyle{\bf\tilde{T}}_{2m}-{\bf\tilde{T}}_{m}={\bf R}_{2m}+{\bf
R}_{4m}-{\bf U}_{4m}-{\bf D}_{4m}-{\bf R}_{m}-{\bf R}_{2m}+{\bf U}_{2m}+{\bf
D}_{2m}$
$\displaystyle\hphantom{{\bf\tilde{T}}_{2m}-{\bf\tilde{T}}_{m}}{}={\bf
R}_{4m}-{\bf R}_{m}-{\bf U}_{4m}-{\bf D}_{4m}+{\bf U}_{2m}+{\bf D}_{2m}={\bf
O}_{4m}+{\bf U}_{2m}+{\bf D}_{2m}$

using (2.8). While if $m$ is odd then

$\displaystyle{\bf\tilde{T}}_{2m}-{\bf\tilde{T}}_{m}={\bf R}_{2m}+{\bf
R}_{4m}-{\bf U}_{4m}-{\bf D}_{4m}-{\bf R}_{m}$
$\displaystyle\hphantom{{\bf\tilde{T}}_{2m}-{\bf\tilde{T}}_{m}}{}={\bf
R}_{2m}+{\bf O}_{4m}={\bf O}_{4m}+{\bf U}_{2m}+{\bf D}_{2m}$

using (2.9).

To establish the proposition if $m$ is even, we have from (2.10)

$\displaystyle{\bf\tilde{T}}_{m}={\bf O}_{2m}+{\bf U}_{m}+{\bf
D}_{m}+{\bf\tilde{T}}_{m/2}=\sum_{r^{2}|m}{\bf
T}_{m/r^{2}}+{\bf\tilde{T}}_{m/2}.$

The result now follows by induction on $\alpha$ where $m=2^{\alpha}\beta$ with
$\beta$ odd. ∎

###  2.2 $\boldsymbol{(2{+})}$-replicability

The operators ${\bf T}_{m}$ and ${\bf\tilde{T}}_{m}$ just defined map the
field of modular functions for $\Gamma_{0}(2)^{+}$ to itself. If $T_{2A}$ is
the normalized Hauptmodul of $\Gamma_{0}(2)^{+}$ this means that ${\bf
T}_{m}(T_{2A})(z)$ is a rational function of $T_{2A}(z)$ and since ${\bf
T}_{m}(T_{2A})(z)$ has no poles in the upper half-plane this rational function
is actually a polynomial. From the power series expansion we can see that it
has to be the $m$-th Faber polynomial of $T_{2A}$. We have just said that

$\displaystyle P_{m}(T_{2A}(\tau))=\sum_{\begin{subarray}{c}ad=m\\\ 0\leq
b<d\end{subarray}}T_{2A}\left(\frac{a\tau+b}{d}\right)+\sum_{\begin{subarray}{c}ad=m\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}T_{2A}\left(\frac{2a\tau+b}{d}\right).$ (2.11)

As discussed in the introduction, we will show that the situation is entirely
analogous to the situation with ${\rm PSL}(2,\mathbb{Z})$, the monster and
“ordinary” replication. For $\Gamma_{0}(2)^{+}$ and the baby monster we will
show that the appropriate generalization is the following:

######  Definition 2.11.

A function $f$ is $(2{+})$-replicable if there are $f^{[n]}$ and
$f^{[n\sqrt{2}]}$, for $n\in\mathbb{N}$, such that

$\displaystyle P_{n,f}(f)=\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{\left[a\right]}\left(\frac{a\tau+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{\left[a\sqrt{2}\right]}\left(\frac{2a\tau+b}{d}\right).$
(2.12)

Equation (2.11) is the $(2{+})$-self-replication property of $T_{2A}$. We make
a few remarks on this definition.

######  Remark 2.12.

Given a $(2{+})$-replicable function $f$, its $(2{+})$-replicates are not
determined uniquely. For example, for $m=2$, equation (2.12) becomes

$\displaystyle
f^{[2]}(2\tau)+f^{[\sqrt{2}]}(\tau)+f^{[\sqrt{2}]}\left(\tau+\frac{1}{2}\right)+f\left(\frac{\tau}{2}\right)+f\left(\frac{\tau+1}{2}\right)=P_{2,f}(f),$

and we can see that $f^{[2]}$ is known when $f$ and $f^{[\sqrt{2}]}$ are
known. Also, the odd-power coefficients of the replicates $f^{[\sqrt{2}n]}$
can be changed freely and identity (2.12) is still true. We will see instances
of $(2{+})$-replicable functions that are $(2{+})$-replicable in different
ways, i.e., have different $(2{+})$-replicates. For example, in Table 3 the
trace functions for the $2\cdot\mathbb{B}$ classes $2e$ and $4d$ are both
$T_{4C}$, the Hauptmodul for $\Gamma_{0}(4)$ which corresponds to the class
$4C$ in the monster. However, for the classes $2e$ and $4d$, $f^{[\sqrt{2}]}$
is given by the monstrous functions $T_{2B}$ and $T_{4C}$ respectively. In
other words there are two $(2{+})$-replication structures which are compatible
with the Hauptmodul $T_{4C}$ and both of these occur for the baby monster. One
might think that for “ordinary” replication there is in principle a similar
lack of uniqueness in the definition of replicates. For example, the monstrous
classes $27A$ and $27B$ have the same trace functions (in fact they are the
only rational classes with this property). But both cube to $9B$. More
generally, as pointed out by Norton [22], the “ordinary” replicates of a
replicable function are unique. So for “ordinary” replication we have a unique
“power map” structure.

######  Remark 2.13.

If a function $f$ is $(2{+})$-replicable then it is replicable with

$\displaystyle f^{(n)}(z)=\begin{cases}f^{[n]}(z),&n\ \mbox{odd},\\\
\displaystyle
f^{[n]}(z)+f^{[\frac{n}{\sqrt{2}}]}\left(\frac{z}{2}\right)+f^{[\frac{n}{\sqrt{2}}]}\left(\frac{z+1}{2}\right),&n\
\mbox{even}.\\\ \end{cases}$

Also, we can see that if $f$ is replicable and if, for every $n$ even, we can
write

$\displaystyle
f^{(n)}(z)=f^{[n]}(z)+f^{[\frac{n}{\sqrt{2}}]}\left(\frac{z}{2}\right)+f^{[\frac{n}{\sqrt{2}}]}\left(\frac{z+1}{2}\right),$

for some $f^{[n]}$, $f^{[\frac{n}{\sqrt{2}}]}$, then $f$ is
$(2{+})$-replicable. For $n$ odd we obviously have $f^{[n]}=f^{(n)}$. This is
shown by the following simple manipulation

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)=\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{odd}\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{even}\\\ 0\leq b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)$
$\displaystyle\hphantom{\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)}{}=\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{odd}\\\ 0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{even}\\\ 0\leq b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}\right)$
$\displaystyle\hphantom{\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)=}{}+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}\frac{a}{2}]}\left(\frac{az+b}{2d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}\frac{a}{2}]}\left(\frac{az+b+d}{2d}\right)$
$\displaystyle\hphantom{\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)}{}=\sum_{\begin{subarray}{c}ad=n\\\
0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{even}\\\ 0\leq
b<2d\end{subarray}}f^{[\sqrt{2}\frac{a}{2}]}\left(\frac{2\left(\frac{a}{2}\right)z+b}{2d}\right)$
$\displaystyle\hphantom{\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{(a)}\left(\frac{az+b}{d}\right)}{}=\sum_{\begin{subarray}{c}ad=n\\\
0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}a]}\left(\frac{2az+b}{d}\right).$

Finally, if $f$ is replicable then $f$ is always $(2{+})$-replicable by
taking, for example, $f^{[n\sqrt{2}]}=0$.

Note that this is in contrast with ordinary replication for which the
replication equations uniquely fix the replicates.

######  Remark 2.14.

Because of Remark 2.13 what we are actually interested in is finding the
possible $(2{+})$-replicables for a given $(2{+})$-replicable function. For
example, we could ask if a $(2{+})$-replicable function is completely
$(2{+})$-replicable in the sense given below. There are examples of
Hauptmoduls that are not complete replicable functions but are completely
$(2{+})$-replicable. As we will see, some examples of these are $T_{4\sim b}$,
$T_{12\sim d}$ among many others (see [24] for the notation).

For the monster, since replicates corresponds to power maps, the replicable
functions satisfy a stronger property called complete replicability. Namely
that the replicate functions are themselves replicable and that taking
replicates is “commutative”. We make a corresponding definition for
$(2{+})$-replicability.

######  Definition 2.15.

We say that a function $f$ is completely $(2{+})$-replicable if

  * –

it is $(2{+})$-replicable, with replicates $f^{[n]}$, $n\in{\rm
Nature}\cup\sqrt{2}{\rm Nature}$, and

  * –

for every $n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$ the function $f^{[n]}$ is
$(2{+})$-replicable with $(2{+})$-replicates
$\big{(}f^{[n]}\big{)}^{[m]}=f^{[mn]}$, for any $m\in{\rm
Nature}\cup\sqrt{2}{\rm Nature}$.

In the following section we show that certain Hauptmoduls are completely
$(2{+})$-replicable. The list will include almost all the Hauptmoduls which
occur in moonshine for the baby monster. The proof of $(2{+})$-replicability
for the remaining baby monstrous moonshine functions will be given at the end
of Section 5.

##  3 Complete $\boldsymbol{(2+)}$-replicability and Hauptmoduls

The following result is Theorem 5.15 in Ferenbaugh’s Ph.D. Thesis [13] and
will be useful in proving that certain Hauptmoduls that are not completely
replicable are completely $(2{+})$-replicable. We refer to [8] and [14] for
the notation on the Atkin–Lehner involutions $W_{e}$ and on groups of the form
$n\,|\,h+e_{1},e_{2},\ldots$ and $n\parallel h+e_{1},e_{2},\ldots$.

######  Theorem 3.1.

Let $p$ be prime and $f$ be the Hauptmodul for some group
$np\,|\,h+e_{1},e_{2}\ldots$ with $p\nmid h$. Then

  * –

if $p$ is one of the $e_{1}$, $e_{2},\ldots$ then
$f^{(p)}(z)=f(z)+f\big{(}\frac{z}{p}\big{)}+\dots+f\big{(}\frac{z+p-1}{p}\big{)}$,

  * –

if $p$ is not one of the $e_{1},e_{2},\ldots$ but $ep$ is, for some $e$ with
$p\\!\nmid\\!e$, then
$f^{(p)}(W_{e}z)=f(W_{p}z)+f\big{(}\frac{z}{p}\big{)}+\dots+f\big{(}\frac{z+p-1}{p}\big{)}$,

  * –

if $ep$ is one of the $e_{1}$, $e_{2},\ldots$ with $p\,|\,e$ then
$f^{(p)}(W_{e}z)=f\big{(}\frac{z}{p}\big{)}+\dots+f\big{(}\frac{z+p-1}{p}\big{)}$.

Remark 2.13 gives motivation to find formulas of type
$A(z)=B(z)+C\big{(}\frac{z}{2}\big{)}+C\big{(}\frac{z+1}{2}\big{)}$. This will
help us finding $(2{+})$-replicates for certain Hauptmoduls. Before we
proceed, we need some notation in order to separate even Atkin–Lehner
involutions from the odd ones.

From now on, any group $n\,|\,h+e,f,\ldots$ will be written as
$n\,|\,h+O_{1}+2^{k}O_{2}$ where $2^{k}$ is the highest power of $2$ that
divides $\frac{n}{h}$, the set $O_{1}$ is the set of odd Atkin–Lehner
involutions of the group and $O_{2}$ is the set of even Atkin–Lehner
involutions of the group divided by $2^{k}$. We always include $1$ in $O_{1}$.

For example, the group $30+6$, $10$, $15$ will be written as
$30+O_{1}+2O_{2}$, with $O_{1}=\\{1,15\\}$ and $O_{2}=\left\\{3,5\right\\}$.

We should also note that the elements of both $O_{1}$ and $O_{2}$ are odd, and
these sets can satisfy $O_{1}=O_{2}$ as, for example, in $6+1$, $2$, $3$, $6$,
where $O_{1}=O_{2}=\\{1,3\\}$.

Recall, $G^{\alpha}$ denotes $\left[\begin{smallmatrix}1&\alpha\\\ 0&1\\\
\end{smallmatrix}\right]G\left[\begin{smallmatrix}1&\alpha\\\ 0&1\\\
\end{smallmatrix}\right]^{-1}$.

######  Remark 3.2.

If the matrix $\left[\begin{smallmatrix}1&\frac{1}{2}\\\ 0&1\\\
\end{smallmatrix}\right]$ does not normalize the genus zero group
$2^{k}N\,|\,h+O_{1}+2^{k}O_{2}$, with $N$ odd, then

$\displaystyle
T_{(2^{k}N|h+O_{1}+2^{k}O_{2})^{\frac{1}{2}}}(z)=-T_{2^{k}N|h+O_{1}+2^{k}O_{2}}\left(z+\frac{1}{2}\right).$

If it normalizes, we have

$\displaystyle
T_{2^{k}N|h+O_{1}+2^{k}O_{2}}\left(z+\frac{1}{2}\right)=-T_{2^{k}N|h+O_{1}+2^{k}O_{2}}(z),$

and this happens exactly when either $h$ is even or $h=1$, $k\geq 2$ and
$O_{2}$ is empty.

Also, this matrix conjugates the groups $2N+O_{1}$ and $4N+O_{1}+4O_{1}$ onto
each other, when $(2,N)=1$.

We make the following observation which will be used several times in what
follows.

######  Remark 3.3.

For a Hauptmodul $f$ of a genus zero group $G$,
$\alpha=\frac{1}{2},\frac{1}{4},\frac{1}{8}$ and $n$ odd we have

$\displaystyle\big{(}f^{\alpha}\big{)}^{(n)}=\big{(}f^{(n)}\big{)}^{\alpha}.$

In particular,
$\big{(}\big{(}f^{\alpha}\big{)}^{(m)}\big{)}^{(n)}=\big{(}f^{\alpha}\big{)}^{(mn)}$,
for $m$, $n$ odd.

######  Lemma 3.4.

We have the following identities:

  1. $1.$

If $(2,N)=1$ then

$\displaystyle
T_{N+O_{1}}(z)=T_{2N+O_{1}+2O_{1}}(z)+T_{2N+O_{1}+2O_{1}}\left(\frac{z}{2}\right)+T_{2N+O_{1}+2O_{1}}\left(\frac{z+1}{2}\right)$

or, equivalently,

$\displaystyle
T_{2N+O_{1}+2O_{1}}(z)=T_{N+O_{1}}(z)+T_{(2N+O_{1}+2O_{1})^{\frac{1}{2}}}\left(\frac{z}{2}\right)+T_{(2N+O_{1}+2O_{1})^{\frac{1}{2}}}\left(\frac{z+1}{2}\right).$

  2. $2.$

If $(2,N)=1$, $k\geq 1$ and $O_{2}$ is not empty then

$\displaystyle
T_{2^{k}N+O_{1}}(z)=T_{2^{k}N+O_{1}+2^{k}O_{2}}(z)+T_{(2^{k+1}N+O_{1}+2^{k+1}O_{2})^{\frac{1}{2}}}\left(\frac{z}{2}\right)$
$\displaystyle\hphantom{T_{2^{k}N+O_{1}}(z)=}{}+T_{(2^{k+1}N+O_{1}+2^{k+1}O_{2})^{\frac{1}{2}}}\left(\frac{z+1}{2}\right).$

In particular, if $k=1$ and $O_{1}=O_{2}$ we have

$\displaystyle
T_{2N+O_{1}}(z)=T_{2N+O_{1}+2O_{1}}(z)+T_{2N+O_{1}}\left(\frac{z}{2}\right)+T_{2N+O_{1}}\left(\frac{z+1}{2}\right).$

  3. $3.$

If $(2,N)=1$, $k\geq 1$ and $O_{2}$ is not empty then

$\displaystyle
T_{2^{k}N+O_{1}+2^{k}O_{2}}(z)=T_{2^{k}N+O_{1}}(z)+T_{2^{k+1}N+O_{1}+2^{k+1}O_{2}}\left(\frac{z}{2}\right)$
$\displaystyle\hphantom{T_{2^{k}N+O_{1}+2^{k}O_{2}}(z)=}{}+T_{2^{k+1}N+O_{1}+2^{k+1}O_{2}}\left(\frac{z+1}{2}\right).$

###### Proof.

The first identity is Theorem 3.1 applied to $f=T_{2N+O_{1}+2O_{1}}$ and
$p=2$. The third identity is also a consequence of Theorem 3.1 applied to
$f=T_{2^{k+1}N+O_{1}+2^{k+1}O_{2}}$ and $p=2$. Since $2^{k}\parallel N$, for
$k\geq 1$, we have

$\displaystyle
T_{2^{k+1}N+O_{1}+2^{k+1}O_{2}}\left(\frac{z}{2}\right)+T_{2^{k+1}N+O_{1}+2^{k+1}O_{2}}\left(\frac{z+1}{2}\right)=T_{2^{k}N+O_{1}}\left(W_{e^{\prime}}z\right),$

where $e^{\prime}=2^{k}e$ for some $e\in O_{2}$. But

$\displaystyle
T_{2^{k}N+O_{1}}(z)+T_{2^{k}N+O_{1}}(W_{e^{\prime}}z)=T_{2^{k}N+O_{1}+2^{k}O_{2}}(z)$

and this proves the third identity. The second identity is the third one
written in a different way. ∎

$g$ | $2N+O_{1}+2O_{2}$ | $4N+O_{1}+4O_{2}$ | $4N+O_{1}+4O_{2}$ | $8N+O_{1}+8O_{2}$
---|---|---|---|---
$g^{[\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$
$g^{[2]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$
$g^{[2\sqrt{2}]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$
$g^{[4]}$ |  | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[4\sqrt{2}]}$ |  |  | $\cdots$ | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  |  | $\cdots$
$g$ | $8N+O_{1}+8O_{2}$ | $16N+O_{1}+16O_{2}$ | $16N+O_{1}+16O_{2}$ | $32N+O_{1}+32O_{2}$
$g^{[\sqrt{2}]}$ | $8N+O_{1}$ | $8N+O_{1}$ | $16N+O_{1}$ | $16N+O_{1}$
$g^{[2]}$ | $4N+O_{1}$ | $8N+O_{1}$ | $8N+O_{1}$ | $16N+O_{1}$
$g^{[2\sqrt{2}]}$ | $4N+O_{1}$ | $4N+O_{1}$ | $8N+O_{1}$ | $8N+O_{1}$
$g^{[4]}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$ | $8N+O_{1}$
$g^{[4\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$
$g^{[8\sqrt{2}]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$
$g^{[16]}$ |  | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[16\sqrt{2}]}$ |  |  | $\cdots$ | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  |  | $\cdots$
$g$ | $2N|2+O_{1}$ | $4N|2+O_{1}+2O_{2}$ | $8N|2+O_{1}+4O_{2}$ | $8N|2+O_{1}+4O_{2}$
$g^{[\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}+4O_{2}$ | $4N|2+O_{1}+2O_{2}$ | $8N+O_{1}+8O_{2}$
$g^{[2]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}+4O_{2}$ | $4N+O_{1}$
$g^{[2\sqrt{2}]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$
$g^{[4]}$ |  | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$
$g^{[4\sqrt{2}]}$ |  | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[8]}$ |  |  | $\cdots$ | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  |  | $\cdots$
$g$ | $16N|2+O_{1}+8O_{2}$ | $16N|2+O_{1}+8O_{2}$ | $32N|2+O_{1}+16O_{2}$ | $32N|2+O_{1}+16O_{2}$
$g^{[\sqrt{2}]}$ | $8N|2+O_{1}+4O_{2}$ | $16N+O_{1}+16O_{2}$ | $16N|2+O_{1}+8O_{2}$ | $32N+O_{1}+32O_{2}$
$g^{[2]}$ | $8N+O_{1}+8O_{2}$ | $8N+O_{1}$ | $16N+O_{1}+16O_{2}$ | $16N+O_{1}$
$g^{[2\sqrt{2}]}$ | $4N+O_{1}$ | $8N+O_{1}$ | $8N+O_{1}$ | $16N+O_{1}$
$g^{[4]}$ | $4N+O_{1}$ | $4N+O_{1}$ | $8N+O_{1}$ | $8N+O_{1}$
$g^{[4\sqrt{2}]}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$ | $8N+O_{1}$
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$
$g^{[8\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $4N+O_{1}$
$g^{[16]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$
$g^{[16\sqrt{2}]}$ |  | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[32]}$ |  |  | $\cdots$ | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  |  | $\cdots$
$g$ | $8N|4+O_{1}+2O_{2}$ | $16N|4+O_{1}+4O_{2}$ | $16N|4+O_{1}+4O_{2}$ | $32N|4+O_{1}+8O_{2}$
$g^{[\sqrt{2}]}$ | $8N|2+O_{1}+4O_{2}$ | $16N|2+O_{1}+8O_{2}$ | $8N|4+O_{1}+2O_{2}$ | $32N|2+O_{1}+16O_{2}$
$g^{[2]}$ | $4N|2+O_{1}+2O_{2}$ | $8N|2+O_{1}+4O_{2}$ | $8N|2+O_{1}+4O_{2}$ | $16N|2+O_{1}+8O_{2}$
$g^{[2\sqrt{2}]}$ | $4N+O_{1}+4O_{2}$ | $8N+O_{1}+8O_{2}$ | $4N|2+O_{1}+2O_{2}$ | $16N+O_{1}+16O_{2}$
$g^{[4]}$ | $2N+O_{1}$ | $4N+O_{1}$ | $4N+O_{1}+4O_{2}$ | $8N+O_{1}$
$g^{[4\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}$ | $2N+O_{1}$ | $8N+O_{1}$
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}$
$g^{[8\sqrt{2}]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}$
$g^{[16]}$ |  | $2N+O_{1}+2O_{1}$ | $\cdots$ | $2N+O_{1}$
$g^{[16\sqrt{2}]}$ |  | $\cdots$ |  | $2N+O_{1}+2O_{1}$
$g^{[32]}$ |  |  |  | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  |  | $\cdots$
$g$ | $4N+O_{1}$ | $4N|2+O_{1}+2O_{1}$ | $4N|2+O_{1}$ | $(4N+O_{1}+4O_{2})^{\frac{1}{2}}$
$g^{[\sqrt{2}]}$ | $(4N+O_{1}+4O_{2})^{\frac{1}{2}}$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ | $(4N+O_{1}+4O_{2})^{\frac{1}{2}}$ | $2N+O_{1}+2O_{2}$
$g^{[2]}$ | $2N+O_{1}+2O_{2}$ | $N+O_{1}$ | $2N+O_{1}+2O_{2}$ | $2N+O_{1}+2O_{1}$
$g^{[2\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[4]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $\cdots$
$\vdots$ | $\cdots$ | $\cdots$ | $\cdots$ |

Table 1: $(2{+})$-replicates of Hauptmoduls.

$g$ | $8N|2+O_{1}$ | $8N|4+O_{1}+2O_{1}$ | $8N|2+O_{1}$ | $8N|4+O_{1}$
---|---|---|---|---
$g^{[\sqrt{2}]}$ | $(8N+O_{1}+8O_{2})^{\frac{1}{2}}$ | $(4N|2+O_{1}+2O_{1})^{\frac{1}{4}}$ | $(8N|2+O_{1}+4O_{1})^{\frac{1}{4}}$ | $(8N|2+O_{1}+4O_{1})^{\frac{1}{4}}$
$g^{[2]}$ | $4N+O_{1}+4O_{2}$ | $4N|2+O_{1}+2O_{1}$ | $4N+O_{1}$ | $4N|2+O_{1}$
$g^{[2\sqrt{2}]}$ | $4N+O_{1}$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ | $(4N+O_{1}+4O_{1})^{\frac{1}{2}}$ | $(4N+O_{1}+4O_{1})^{\frac{1}{2}}$
$g^{[4]}$ | $2N+O_{1}$ | $N+O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[4\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$\vdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$
$g$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ | $N+O_{1}$ | $(8N+O_{1}+8O_{2})^{\frac{1}{2}}$ | $(8N+O_{1}+8O_{2})^{\frac{1}{2}}$
$g^{[\sqrt{2}]}$ | $N+O_{1}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}+4O_{2}$ | $8N+O_{1}+8O_{2}$
$g^{[2]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}$ | $4N+O_{1}$
$g^{[2\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $\cdots$ | $2N+O_{1}$ | $2N+O_{1}$
$g^{[4]}$ | $\cdots$ |  | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[4\sqrt{2}]}$ |  |  | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$\vdots$ |  |  | $\cdots$ | $\cdots$
$g$ | $(16N+O_{1}+16O_{2})^{\frac{1}{2}}$ | $(32N+O_{1}+32O_{2})^{\frac{1}{2}}$ | $(4N|2+O_{1}+2O_{1})^{\frac{1}{4}}$ | $(8N|2+O_{1}+4O_{2})^{\frac{1}{4}}$
$g^{[\sqrt{2}]}$ | $8N+O_{1}+8O_{2}$ | $16N+O_{1}+16O_{2}$ | $4N|2+O_{1}+2O_{1}$ | $4N|2+O_{1}$
$g^{[2]}$ | $8N+O_{1}$ | $16N+O_{1}$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ | $(4N+O_{1}+4O_{2})^{\frac{1}{2}}$
$g^{[2\sqrt{2}]}$ | $4N+O_{1}$ | $8N+O_{1}$ | $N+O_{1}$ | $2N+O_{1}+2O_{2}$
$g^{[4]}$ | $4N+O_{1}$ | $8N+O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[4\sqrt{2}]}$ | $2N+O_{1}$ | $4N+O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $4N+O_{1}$ | $\cdots$ | $\cdots$
$g^{[8\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}$ |  |
$g^{[16]}$ | $\cdots$ | $2N+O_{1}+2O_{1}$ |  |
$g^{[16\sqrt{2}]}$ |  | $2N+O_{1}+2O_{1}$ |  |
$\vdots$ |  | $\cdots$ |  |
$g$ | $(16N|2+O_{1}+8O_{1})^{\frac{1}{4}}$ | $(8N|4+O_{1}+2O_{1})^{\frac{1}{8}}$ | $(8N|4+O_{1}+2O_{1})^{\frac{1}{8}}$ |
$g^{[\sqrt{2}]}$ | $8N|2+O_{1}$ | $8N|4+O_{1}+2O_{1}$ | $8N|2+O_{1}+2O_{1}$ |
$g^{[2]}$ | $(8N+O_{1}+8O_{1})^{\frac{1}{2}}$ | $(4N|2+O_{1}+2O_{1})^{\frac{1}{4}}$ | $(4N|2+O_{1}+2O_{1})^{\frac{1}{4}}$ |
$g^{[2\sqrt{2}]}$ | $4N+O_{1}+4O_{1}$ | $4N|2+O_{1}+2O_{1}$ | $4N|2+O_{1}+2O_{1}$ |
$g^{[4]}$ | $4N+O_{1}$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ | $(2N+O_{1}+2O_{1})^{\frac{1}{2}}$ |
$g^{[4\sqrt{2}]}$ | $2N+O_{1}$ | $N+O_{1}$ | $N+O_{1}$ |
$g^{[8]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ |
$g^{[8\sqrt{2}]}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ | $2N+O_{1}+2O_{1}$ |
$g^{[16]}$ | $\cdots$ | $\cdots$ | $\cdots$ |

Table 2: $(2{+})$-replicates of Hauptmoduls (continued).

######  Theorem 3.5.

Every function $g$ of type as given in Tables 1 and 2 is $(2{+})$-replicable.
The $\big{[}2^{\frac{n}{2}}\big{]}$-replicates, for $n$ a non-negative
integer, are the ones given in those tables and replicates
$g^{[2^{\frac{n}{2}}m]}$, for $n$ a non-negative integer and $m$ a positive
odd integer, are given by
$g^{[2^{\frac{n}{2}}m]}=\big{(}g^{[2^{\frac{n}{2}}]}\big{)}^{(m)}$.

###### Proof.

We start by proving the result for completely replicable functions and then
use this to complete the proof for conjugates of completely replicable
functions. For the cases where the function is completely replicable we know
how replication works and for each $g^{(2^{m})}$ we will either apply Lemma
3.4 or, in view of Remark 2.13, take $g^{(2^{m})}$ as $g^{[2^{m}]}$ and any
function satisfying the condition in Remark 3.2 as
$g^{[\frac{2^{m}}{\sqrt{2}}]}$. This defines the $(2+)$-replicates $g^{[m]}$,
with $m$ a power of $\sqrt{2}$. Having done so, and because of Remark 2.13
again, it is enough in order to finish the proof in this case to show that,
for all positive $m$ and odd $n$, we have

$\displaystyle
g^{(2^{m}n)}(z)=\left(g^{[2^{m}]}\right)^{(n)}(z)+\left(g^{[\frac{2^{m}}{\sqrt{2}}]}\right)^{(n)}\left(\frac{z}{2}\right)+\left(g^{[\frac{2^{m}}{\sqrt{2}}]}\right)^{(n)}\left(\frac{z+1}{2}\right).$
(3.1)

This is a case by case check on every $g^{(2^{m})}$.

1\. Our first case is when we are using some identity from Lemma 3.4 to
decompose $g^{(2^{m})}$. In this case, we must have
$g^{(2^{m})}=2^{k}N+O_{1}+2^{k}O_{2}$, i.e., $h=1$, and
$g^{(2^{m}n)}=\left(g^{(2^{m})}\right)^{(n)}=2^{k}N^{\prime}+O^{\prime}_{1}+2^{k}O^{\prime}_{2}$
where $N^{\prime}=N/\gcd(N,n)$, $O^{\prime}_{1}=\\{e\in O_{1}\,|\,e\mbox{
divides }N^{\prime}\\}$ and $O^{\prime}_{2}=\\{e\in O_{2}\,|\,e\mbox{ divides
}N^{\prime}\\}$. We fix an odd $n$ and divide this case further into three
different subcases.

i) If $O_{2}=\varnothing$ (which implies $O^{\prime}_{2}=\varnothing$) then
the possibilities, depending on $k$, for $g^{[2^{m}]}$ and
$g^{[\frac{2^{m}}{\sqrt{2}}]}$ when we apply Lemma 3.4 to decompose
$g^{(2^{m})}$ are

$\displaystyle\begin{array}[]{|c|c|c|c|}\hline\cr&g^{(2^{m})}&g^{[2^{m}]}&g^{[\frac{2^{m}}{\sqrt{2}}]}\\\
\hline\cr k=0&N+O_{1}&2N+O_{1}+2O_{1}&2N+O_{1}+2O_{1}\\\ \hline\cr
k=1&2N+O_{1}&2N+O_{1}+2O_{1}&2N+O_{1}\\\ \hline\cr k\geq
1&2^{k}N+O_{1}&2^{k}N+O_{1}+2^{k}O_{2}&\left(2^{k+1}N+O_{1}+2^{k+1}O_{2}\right)^{\frac{1}{2}}\\\
\hline\cr\end{array}$

Applying $(n)$-replication to every entry in the table we obtain

$\displaystyle\begin{array}[]{|c|c|c|c|}\hline\cr&g^{(2^{m}n)}&\big{(}g^{[2^{m}]}\big{)}^{(n)}&\big{(}g^{[\frac{2^{m}}{\sqrt{2}}]}\big{)}^{(n)}\\\
\hline\cr
k=0&N^{\prime}+O^{\prime}_{1}&2N^{\prime}+O^{\prime}_{1}+2O^{\prime}_{1}&2N^{\prime}+O^{\prime}_{1}+2O^{\prime}_{1}\\\
\hline\cr
k=1&2N^{\prime}+O^{\prime}_{1}&2N^{\prime}+O^{\prime}_{1}+2O^{\prime}_{1}&2N^{\prime}+O^{\prime}_{1}\\\
\hline\cr k\geq
1&2^{k}N^{\prime}+O^{\prime}_{1}&2^{k}N^{\prime}+O^{\prime}_{1}+2^{k}O^{\prime}_{2}&\left(2^{k+1}N^{\prime}+O^{\prime}_{1}+2^{k+1}O^{\prime}_{2}\right)^{\frac{1}{2}}\\\
\hline\cr\end{array}$

We can see that every row in the table corresponds to some identity from Lemma
3.4 and identity (3.1) is satisfied in this case.

ii) If $O_{2}\neq\varnothing$ and $O^{\prime}_{2}\neq\varnothing$ then the
possibilities, depending on $k$, for $g^{[2^{m}]}$ and
$g^{[\frac{2^{m}}{\sqrt{2}}]}$ when we apply Lemma 3.4 to decompose
$g^{(2^{m})}$ are

$\displaystyle\begin{array}[]{|c|c|c|c|}\hline\cr&g^{(2^{m})}&g^{[2^{m}]}&g^{[\frac{2^{m}}{\sqrt{2}}]}\\\
\hline\cr k=1&2N+O_{1}+2O_{1}&N+O_{1}&(2N+O_{1}+2O_{1})^{\frac{1}{2}}\\\
\hline\cr k\geq
1&2^{k}N+O_{1}+2^{k}O_{2}&2^{k}N+O_{1}&2^{k+1}N+O_{1}+2^{k+1}O_{2}\\\
\hline\cr\end{array}$

Applying $(n)$-replication to every entry in the table we obtain

$\displaystyle\begin{array}[]{|c|c|c|c|}\hline\cr&g^{(2^{m}n)}&\big{(}g^{[2^{m}]}\big{)}^{(n)}&\big{(}g^{[\frac{2^{m}}{\sqrt{2}}]}\big{)}^{(n)}\\\
\hline\cr
k=1&2N^{\prime}+O^{\prime}_{1}+2O^{\prime}_{1}&N^{\prime}+O^{\prime}_{1}&(2N^{\prime}+O^{\prime}_{1}+2O^{\prime}_{1})^{\frac{1}{2}}\\\
\hline\cr k\geq
1&2^{k}N^{\prime}+O^{\prime}_{1}+2^{k}O^{\prime}_{2}&2^{k}N^{\prime}+O^{\prime}_{1}&2^{k+1}N^{\prime}+O^{\prime}_{1}+2^{k+1}O^{\prime}_{2}\\\
\hline\cr\end{array}$

We can see that every row in the table corresponds to some identity from Lemma
3.4 and identity (3.1) is satisfied in this case.

iii) If $O_{2}\neq\varnothing$ and $O^{\prime}_{2}=\varnothing$ then the
possibilities, depending on $k$, for $g^{[2^{m}]}$ and
$g^{[\frac{2^{m}}{\sqrt{2}}]}$ when we apply Lemma 3.4 to decompose
$g^{(2^{m})}$ are

$\displaystyle\begin{array}[]{|c|c|c|c|}\hline\cr&g^{(2^{m})}&g^{[2^{m}]}&g^{[\frac{2^{m}}{\sqrt{2}}]}\\\
\hline\cr k=1&2N+O_{1}+2O_{1}&N+O_{1}&(2N+O_{1}+2O_{1})^{\frac{1}{2}}\\\
\hline\cr k\geq
1&2^{k}N+O_{1}+2^{k}O_{2}&2^{k}N+O_{1}&2^{k+1}N+O_{1}+2^{k+1}O_{2}\\\
\hline\cr\end{array}$

Applying $(n)$-replication to every entry in the table we obtain

$\displaystyle\begin{array}[]{|c|c|c|c|c|}\hline\cr&g^{(2^{m}n)}&\big{(}g^{[2^{m}]}\big{)}^{(n)}&\big{(}g^{[\frac{2^{m}}{\sqrt{2}}]}\big{)}^{(n)}\\\
\hline\cr k=1&2N^{\prime}&N^{\prime}&(2N^{\prime})^{\frac{1}{2}}\\\ \hline\cr
k\geq
1&2^{k}N^{\prime}+O^{\prime}_{1}&2^{k}N^{\prime}+O^{\prime}_{1}&2^{k+1}N^{\prime}+O^{\prime}_{1}\\\
\hline\cr\end{array}$

We can see that the first row corresponds to an identity from Lemma 3.4 and in
the last row the Hauptmodul $2^{k+1}N^{\prime}+O^{\prime}_{1}$ satisfies the
condition of Remark 3.2. Identity (3.1) is satisfied in both cases.

2\. If we are taking $g^{[2^{m}]}=g^{(2^{m})}$ and
$g^{[\frac{2^{m}}{\sqrt{2}}]}$ to be any Hauptmodul satisfying the condition
in Remark 3.2 then identity (3.1) is trivially satisfied because odd
replication preserves the property of Remark 3.2.

This completes the proof in the case $g$ is completely replicable.

The proof also shows that in Tables 1 and 2 we can take any
$2^{k}N\,\parallel\,h+O_{1}+2^{k}O_{2}$ instead of
$2^{k}N\,|\,h+O_{1}+2^{k}O_{2}$ and the result is still valid.

For the remaining cases in the tables, $g$ is not completely replicable, but
its invariance group is conjugate by $\left[\begin{smallmatrix}1&\alpha\\\
0&1\\\ \end{smallmatrix}\right]$, with $\alpha=\frac{1}{2},\frac{1}{4}$ or
$\frac{1}{8}$, to some group whose Hauptmodul $f$ is completely replicable. We
shall give the proof for the case $\alpha=\frac{1}{2}$ as the other two cases
are similar.

Let $g=T_{G}$ and $f=T_{G^{\alpha}}$ so that
$g(z)=-f\big{(}z+\frac{1}{2}\big{)}$. Since we are assuming $f$ is completely
replicable it follows, as remarked above, that $f$ is $(2{+})$-replicable. So
for some – not necessarily unique – $(2{+})$-replicates we have

$\displaystyle P_{n,f}(f(z))=\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}a]}\left(\frac{2az+b}{d}\right).$ (3.2)

Our aim is to use the relationship between $f$ and $g$ to show that (3.2)
implies that $g$ is $(2{+})$-replicable after a suitable choice of
$(2{+})$-replicates for $f$.

Now by the definition of the Faber polynomials, for any series
$h(z)=\frac{1}{q}+\cdots$ we have
$P_{n,h}\big{(}h\big{(}z+\frac{1}{2}\big{)}\big{)}=(-1)^{n}P_{n,h^{\frac{1}{2}}}\big{(}h^{\frac{1}{2}}(z)\big{)}$,
where $h^{\frac{1}{2}}(z)=-h\big{(}z+\frac{1}{2}\big{)}$.

In particular applying this to $f$ and $g$ gives
$P_{n,f}\big{(}f\big{(}z+\frac{1}{2}\big{)}\big{)}=(-1)^{n}P_{n,g}(g)$. So for
$n$ odd, substituting $z$ by $z+\frac{1}{2}$ in (3.2) gives

$\displaystyle P_{n,g}(g(z))=-\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}+\frac{a}{2d}\right)=\sum_{\begin{subarray}{c}ad=n\\\
0\leq
b<d\end{subarray}}(f^{[a]})^{\frac{1}{2}}\left(\frac{az+b+\frac{a-d}{2}}{d}\right)$
$\displaystyle\hphantom{P_{n,g}(g(z))=}{}=\sum_{\begin{subarray}{c}ad=n\\\
0\leq b<d\end{subarray}}(f^{[a]})^{\frac{1}{2}}\left(\frac{az+b}{d}\right).$

So these replication identities for $g$ are satisfied by choosing
$g^{[a]}=\big{(}f^{[a]}\big{)}^{\frac{1}{2}}$ for $a$ odd.

For $n$ even, we again substitute $z$ by $z+\frac{1}{2}$ in (3.2). Arranging
the resulting right hand side by the highest power of 2 which divides $a$
gives

$\displaystyle P_{n,g}(g(z))=\sum_{\begin{subarray}{c}ad=n\\\ a\ \text{odd}\\\
0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}+\frac{a}{2d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{odd}\\\ d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}a]}\left(\frac{2az+b}{d}\right)$
$\displaystyle\hphantom{P_{n,g}(g(z))=}{}+\sum_{i=1}^{\infty}\left(\sum_{\begin{subarray}{c}ad=\frac{n}{2^{i}}\\\
a\ \text{odd}\\\ 0\leq
b<d\end{subarray}}f^{[2^{i}a]}\left(\frac{2^{i}az+b}{d}\right)+\sum_{\begin{subarray}{c}2^{i}ad=n\\\
a\ \text{odd}\\\ d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[2^{i}\sqrt{2}a]}\left(\frac{2^{i+1}az+b}{d}\right)\right).$
(3.3)

So if we set $g^{[2^{\frac{n}{2}}m]}=f^{[2^{\frac{n}{2}}m]}$ for $n\geq 2$
these replication identities for $g$ will be satisfied provided we can show
that the $i=0$ term

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ a\ \text{odd}\\\ 0\leq
b<d\end{subarray}}f^{[a]}\left(\frac{az+b}{d}+\frac{a}{2d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{odd}\\\ d\ \text{even}\\\ 0\leq
b<d\end{subarray}}f^{[\sqrt{2}a]}\left(\frac{2az+b}{d}\right)$ (3.4)

has the form

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ a\ \text{odd}\\\ 0\leq
b<d\end{subarray}}g^{[a]}\left(\frac{az+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
a\ \text{odd}\\\ d\ \text{even}\\\ 0\leq
b<d\end{subarray}}g^{[\sqrt{2}a]}\left(\frac{2az+b}{d}\right)$ (3.5)

with $g^{[a]}$ as given above and for suitable choices of the
$g^{[\sqrt{2}a]}$.

The proof is by considering cases. First consider
$g=T_{(2N+O_{1}+2O_{1})^{\frac{1}{2}}}$, thus $f=T_{2N+O_{1}+2O_{1}}$. Then
each $f^{[a]}$ in (3.4) is of type $T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}$,
where $M=N/\gcd(N,a)$ and $O^{\prime}_{1}=\\{e\in O_{1}\,|\,e\mbox{ divides
}M\\}$ and we know from Theorem 3.1 that in this case

$\displaystyle
T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{z}{2}\right)+T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{z+1}{2}\right)+T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}(z)=T_{M+O^{\prime}_{1}}(z).$

In particular, substituting $z$ by $\frac{2az+b}{d}$ and summing over $b$ we
have

$\displaystyle\sum_{0\leq
b<2d}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{az}{d}+\frac{b}{2d}\right)+\sum_{0\leq
b<d}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=\sum_{0\leq
b<d}T_{M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$

or, equivalently,

$\displaystyle\sum_{0\leq
b<d}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{az+b}{d}+\frac{1}{2d}\right)+\sum_{0\leq
b<d}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=-\sum_{0\leq
b<d}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\left(\frac{az+b}{d}\right)+\sum_{0\leq
b<d}T_{M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=\sum_{0\leq
b<d}T_{(2M+O^{\prime}_{1}+2O^{\prime}_{1})^{\frac{1}{2}}}\left(\frac{az+b}{d}+\frac{1}{2}\right)+\sum_{0\leq
b<d}T_{M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=\sum_{0\leq
b<d}T_{(2M+O^{\prime}_{1}+2O^{\prime}_{1})^{\frac{1}{2}}}\left(\frac{az+b}{d}\right)+\sum_{0\leq
b<d}T_{M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right).$ (3.6)

As mentioned above, the $(2{+})$-replicates of $f$ in (3.2) are not
necessarily unique and we are free to make any convenient choices. So we now
choose $f^{[\sqrt{2}a]}=T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}$. Identity (3.6)
then shows that (3.4) has the form (3.5), with
$g^{[a]}=\big{(}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\big{)}^{(\frac{1}{2})}$
and $g^{[\sqrt{2}a]}=T_{M+O^{\prime}_{1}}$, as required.

For the second part of the theorem we just note that when $n\geq 2$ we have
$g^{[2^{\frac{n}{2}}m]}=f^{[2^{\frac{n}{2}}m]}=\big{(}f^{[2^{\frac{n}{2}}]}\big{)}^{(m)}=\big{(}g^{[2^{\frac{n}{2}}]}\big{)}^{(m)}$,
when $n=1$ we have
$g^{[\sqrt{2}m]}=T_{M+O^{\prime}_{1}}=\big{(}T_{N+O_{1}}\big{)}^{(m)}=\big{(}g^{[\sqrt{2}]}\big{)}^{(m)}$
and when $n=0$ we have

$\displaystyle
g^{[m]}=\big{(}T_{2M+O^{\prime}_{1}+2O^{\prime}_{1}}\big{)}^{\frac{1}{2}}=\big{(}\big{(}T_{2N+O_{1}+2O_{1}}\big{)}^{(m)}\big{)}^{\frac{1}{2}}=\big{(}\big{(}T_{2N+O_{1}+2O_{1}}\big{)}^{\frac{1}{2}}\big{)}^{(m)}=g^{(m)}.$

This finishes the proof for functions of the form
$T_{(2N+O_{1}+2O_{1})^{\frac{1}{2}}}$.

We now consider $g$ of the form $T_{(2^{k}N+O_{1}+2^{k}O_{2})^{\frac{1}{2}}}$,
with $k\geq 2$. In this case, the $f^{[a]}$ in (3.3) are equal to
$T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}$, where $M=N/\gcd(N,a)$,
$O^{\prime}_{1}=\\{e\in O_{1}\,|\,e\mbox{divides }M\\}$ and
$O^{\prime}_{2}=\\{e\in O_{2}\,|\,e\mbox{ divides }M\\}$ and applying Lemma
3.4 to this function we have

$\displaystyle
T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\left(\frac{z}{2}\right)+T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\left(\frac{z+1}{2}\right)+T_{2^{k-1}M+O^{\prime}_{1}}(z)=T_{2^{k-1}M+O^{\prime}_{1}+2^{k-1}O^{\prime}_{2}}(z).$

Substituting $z$ by $\frac{2az+b}{d}$ and summing over $b$ we obtain

$\displaystyle\begin{split}&\sum_{0\leq
b<2d}T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\left(\frac{az}{d}+\frac{b}{2d}\right)+\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)\\\
&\qquad{}=\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}+2O^{\prime}_{2}}\left(\frac{2az+b}{d}\right)\end{split}$

and, in particular,

$\displaystyle\sum_{0\leq
b<d}{}T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\left(\frac{az+b}{d}+\frac{1}{2d}\right)+\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=-\sum_{0\leq
b<d}T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\left(\frac{az+b}{d}\right)+\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}+2^{k-1}O^{\prime}_{2}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=\sum_{0\leq
b<d}T_{(2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2})^{\frac{1}{2}}}\left(\frac{az+b}{d}+\frac{1}{2}\right)+\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}+2^{k-1}O^{\prime}_{2}}\left(\frac{2az+b}{d}\right)$
$\displaystyle\qquad{}=\sum_{0\leq
b<d}T_{(2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2})^{\frac{1}{2}}}\left(\frac{az+b}{d}\right)+\sum_{0\leq
b<d}T_{2^{k-1}M+O^{\prime}_{1}+2^{k-1}O^{\prime}_{2}}\left(\frac{2az+b}{d}\right).$

As mentioned above, the $(2{+})$-replicates of $f$ in (3.2) are not
necessarily unique and we are free to make any convenient choices. So we now
choose $f^{[\sqrt{2}a]}$ of the form $T_{2^{k-1}M+O^{\prime}_{1}}$. Identity
(3.6) then shows that (3.4) has the form (3.5), with
$g^{[a]}=\big{(}T_{2M+O^{\prime}_{1}+2O^{\prime}_{2}}\big{)}^{\frac{1}{2}}$
and $g^{[\sqrt{2}a]}=T_{M+O^{\prime}_{1}}$, as required.

For the second part of the theorem we just note that when $n\geq 2$ we have
$g^{[2^{\frac{n}{2}}m]}=f^{[2^{\frac{n}{2}}m]}=\big{(}f^{[2^{\frac{n}{2}}]}\big{)}^{(m)}\\!=\big{(}g^{[2^{\frac{n}{2}}]}\big{)}^{(m)}$,
when $n=1$ we have
$g^{[\sqrt{2}m]}=T_{2^{k-1}M+O^{\prime}_{1}+2^{k-1}O^{\prime}_{2}}=\big{(}T_{2^{k-1}N+O_{1}+2^{k}O_{2}}\big{)}^{(m)}\\!$
$=\big{(}g^{[\sqrt{2}]}\big{)}^{(m)}$ and when $n=0$ we have
$g^{[m]}=\big{(}T_{2^{k}M+O^{\prime}_{1}+2^{k}O^{\prime}_{2}}\big{)}^{\frac{1}{2}}=\big{(}\big{(}T_{2^{k}N+O_{1}+2^{k}O_{2}}\big{)}^{(m)}\big{)}^{\frac{1}{2}}=\big{(}\big{(}T_{2^{k}N+O_{1}+2^{k}O_{2}}\big{)}^{\frac{1}{2}}\big{)}^{(m)}=g^{(m)}$.

The remaining cases can be done similarly. For Hauptmoduls of groups of the
form $G^{\frac{1}{4}}$ (resp. $G^{\frac{1}{8}}$) it is the summand
corresponding to $i=1$ (resp. $i=2$) that matters. We just note that the
replicates $f^{[\sqrt{2}a]}$ (resp. $f^{[\sqrt{2}a]}$ and $f^{[2\sqrt{2}a]}$)
with $a$ odd, have a power series expansion with coefficients of even powers
of $q$ equal to zero. This means that any other function with the same
property will work as well. We just chose those particular ones because they
give complete $(2{+})$-replicability. ∎

######  Corollary 3.6.

The functions $g$ on Tables 1 and 2 are completely $(2{+})$-replicable.

###### Proof.

We begin the proof showing that we have complete $(2{+})$-replicability for
powers of $\sqrt{2}$.

By complete $(2{+})$-replicability for powers of $\sqrt{2}$ we mean that, for
any $(2{+})$-replicate of $g$, $g^{[2^{\frac{m}{2}}n]}$, if we choose
$\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[\sqrt{2}]}=g^{[2^{\frac{m+1}{2}}n]}$
and apply the $(2{+})$-replication rules from the theorem/tables we will
necessarily have

$\displaystyle\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}]}=g^{[2^{\frac{m+m^{\prime}}{2}}n]},\qquad\text{for
all}\ m^{\prime}\geq 2.$ (3.7)

For doing this, we just have to see that after removing the first few entries
of any column or applying $(m)$-replication, with $m$ odd, to a full column,
the list we obtain is still a full column in the tables. This is a case by
case check and is what guarantees that any replicate $g^{[2^{\frac{k}{2}}n]}$,
together with $g^{[2^{\frac{k+1}{2}}n]}$ as its $[\sqrt{2}]$-replicate, will
have $(2{+})$-replicates as expected

$\displaystyle\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}n^{\prime}]}=\big{(}\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}]}\big{)}^{(n^{\prime})}\quad\text{(by
definition})$
$\displaystyle\hphantom{\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}n^{\prime}]}}{}=\big{(}g^{[2^{\frac{m+m^{\prime}}{2}}n]}\big{)}^{(n^{\prime})}\quad\text{(from
(\ref{commutation})})$
$\displaystyle\hphantom{\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}n^{\prime}]}}{}=\big{(}\big{(}g^{[2^{\frac{m+m^{\prime}}{2}}]}\big{)}^{(n)}\big{)}^{(n^{\prime})}\quad\text{(by
definition)}$
$\displaystyle\hphantom{\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}n^{\prime}]}}{}=\big{(}g^{[2^{\frac{m+m^{\prime}}{2}}]}\big{)}^{(nn^{\prime})}\quad\text{(with
$g^{[2^{\frac{m+m^{\prime}}{2}}]}$ comp.\ repl.\ or not)}$
$\displaystyle\hphantom{\big{(}g^{[2^{\frac{m}{2}}n]}\big{)}^{[2^{\frac{m^{\prime}}{2}}n^{\prime}]}}{}=g^{[2^{\frac{m+m^{\prime}}{2}}nn^{\prime}]}\quad\text{(by
definition)}.$ ∎

##  4 Complete $\boldsymbol{(2{+})}$-replicability
and generalized Mahler recurrence relations

In this Section we extend some of Martin’s results from [21] to
$(2{+})$-replication. Namely, we adapt his proof to show that the coefficients
of completely $(2{+})$-replicable functions satisfy recurrence relations very
similar to those of completely replicable functions.

We consider $\mathcal{L}$ a field extension of $\mathbb{Q}$ containing all
roots of unity, the ring
$K\\!=\\!\mathcal{L}\big{[}\ldots,x_{m}^{[n]},\ldots\big{]}$, $m\in{\rm
Nature}$, $n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$, the series

$\displaystyle h^{[r]}(q)=\frac{1}{q}+\sum_{m=1}^{\infty}x_{m}^{[r]}q^{m}$

for $r\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$, and the polynomials
$P_{k,r}(t)$ defined inductively by $P_{1,r}(t)=t$ and
$P_{k,r}(t)=tP_{k-1,r}(t)-\sum\limits_{s=1}^{k-2}x_{s}^{[r]}P_{k-s-1,r}(t)-kx_{k-1}^{[k]}$.
These are the same recurrence relations that Faber polynomials satisfy.

We fix $r\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$ and consider the set of
equations indexed by $k\geq 1$

$\displaystyle\sum_{\begin{subarray}{c}ad=k\\\ 0\leq
b<d\end{subarray}}h^{[ra]}\big{(}e^{2\pi
i\frac{b}{d}}q^{\frac{a}{d}}\big{)}+\sum_{\begin{subarray}{c}ad=k\\\ d\
\text{even}\\\ 0\leq b<d\end{subarray}}h^{[ra\sqrt{2}]}\big{(}e^{2\pi
i\frac{b}{d}}q^{\frac{2a}{d}}\big{)}=P_{k,r}\big{(}h^{[r]}(q)\big{)}.$ (4.1)

These equations give an infinite set of identities in $K$ by equating the
coefficients of equal powers of $q$ in both sides of the each equation. We
denote by $I^{[r]}$ the ideal in $K$ generated by them and write $I$ for the
ideal in $K$ generated by $\bigcup\limits_{r\in{\rm Nature}\cup\sqrt{2}{\rm
Nature}}I^{[r]}$.

If $f(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}q^{k}$ is completely
$(2{+})$-replicable with replicates
$f^{[n]}(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}^{[n]}q^{k}$, $n\in{\rm
Nature}\cup\sqrt{2}{\rm Nature}$ then they satisfy equations (4.1) with
$h^{[n]}(q)$ and $h^{[\sqrt{2}n]}(q)$ replaced by $f^{[n]}(q)$ and
$f^{[\sqrt{2}n]}(q)$, respectively. This means that every completely
$(2{+})$-replicable function $f$ induces a non-trivial homomorphism
$E_{f}\colon K\longrightarrow\mathbb{C}$ with
$E_{f}\big{(}x_{k}^{[n]}\big{)}=a_{k}^{[n]}$, whose kernel contains $I$.

For $u\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$ we define a
$\mathcal{L}$-algebra endomorphism $\psi_{u}$ of $K$ letting it fix every
element of $\mathcal{L}$ and mapping $x_{m}^{[n]}$ to $x_{m}^{[nu]}$. Since
the equations defining $I^{[r]}$ and $I^{[ru]}$ have the same form, it is
clear that $\psi_{u}(I^{[r]})=I^{[ru]}$ for any $r$. Consequently,
$\psi_{u}(I)\subseteq I$ and we think of $\psi_{u}$ as an
$\mathcal{L}$-algebra endomorphism of the quotient $K/I$.

For each $M\geq 1$ we set
$R_{M}=K/I\big{[}\big{[}q^{\frac{1}{2M}}\big{]}\big{]}$ and also

$\displaystyle\Delta=\left\\{\left[\begin{matrix}a&b\\\
0&d\end{matrix}\right]|\,a,b,d\in{\rm
Nature}\right\\}\cup\left\\{\left[\begin{matrix}\sqrt{2}a&b/\sqrt{2}\\\
0&\sqrt{2}d\end{matrix}\right]|\,a,b,d\in{\rm Nature}\right\\}.$

We define an action of $\Delta$ in $R_{M}$ in the following way. For
$\alpha=\left[\begin{smallmatrix}u&v\\\ 0&y\\\
\end{smallmatrix}\right]\in\Delta$ we set $e\parallel\alpha=e$,
$x_{m}^{[r]}\parallel\alpha=\psi_{u}\big{(}x_{m}^{[r]}\big{)}=x_{m}^{[ru]}$
and $q^{\frac{1}{M}}\parallel\alpha=e^{2\pi i\frac{v}{My}}q^{\frac{u}{My}}$.
Then we extend this map to an $\mathcal{L}$-homomorphism from
$K\big{[}\big{[}q^{\frac{1}{M}}\big{]}\big{]}$ to
$K\big{[}\big{[}q^{\frac{1}{My}}\big{]}\big{]}$. Though the context should
make it clear, we warn the reader that in this Chapter only the symbol
$\parallel$ has this meaning.

######  Remark 4.1.

The ideal $I$ of $K$ is stable under $\parallel\alpha$ for every
$\alpha\in\Delta$.

######  Remark 4.2.

For every $\alpha,\beta\in\Delta$ and $h(q)\in R_{M}$ we have

$\displaystyle(h(q)\parallel\alpha)\parallel\beta=h(q)\parallel\alpha\beta.$

######  Remark 4.3.

If $\alpha=\left[\begin{smallmatrix}u&x\\\ 0&y\\\
\end{smallmatrix}\right]\in\Delta$ then
$h^{[r]}(q)\parallel\alpha=h^{[ru]}\big{(}e^{2\pi
i\frac{x}{y}}q^{\frac{u}{y}}\big{)}$.

######  Remark 4.4.

For every positive integer $j$,
$\big{(}h^{[r]}(q)\big{)}^{j}\parallel\alpha=\big{(}h^{[r]}(q)\parallel\alpha\big{)}^{j}$
as $\parallel\alpha$ is a ring homomorphism.

######  Definition 4.5.

Let $n$ be a positive integer and $h(q)\in R_{1}$. We define ${\bf T}_{n}$ by

$\displaystyle{\bf T}_{n}(h(q))=\sum_{\begin{subarray}{c}uy=n\\\ 0\leq
v<y\end{subarray}}h(q)\parallel\left[\begin{matrix}u&v\\\ 0&y\\\
\end{matrix}\right]+\sum_{\begin{subarray}{c}uy=\frac{n}{2}\\\ 0\leq
v<2y\end{subarray}}h(q)\parallel\left[\begin{matrix}\sqrt{2}u&\dfrac{v}{\sqrt{2}}\\\
0&\sqrt{2}y\\\ \end{matrix}\right]$

and call ${\bf T}_{n}$ a generalized Hecke operator. Both sums are over
positive integer numbers which makes the second sum be zero if $n$ is odd.

With this definition equation (4.1) becomes

$\displaystyle{\bf
T}_{k}\big{(}h^{[r]}(q)\big{)}=P_{k,r}\big{(}h^{[r]}(q)\big{)}.$ (4.2)

######  Definition 4.6.

Let $n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$ and $h(q)\in R_{1}$. We denote
by $\Psi_{n}$ the mapping
$\Psi_{n}h(q)=h(q)\parallel\left[\begin{smallmatrix}n&0\\\
0&n\end{smallmatrix}\right]$.

######  Proposition 4.7.

Let $l_{1}$ and $l_{2}$ be relatively prime positive integers and $h(q)\in
R_{1}$. Then

$\displaystyle{\bf T}_{l_{1}}{\bf T}_{l_{2}}h(q)={\bf T}_{l_{1}l_{2}}h(q).$

In particular, ${\bf T}_{l_{1}}$ and ${\bf T}_{l_{2}}$ commute.

###### Proof.

This comes from the fact that

$\displaystyle\left\\{\left[\begin{matrix}u&v\\\ 0&y\\\
\end{matrix}\right]\left[\begin{matrix}u^{\prime}&v^{\prime}\\\
0&y^{\prime}\\\
\end{matrix}\right]|\,uy=l_{1},\,u^{\prime}y^{\prime}=l_{2},\,0\leq
v<y,\,0\leq v^{\prime}<y^{\prime},\,(yy^{\prime}\ \text{even})\right\\}$
$\displaystyle\qquad=\left\\{\left[\begin{matrix}u^{\prime\prime}&v^{\prime\prime}\\\
0&y^{\prime\prime}\\\
\end{matrix}\right]|\,u^{\prime\prime}v^{\prime\prime}=l_{1}l_{2},\,0\leq
v^{\prime\prime}<y^{\prime\prime},\,(y^{\prime\prime}\ \text{even})\right\\}.$
∎

######  Proposition 4.8.

If $p$ is an odd prime then

$\displaystyle{\bf T}_{p^{n}}{\bf T}_{p}(h(q))={\bf T}_{p^{n+1}}(h(q))+p{\bf
T}_{p^{n-1}}\Psi_{p}(h(q)).$

If $p=2$ then

$\displaystyle{\bf T}_{2^{n}}{\bf T}_{2}(h(q))={\bf T}_{2^{n+1}}(h(q))+2{\bf
T}_{2^{n}}\Psi_{\sqrt{2}}(h(q))+2{\bf T}_{2^{n-1}}\Psi_{2}(h(q)).$

###### Proof.

The case where $p$ is odd is true as the operators ${\bf T}_{p^{n}}$ agree
with the Hecke operator for ${\rm PSL}_{2}(\mathbb{Z})$. For the case $p=2$ we
see that ${\bf T}_{2^{n}}{\bf T}_{2}(h(q))$ equals

$\displaystyle h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,1\\\ 0\leq
b<2^{i}\end{subarray}}\left[\begin{matrix}2^{1-i}&b\\\
0&2^{i}\end{matrix}\right]+\sum_{0\leq
b<2}\left[\begin{matrix}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&\sqrt{2}\end{matrix}\right]\right)\left(\sum_{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n-i}&b\\\
0&2^{i}\end{matrix}\right]\right)$
$\displaystyle\qquad{}+h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,1\\\
0\leq b<2^{i}\end{subarray}}\left[\begin{matrix}2^{1-i}&b\\\
0&2^{i}\end{matrix}\right]+\sum_{0\leq
b<2}\left[\begin{matrix}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&\sqrt{2}\end{matrix}\right]\right)\left(\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\
0\leq
b<2^{i+1}\end{subarray}}\left[\begin{matrix}2^{n-1-i}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&2^{i}\sqrt{2}\\\ \end{matrix}\right]\right)$
$\displaystyle{}=h(q)\parallel\\!\left(\sum_{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq b<2^{i}\end{subarray}}\begin{bmatrix}2^{n+1-i}&2b\\\
0&2^{i}\end{bmatrix}+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i}\end{subarray}}\begin{bmatrix}2^{n-i}\sqrt{2}&\sqrt{2}b\\\
0&2^{i}\sqrt{2}\end{bmatrix}\right.+\sum_{{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq
b<2^{i}\end{subarray}}}\begin{bmatrix}2^{n-i}\sqrt{2}&\sqrt{2}b+\frac{2^{i}}{\sqrt{2}}\\\
0&2^{i}\sqrt{2}\end{bmatrix}$
$\displaystyle\qquad{}+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i}\end{subarray}}\begin{bmatrix}2^{n-i}&b\\\
0&2^{i+1}\end{bmatrix}+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i}\end{subarray}}\begin{bmatrix}2^{n-i}&b+2^{i}\\\
0&2^{i+1}\end{bmatrix}+\sum_{{{\begin{subarray}{c}{i=0,\ldots,n-1}\\\ {0\leq
b<2^{i+1}}\end{subarray}}}}\begin{bmatrix}2^{n-i}\sqrt{2}&\sqrt{2}b\\\
0&2^{i}\sqrt{2}\end{bmatrix}$
$\displaystyle\qquad{}+\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-i}&b\\\
0&2^{i+1}\end{bmatrix}+\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-i}&b+2^{i}\\\
0&2^{i+1}\end{bmatrix}$
$\displaystyle\qquad{}+\sum_{{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+1}\end{subarray}}}\begin{bmatrix}2^{n-1-i}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&2^{i+1}\sqrt{2}\end{bmatrix}+\left.\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\
0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-1-i}\sqrt{2}&\frac{b}{\sqrt{2}}+2^{i}b\\\
0&2^{i+1}\sqrt{2}\\\ \end{bmatrix}\right)$
$\displaystyle{}=h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n+1-i}&2b\\\
0&2^{i}\end{matrix}\right]+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n-i}\sqrt{2}&\sqrt{2}b\\\
0&2^{i}\sqrt{2}\end{matrix}\right]\right.$
$\displaystyle\qquad{}+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n-i}\sqrt{2}&\sqrt{2}b+\frac{2^{i}}{\sqrt{2}}\\\
0&2^{i}\sqrt{2}\end{matrix}\right]+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq b<2^{i+1}\end{subarray}}\left[\begin{matrix}2^{n-i}&b\\\
0&2^{i+1}\end{matrix}\right]$
$\displaystyle\qquad{}+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n-i}\sqrt{2}&\sqrt{2}b\\\
0&2^{i}\sqrt{2}\end{matrix}\right]+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\
0\leq b<2^{i+1}\end{subarray}}\left[\begin{matrix}2^{n-i}&b\\\
0&2^{i+1}\end{matrix}\right]$
$\displaystyle\left.\qquad{}+\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+2}\end{subarray}}\left[\begin{matrix}2^{n-1-i}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&2^{i+1}\sqrt{2}\end{matrix}\right]\right).$

Now, the first and fourth summands equal

$\displaystyle h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,\ldots,n+1\\\
0\leq b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n+1-i}&b\\\
0&2^{i}\end{matrix}\right]+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i}\end{subarray}}\left[\begin{matrix}2^{n-i}&2b\\\ 0&2^{i}\\\
\end{matrix}\right]\right),$

the second, third and last summands equal

$\displaystyle h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,\ldots,n\\\
0\leq
b<2^{i+1}\end{subarray}}\left[\begin{matrix}2^{n-i}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&2^{i}\sqrt{2}\end{matrix}\right]+2\sum_{\begin{subarray}{c}i=0,\ldots,n-2\\\
0\leq b<2^{i+1}\end{subarray}}\left[\begin{matrix}2^{n-2-i}&\sqrt{2}b\\\
0&2^{i+1}\sqrt{2}\end{matrix}\right]\right),$

and this shows that

$\displaystyle{\bf T}_{2^{n}}{\bf
T}_{2}(h(q))=h(q)\parallel\left(\sum_{\begin{subarray}{c}i=0,\ldots,n+1\\\
0\leq b<2^{i}\end{subarray}}\begin{bmatrix}2^{n+1-i}&b\\\
0&2^{i}\end{bmatrix}\right.+\sum_{\begin{subarray}{c}i=0,\ldots,n\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-i}\sqrt{2}&\frac{b}{\sqrt{2}}\\\
0&2^{i}\sqrt{2}\end{bmatrix}$ $\displaystyle\hphantom{{\bf T}_{2^{n}}{\bf
T}_{2}(h(q))=}{}+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-i}\sqrt{2}&\sqrt{2}b\\\
0&2^{i}\sqrt{2}\end{bmatrix}+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-i}&b\\\ 0&2^{i+1}\end{bmatrix}$
$\displaystyle\left.\hphantom{{\bf T}_{2^{n}}{\bf
T}_{2}(h(q))=}{}+2\sum_{\begin{subarray}{c}i=0,\ldots,n-1\\\ 0\leq
b<2^{i}\end{subarray}}\begin{bmatrix}2^{n-i}&2b\\\
0&2^{i+1}\end{bmatrix}+2\sum_{\begin{subarray}{c}i=0,\ldots,n-2\\\ 0\leq
b<2^{i+1}\end{subarray}}\begin{bmatrix}2^{n-2-i}&\sqrt{2}b\\\
0&2^{i+1}\sqrt{2}\end{bmatrix}\right)$ $\displaystyle\hphantom{{\bf
T}_{2^{n}}{\bf T}_{2}(h(q))}{}={\bf T}_{2^{n+1}}(h(q))+2{\bf
T}_{2^{n}}\Psi_{\sqrt{2}}(h(q))+2{\bf T}_{2^{n-1}}\Psi_{2}(h(q)),$

and the theorem is proven. ∎

######  Corollary 4.9.

The algebra generated by the operators ${\bf T}_{n}$, for $n\in{\rm Nature}$,
is commutative.

Let $l$ be a fixed prime. We set $Q_{k}={\bf T}_{l}\big{(}h(q)^{k}\big{)}$ for
$k\geq 1$, and for $b\in\frac{K}{I}$ we use $b^{[l]}$ to denote
$b\parallel\left[\begin{smallmatrix}l&\ast\\\ 0&\ast\end{smallmatrix}\right]$.
We set also $Q_{0}=\begin{cases}2l+1,&\text{if }l=2,\\\ l+1,&\text{if }l\neq
2\end{cases}$ and define ${\bf T}_{\frac{k}{l}}$ as the operator that sends
$R_{1}$ to zero if $l$ does not divide $k$. We use the same notation to denote
both $P_{k,r}(t)$ and its image in $\frac{K}{I}[t]$.

######  Proposition 4.10.

For $k\in{\rm Nature}$ write
$P_{k,r}(t)=t^{k}+\sum\limits_{i=1}^{k}b_{k,i}t^{k-i}\in\frac{K}{I}[t]$. If
$l$ is odd then

$\displaystyle
Q_{k}+\sum_{i=1}^{k}b_{k,i}Q_{k-i}+\sum_{i=1}^{k}\big{(}b_{k,i}^{[l]}-b_{k,i}\big{)}h^{[l]}(q^{l})^{k-i}=P_{kl}(h(q))+k{\bf
T}_{\frac{k}{l}}\Psi_{l}(h(q)),$

and if $l=2$ we have

$\displaystyle
Q_{k}+\\!\sum_{i=1}^{k}\\!b_{k,i}Q_{k-i}\\!+\\!\sum_{i=1}^{k}\\!\big{(}b_{k,i}^{[\sqrt{2}]}\\!-b_{k,i}\big{)}\big{(}h^{[\sqrt{2}]}(q)^{k-i}\\!+h^{[\sqrt{2}]}(-q)^{k-i}\big{)}\\!+\\!\sum_{i=1}^{k}\\!\big{(}b_{k,i}^{[2]}\\!-b_{k,i}\big{)}h^{[2]}\big{(}q^{2}\big{)}^{k-i}$
$\displaystyle\qquad{}=\begin{cases}P_{2k,1}(h(q)),&\mbox{if
}2\\!\nmid\\!k,\\\ P_{2k,1}(h(q))+2{\bf T}_{k}\Psi_{\sqrt{2}}(h(q))+2{\bf
T}_{\frac{k}{2}}\Psi_{2}(h(q)),&\mbox{if }2\,|\,k\\\ \end{cases}$

###### Proof.

The case where $l$ is odd can be found in [21]. When $l=2$ we apply ${\bf
T}_{2}$ to both sides of equation (4.2)

$\displaystyle{\bf T}_{2}{\bf T}_{k}(h(q))={\bf
T}_{2}\left(h(q)^{k}+\sum_{i=1}^{k}b_{k,i}h(q)^{k-i}\right),$

and the following manipulation

$\displaystyle{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=b_{k,i}h(q)^{k-i}\parallel\left[\begin{matrix}2&0\\\
0&1\end{matrix}\right]+b_{k,i}h(q)^{k-i}\parallel\left[\begin{matrix}\sqrt{2}&0\\\
0&\sqrt{2}\end{matrix}\right]$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=}{}+b_{k,i}h(q)^{k-i}\parallel\left[\begin{matrix}\sqrt{2}&\frac{1}{\sqrt{2}}\\\
0&\sqrt{2}\end{matrix}\right]+b_{k,i}h(q)^{k-i}\parallel\left[\begin{matrix}1&0\\\
0&2\end{matrix}\right]$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=}{}+b_{k,i}h(q)^{k-i}\parallel\left[\begin{matrix}1&1\\\
0&2\end{matrix}\right]$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}}{}=(b^{[2]}_{k,i}-b_{k,i})\left(h(q)^{k-i}\parallel\left[\begin{matrix}2&0\\\
0&1\end{matrix}\right]\right)$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=}{}+(b^{[\sqrt{2}]}_{k,i}-b_{k,i})\left(h(q)^{k-i}\parallel\left[\begin{matrix}\sqrt{2}&0\\\
0&\sqrt{2}\end{matrix}\right]\right)$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=}{}+(b^{[\sqrt{2}]}_{k,i}-b_{k,i})\left(h(q)^{k-i}\parallel\left[\begin{matrix}\sqrt{2}&\frac{1}{\sqrt{2}}\\\
0&\sqrt{2}\end{matrix}\right]\right)+b_{k,i}{\bf
T}_{2}\big{(}h(q)^{k-i}\big{)}$ $\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}}{}=\big{(}b^{[2]}_{k,i}-b_{k,i}\big{)}h^{[2]}\big{(}q^{2}\big{)}^{k-i}+\big{(}b^{[\sqrt{2}]}_{k,i}-b_{k,i}\big{)}$
$\displaystyle\hphantom{{\bf
T}_{2}\big{(}b_{k,i}h(q)^{k-i}\big{)}=}{}\times\big{(}h^{[\sqrt{2}]}(q)^{k-i}+h^{[\sqrt{2}]}(-q)^{k-i}\big{)}+b_{k,i}Q_{k-i}$

shows that ${\bf T}_{2}{\bf T}_{k}(h(q))$ is equal to

$\displaystyle
Q_{k}+\sum_{i=1}^{k}b_{k,i}Q_{k-i}+\sum_{i=1}^{k}\left(\big{(}b^{[2]}_{k,i}-b_{k,i}\big{)}h^{[2]}\big{(}q^{2}\big{)}^{k-i}+\big{(}b^{[\sqrt{2}]}_{k,i}-b_{k,i}\big{)}\right.$
$\displaystyle\left.\qquad{}\times\big{(}h^{[\sqrt{2}]}(q)^{k-i}+h^{[\sqrt{2}]}(-q)^{k-i}\big{)}\right).$

If $(2,k)=1$ then ${\bf T}_{2}{\bf T}_{k}(h(q))={\bf
T}_{2k}(h(q))=P_{2k,1}(h(q))$ and if $2^{r}$ is the exact power of $2$ that
divides $k$ we have that

$\displaystyle{\bf T}_{2}{\bf T}_{k}(h(q))={\bf T}_{\frac{k}{2^{r}}}{\bf
T}_{2^{r}}{\bf T}_{2}={\bf T}_{\frac{k}{2^{r}}}\big{(}{\bf
T}_{2^{r+1}}(h(q))+2{\bf T}_{2^{r}}\Psi_{\sqrt{2}}(h(q))+2{\bf
T}_{2^{r-1}}\Psi_{2}(h(q))\big{)}$ $\displaystyle\hphantom{{\bf T}_{2}{\bf
T}_{k}(h(q))}{}={\bf T}_{2k}(h(q))+2{\bf T}_{k}\Psi_{\sqrt{2}}(h(q))+2{\bf
T}_{\frac{k}{2}}\Psi_{2}(h(q)).$

This proves the assertion of the theorem. ∎

When $l=2$ the $Q_{k}$ are simply the power sum symmetric functions on
$h^{[2]}\big{(}q^{2}\big{)}$, $h^{[\sqrt{2}]}(q)$, $h^{[\sqrt{2}]}(-q)$,
$h(q)$, $h(-q)$. By induction, using the previous result, one shows that every
$Q_{j}$ is a polynomial in $h^{[2]}(q)$, $h^{[2]}\big{(}q^{2}\big{)}$,
$h^{[\sqrt{2}]}(q)$, $h^{[\sqrt{2}]}(-q)$ and $h(q)$. Let

$\displaystyle\sigma_{n}=\sum_{1\leq i_{1}<i_{2}<\cdots<i_{n}\leq
5}x_{i_{1}}\cdots x_{i_{n}}$

be the elementary symmetric functions in the indeterminates $x_{1}$, $x_{2}$,
$x_{3}$, $x_{4}$, $x_{5}$. We know that the elementary symmetric function are
polynomials in the power sum symmetric functions and from this we conclude
that the elementary symmetric functions on $h^{[2]}\big{(}q^{2}\big{)}$,
$h^{[\sqrt{2}]}(q)$, $h^{[\sqrt{2}]}(-q)$, $h(q)$, $h(-q)$ are polynomials in
$h^{[2]}(q)$, $h^{[2]}\big{(}q^{2}\big{)}$, $h^{[\sqrt{2}]}(q)$,
$h^{[\sqrt{2}]}(-q)$ and $h(q)$. We use these facts in the proof of the
following proposition.

######  Proposition 4.11.

If $f(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}q^{k}$ is completely
$(2{+})$-replicable with replicates
$f^{[n]}(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}^{[n]}q^{k}$, for
$n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$, then

$\displaystyle\begin{split}&\sigma_{2}\left(f^{[2]}(2z),f^{[\sqrt{2}]}(z),f^{[\sqrt{2}]}\left(z+\frac{1}{2}\right),f\left(\frac{z}{2}\right),f\left(\frac{z+1}{2}\right)\right)\\\
&\qquad{}=2a_{2}f(z)-f^{[2]}(z)+2\left(a_{4}-a_{1}\right)+2a_{2}^{[\sqrt{2}]}-\big{(}f^{[\sqrt{2}]}(z)\big{)}^{2}.\end{split}$

###### Proof.

We take $h^{[n]}(q)$, for $n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$ as before
and start by noticing that

$\displaystyle\sigma_{2}\left(h^{[2]}(2z),h^{[\sqrt{2}]}(z),h^{[\sqrt{2}]}\left(z+\frac{1}{2}\right),h\left(\frac{z}{2}\right),h\left(\frac{z+1}{2}\right)\right)=\frac{1}{2}\big{(}Q_{1}^{2}-Q_{2}\big{)}.$

Then we use Proposition 4.10 to express $Q_{1}$ and $Q_{2}$ as polynomials in
$h^{[2]}(q)$, $h^{[2]}\big{(}q^{2}\big{)}$, $h^{[\sqrt{2}]}(q)$,
$h^{[\sqrt{2}]}(-q)$ and $h(q)$.

Now,

$\displaystyle Q_{1}={\bf T}_{1}(h(q))=P_{1}(h(q))=h^{2}(q)-2x_{1},$

and from Proposition 4.10

$\displaystyle
Q_{2}=P_{4}(h(q))+2P_{2}\big{(}h^{[\sqrt{2}]}(q)\big{)}\\!+2h^{[2]}(q)-\big{(}{-}10x_{1}+2\big{(}{-}2x_{1}^{[\sqrt{2}]}\\!+2x_{1}\big{)}+\big{(}{-}2x_{1}^{[2]}\\!+2x_{1}\big{)}\big{)},$

because $b_{2,1}=b_{2,1}^{[\sqrt{2}]}=b_{2,1}^{[2]}=0$ and $b_{2,1}=-2x_{1}$,
$b_{2,1}^{[\sqrt{2}]}=-2x_{1}^{[\sqrt{2}]}$, $b_{2,1}^{[2]}=-2x_{1}^{[2]}$,
and $\frac{1}{2}\big{(}Q_{1}^{2}-Q_{2}\big{)}$ becomes

$\displaystyle\frac{1}{2}(h^{4}(q)-4x_{1}h^{2}(q)+4x_{1}^{2}-\big{(}h^{4}(q)-4x_{1}h^{2}(q)-4x_{2}h(q)-4x_{3}+2x_{1}^{2}$
$\displaystyle\qquad\quad{}+2h^{[\sqrt{2}]}(q)^{2}-4x_{1}^{[\sqrt{2}]}+2h^{[2]}(q)+10x_{1}+4x_{1}^{[\sqrt{2}]}-4x_{1}+2x_{1}^{[2]}-2x_{1}\big{)}$
$\displaystyle\qquad{}=2x_{2}h(q)-\big{(}h^{[\sqrt{2}]}(q)\big{)}^{2}-h^{[2]}(q)+x_{1}^{2}+2x_{3}-2x_{1}-x_{1}^{[2]}.$

Applying the homomorphism $E_{f}$ defined at the beginning of this section we
get

$\displaystyle\sigma_{2}\left(f^{[2]}(2z),f^{[\sqrt{2}]}(z),f^{[\sqrt{2}]}\left(z+\frac{1}{2}\right),f\left(\frac{z}{2}\right),f\left(\frac{z+1}{2}\right)\right)$
$\displaystyle\qquad{}=2a_{2}f(q)-\big{(}f^{[\sqrt{2}]}(q)\big{)}^{2}-f^{[2]}(q)+a_{1}^{2}+2a_{3}-2a_{1}-a_{1}^{[2]}.$

Equating the coefficient of $q^{2}$ in both sides of the equation

$\displaystyle
f^{[2]}\big{(}q^{2}\big{)}+f^{[\sqrt{2}]}(q)+f^{[\sqrt{2}]}(-q)+f\big{(}q^{\frac{1}{2}}\big{)}+f\big{(}{-}q^{\frac{1}{2}}\big{)}=P_{2}(f(q)),$

we see that $a_{1}^{2}+2a_{3}-a_{1}^{[2]}=2a_{4}+2a_{2}^{[\sqrt{2}]}$ and this
concludes the proof. ∎

######  Theorem 4.12.

If $f(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}q^{k}$ is completely
$(2{+})$-replicable with replicates
$f^{[n]}(q)=\frac{1}{q}+\sum\limits_{k=1}^{\infty}a_{k}^{[n]}q^{k}$, for
$n\in{\rm Nature}\cup\sqrt{2}{\rm Nature}$, then their coefficients satisfy
the following recurrence relation:

$\displaystyle 1)\quad$ $\displaystyle
a_{4k}=a_{2k+1}+\sum_{j=1}^{k-1}a_{j}a_{2k-j}+\frac{1}{2}\big{(}a_{k}^{2}-a_{k}^{[2]}\big{)}-a_{2k}^{[\sqrt{2}]},$
$\displaystyle 2)\quad$ $\displaystyle
a_{4k+1}=a_{2k+3}-a_{2}a_{2k}+\sum_{j=1}^{k}a_{j}a_{2k+2-j}+\sum_{j=1}^{k-1}a_{j}^{[2]}a_{2k-2j}^{[\sqrt{2}]}+2\sum_{j=1}^{k-1}a_{4j}a_{2k-2j}^{[\sqrt{2}]}$
$\displaystyle\hphantom{a_{4k+1}=}{}+\sum_{j=1}^{k-1}a_{4j}a_{k-j}^{[2]}+\sum_{j=1}^{2k-1}(-1)^{j}a_{j}a_{4k-j}+\sum_{j=1}^{k-1}a_{2j}^{[\sqrt{2}]}a_{2k-2j}^{[\sqrt{2}]}$
$\displaystyle\hphantom{a_{4k+1}=}{}+\frac{1}{2}\big{(}a_{k+1}^{2}-a_{k+1}^{[2]}+a_{2k}^{2}+a_{2k}^{[2]}\big{)},$
$\displaystyle 3)\quad$ $\displaystyle
a_{4k+2}=\sum_{j=1}^{k}a_{j}a_{2k+1-j}+a_{2k+2},$ $\displaystyle 4)\quad$
$\displaystyle
a_{4k+3}=a_{2k+4}-a_{2}a_{2k+1}-\frac{1}{2}\big{(}a_{2k+1}^{2}-a_{2k+1}^{[2]}\big{)}+\sum_{j=1}^{2k}(-1)^{j}a_{j}a_{4k+2-j}+a_{2k+2}^{[\sqrt{2}]}$
$\displaystyle\hphantom{a_{4k+3}=}{}+\sum_{j=1}^{k}a_{4j-2}a_{k+1-j}^{[2]}+\sum_{j=1}^{k+1}a_{j}a_{2k+3-j}+2\sum_{j=1}^{2k}a_{2j}a_{2k+1-j}^{[\sqrt{2}]}+\sum_{j=1}^{k}a_{j}^{[\sqrt{2}]}a_{2k+1-j}^{[\sqrt{2}]}.$

###### Proof.

This is a consequence of the following two identities

$\displaystyle\sigma_{1}\left(f^{[2]}(2z),f^{[\sqrt{2}]}(z),f^{[\sqrt{2}]}\left(z+\frac{1}{2}\right),f\left(\frac{z}{2}\right),f\left(\frac{z+1}{2}\right)\right)=P_{2,f}(f(z)),$
$\displaystyle\sigma_{2}\left(f^{[2]}(2z),f^{[\sqrt{2}]}(z),f^{[\sqrt{2}]}\left(z+\frac{1}{2}\right),f\left(\frac{z}{2}\right),f\left(\frac{z+1}{2}\right)\right)$
$\displaystyle\qquad{}=2a_{2}f(z)-f^{[2]}(z)+2\left(a_{4}-a_{1}\right)+2a_{2}^{[\sqrt{2}]}-\big{(}f^{[\sqrt{2}]}(z)\big{)}^{2}.$
∎

##  5 The baby monster Lie algebra and $\boldsymbol{(2{+})}$-replication

The moonshine module $V^{\natural}$ was constructed in [15] by Frenkel,
Lepowsky and Meurman. This is a vertex operator algebra that has $\mathbb{M}$,
the monster group, as symmetry group A vertex operator algebra is an intricate
algebraic structure and we refer to [20] for the definition and the basics of
its theory. The moonshine module has a grading
$V^{\natural}=\bigoplus_{n\geq-1}V^{\natural}_{(n)}$ and its graded dimension
$\sum\limits_{n\geq-1}\big{(}\dim V^{\natural}_{(n)}\big{)}q^{n}$ is the
$J$-function. Borcherds ([4]) uses this vertex operator algebra to prove the
moonshine conjectures in the following way. First, he shows that the
McKay–Thompson series for $V^{\natural}$, i.e.,
$T_{g}=\sum\limits_{n\geq-1}\operatorname{Tr}\big{(}g|V^{\natural}_{(n)}\big{)}q^{n}$
are completely replicable functions. To do this, he uses $V^{\natural}$ to
build a generalized Kac–Moody algebra, the monster Lie algebra, whose twisted
denominator identity is essentially the statement that the $n$-th replicate of
a $T_{g}$ is $T_{g^{n}}$. Knowing that Hauptmoduls for genus-zero congruence
groups are also completely replicable functions and that completely replicable
functions satisfy some recurrence relations that determine a function from the
first 5 coefficients of the function and its replicates, he was able to show
that every McKay–Thompson series is indeed a Hauptmodul for some genus-zero
congruence groups by just comparing the first few coefficients of the
functions involved.

In [18], Höhn shows that there is a vertex algebra $W$ where
$2\cdot\mathbb{B}$ acts as a symmetry group. This group is a central extension
of $\mathbb{B}$, the baby monster group, and it arises as the centralizer of
an element of class $2A$ in $\mathbb{M}$. This vertex operator algebra plays
for $2\cdot\mathbb{B}$ the role that $V^{\natural}$ plays for the $\mathbb{M}$
and it was used to prove the generalized moonshine conjectures for the case of
the baby monster, i.e., when $g$ in $T_{g,h}$ (see [23] for a precise
statement of the generalized moonshine conjectures and what $T_{g,h}$ is) is
an involution of type $2A$ in $\mathbb{M}$. In this section, we use $t$ to
represent a (fixed) element in class $2A$ in $\mathbb{M}$. More precisely,
what Höhn states in [18] is the following. If

$\displaystyle
V^{\natural}(t)=\bigoplus_{n\geq-1}V^{\natural}_{(\frac{n}{2})}(t)$

is the $t$-twisted module and $h$ is an element in the centralizer of $t$ in
$\mathbb{M}$ then the McKay–Thompson series

$\displaystyle
T_{t,g}=\sum_{n\geq-1}\operatorname{Tr}\big{(}g|V^{\natural}_{\left(\frac{n}{2}\right)}(t)\big{)}q^{n}$

is the Hauptmodul for some genus zero congruence subgroup.

We give a very brief sketch of the results from [18]. From the decompositions
of $V^{\natural}=V^{00}\bigoplus V^{01}$ and $V^{\natural}(t)=V^{10}\bigoplus
V^{11}$ of $+1$ and $-1$ eigenspaces for $t$, Höhn builds the vertex algebra
$W$ mentioned above on which $2\cdot\mathbb{B}$ acts. Using this vertex
algebra $W$ a Lie algebra $g_{\mathbb{B}}^{\natural}$, the baby monster Lie
algebra, is constructed too. This is a
$\frac{1}{2}\mathbb{Z}\times\frac{1}{2}\mathbb{Z}$-graded Lie algebra that has
an action of $2\cdot\mathbb{B}$ in it that respects the grading and its
$\big{(}\frac{m}{2},\frac{n}{2}\big{)}$ piece is isomorphic to
$V^{[m,n]}_{(2mn)}$ ($[\cdot,\cdot]$ represents reduction ${\rm mod}\,2$).
Also, $V^{10}$ is isomorphic to $V^{01}$ as $2\cdot\mathbb{B}$-modules. Höhn
shows that $W$ is a generalized Kac–Moody algebra with twisted denominator
formula

$\displaystyle\sum_{m\in\mathbb{Z}}\operatorname{Tr}\big{(}g|V^{[m,1]}_{\left(\frac{m}{2}\right)}\big{)}p^{\frac{m}{2}}-\sum_{n\in\mathbb{Z}}\operatorname{Tr}\big{(}g|V^{[1,n]}_{\left(\frac{n}{2}\right)}\big{)}q^{\frac{n}{2}}$
$\displaystyle\qquad{}=p^{-\frac{1}{2}}\exp\left(-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[m,n]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{\frac{im}{2}}q^{\frac{in}{2}}}{i}\right).$
(5.1)

We can now state our main result.

######  Theorem 5.1.

The McKay–Thompson series $T_{t,g}(z)$, for $g\in 2\cdot\mathbb{B}$, are
completely $(2{+})$-replicable with replicates $T_{t,g}^{[n]}=T_{t,g^{n}}$ and
$T_{t,g}^{[\sqrt{2}n]}=T_{1,g^{n}t}$.

###### Proof.

From the twisted denominator identity for the baby monster Lie algebra (5.1)
we have, after substituting $p^{\frac{1}{2}}$ for $p$ and $q^{\frac{1}{2}}$
for $q$,

$\displaystyle\sum_{m\in\mathbb{Z}}\operatorname{Tr}\big{(}g|V^{[m,1]}_{\left(\frac{m}{2}\right)}\big{)}p^{m}-\sum_{n\in\mathbb{Z}}\operatorname{Tr}\big{(}g|V^{[1,n]}_{\left(\frac{n}{2}\right)}\big{)}q^{n}=p^{-1}\exp\left(-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[m,n]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}\right).$

The right side of this equation is $p^{-1}\exp(Z)$, where

$\displaystyle Z=-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[m,n]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}.$

Because of the isomorphism between $V^{01}$ and $V^{10}$,

$\displaystyle Z=-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{or}\ n\
\text{odd}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[1,mn]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{and}\ n\
\text{even}\end{subarray}}\operatorname{Tr}\big{(}g^{i}t|V^{00}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}$
$\displaystyle\hphantom{Z}{}=-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{or}\ n\
\text{odd}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[1,mn]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}$
$\displaystyle\hphantom{Z=}{}-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{and}\ n\
\text{even}\end{subarray}}\left(\operatorname{Tr}\big{(}g^{i}t|V^{\natural}_{\left(\frac{mn}{2}\right)}\big{)}-\operatorname{Tr}\big{(}g^{i}t|V^{01}_{\left(\frac{mn}{2}\right)}\big{)}\right)\frac{p^{im}q^{in}}{i}$
$\displaystyle\hphantom{Z}{}=-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{or}\ n\
\text{odd}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[1,mn]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}$
$\displaystyle\hphantom{Z=}{}-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\\\ m\ \text{and}\ n\
\text{even}\end{subarray}}\left(\operatorname{Tr}\big{(}g^{i}t|V^{\natural}_{\left(\frac{mn}{2}\right)}\big{)}+\operatorname{Tr}\big{(}g^{i}|V^{01}_{\left(\frac{mn}{2}\right)}\big{)}\right)\frac{p^{im}q^{in}}{i}$
$\displaystyle\hphantom{Z}{}=-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\end{subarray}}\operatorname{Tr}\big{(}g^{i}|V^{[1,mn]}_{\left(\frac{mn}{2}\right)}\big{)}\frac{p^{im}q^{in}}{i}-\sum_{i>0}\sum_{\begin{subarray}{c}m\in\mathbb{Z}^{+}\\\
n\in\mathbb{Z}\end{subarray}}\operatorname{Tr}\big{(}g^{i}t|V^{\natural}_{(2mn)}\big{)}\frac{p^{2im}q^{2in}}{i}$
$\displaystyle\hphantom{Z}{}=-\sum_{n=1}^{+\infty}\frac{1}{n}\left(\sum_{ad=n}d\cdot\sum_{k\in\mathbb{Z}}\operatorname{Tr}\big{(}g^{a}|V_{\left(\frac{kd}{2}\right)}^{[1,kd]}\big{)}q^{ak}+\sum_{\begin{subarray}{c}ad=n\\\
d\
\text{even}\end{subarray}}d\cdot\sum_{k\in\mathbb{Z}}\operatorname{Tr}\big{(}g^{a}t|V_{\left(\frac{kd}{2}\right)}^{\natural}\big{)}q^{2ak}\right)p^{n}$
$\displaystyle\hphantom{Z}{}=-\sum_{n=1}^{+\infty}\frac{1}{n}\left(\sum_{\begin{subarray}{c}ad=n\\\
0\leq
b<d\end{subarray}}T_{t,g^{a}}\left(\frac{a\tau+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}T_{1,g^{a}t}\left(\frac{2a\tau+b}{d}\right)\right)p^{n}.$

Since

$\displaystyle
T_{t,g}(p)-T_{t,g}(q)=p^{-1}\exp\left(-\sum_{n=1}^{+\infty}\frac{1}{n}P_{n}(T_{t,g}(q))p^{n}\right),$

where $P_{n}$ is the $n$-th Faber polynomial, we conclude that, for all
$n\in\mathbb{N}$,

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}T_{t,g^{a}}\left(\frac{a\tau+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}T_{1,g^{a}t}\left(\frac{2a\tau+b}{d}\right)=P_{n}(T_{t,g}(q)),$

and we get that $T_{t,g}$ is $(2{+})$-replicable with $(2{+})$-replicates
given as stated in the theorem.

By substituting $g$ by $g^{n}$ we see that $T_{t,g^{n}}$ is
$(2{+})$-replicable with replicates
$T_{t,g^{n}}^{[m]}=T_{t,g^{mn}}=T_{t,g}^{[mn]}$ and
$T_{t,g^{n}}^{[\sqrt{2}m]}=T_{1,g^{mn}t}=T_{t,g}^{[\sqrt{2}nm]}$, i.e.
$\big{(}T_{t,g}^{[n]}\big{)}^{[m]}=T_{t,g}^{[mn]}$ and
$\big{(}T_{t,g}^{[n]}\big{)}^{[\sqrt{2}m]}=T_{t,g}^{[\sqrt{2}nm]}$. To
complete the proof it remains to see that $T_{t,g}^{[\sqrt{2}n]}=T_{1,g^{n}t}$
is $(2{+})$-replicable with replicates
$\big{(}T_{t,g}^{[\sqrt{2}n]}\big{)}^{[m]}=T_{1,g^{nm}t}$ and
$\big{(}T_{t,g}^{[\sqrt{2}n]}\big{)}^{[\sqrt{2}m]}=T_{t,g^{2nm}}$, for
$m\in{\rm Nature}$. Equivalently, what we need to prove is that, for every
$m\in{\rm Nature}$,

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}T_{1,g^{ma}t}\left(\frac{a\tau+b}{d}\right)+\sum_{\begin{subarray}{c}ad=n\\\
d\ \text{even}\\\ 0\leq
b<d\end{subarray}}T_{t,g^{2ma}}\left(\frac{2a\tau+b}{d}\right)=P_{n}\big{(}T_{1,g^{m}t}(q)\big{)}.$

But since $T_{1,g^{m}t}$ is a monstrous function we know that

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}T_{1,(g^{m}t)^{a}}\left(\frac{a\tau+b}{d}\right)=P_{n}\big{(}T_{1,g^{m}t}(q)\big{)}.$

But now,

$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}T_{1,g^{ma}t}\left(\frac{a\tau+b}{d}\right)=n\sum_{k\in{\rm
Nature}}\left(\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)}\right)q^{k},$
$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ d\ \text{even}\\\ 0\leq
b<d\end{subarray}}T_{t,g^{2ma}}\left(\frac{2a\tau+b}{d}\right)=n\sum_{k\in{\rm
Nature}}\left(\sum_{\begin{subarray}{c}a|(n,k)\\\ \frac{n}{a}\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{[1,\frac{kn}{a^{2}}]}_{\left(\frac{nk}{2a^{2}}\right)}\big{)}\right)q^{2k},$
$\displaystyle\sum_{\begin{subarray}{c}ad=n\\\ 0\leq
b<d\end{subarray}}T_{1,(g^{m}t)^{a}}\left(\frac{a\tau+b}{d}\right)=n\sum_{k\in{\rm
Nature}}\left(\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)}\right)q^{k},$

and what we have to show is

$\displaystyle\sum_{k\in{\rm
Nature}}\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)}q^{k}+\sum_{k\in{\rm
Nature}}\sum_{\begin{subarray}{c}a|(n,k)\\\ \frac{n}{a}\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{[1,\frac{kn}{a^{2}}]}_{\left(\frac{nk}{2a^{2}}\right)}\big{)}q^{2k}$
$\displaystyle\qquad{}=\sum_{k\in{\rm
Nature}}\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)}q^{k}.$

This means that for $k$ odd we have to show that

$\displaystyle\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)}=\sum_{a|(n,k)}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{nk}{a^{2}}\right)}\big{)},$

which is true because $t$ and $g$ commute and every $a$ in the sum, being a
divisor of $k$, is odd too.

For $k=2k^{\prime}$ even we have to show that

$\displaystyle\sum_{a|(n,2k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{2nk^{\prime}}{a^{2}}\right)}\big{)}+\sum_{\begin{subarray}{c}a|(n,k^{\prime})\\\
\frac{n}{a}\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{[1,\frac{k^{\prime}n}{a^{2}}]}_{\left(\frac{nk^{\prime}}{2a^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{a|(n,2k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{2nk^{\prime}}{a^{2}}\right)}\big{)}.$

This identity is clearly true for $n$ odd and for $n=2n^{\prime}$ even it
becomes, because of the isomorphism between $V^{01}$ and $V^{10}$,

$\displaystyle\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}+\sum_{\begin{subarray}{c}a|(2n^{\prime},k^{\prime})\\\
\frac{2n^{\prime}}{a}\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{01}_{\left(\frac{n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$

and this is now easy to prove

$\displaystyle\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}+\sum_{\begin{subarray}{c}a|(2n^{\prime},k^{\prime})\\\
\frac{2n^{\prime}}{a}\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{01}_{\left(\frac{n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}+\sum_{a|(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{2ma}|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{(2a)^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}-\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{even}\end{subarray}}\frac{2}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{00}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}+\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad\quad{}+\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{odd}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}-\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{even}\end{subarray}}\frac{2}{a}\operatorname{Tr}\big{(}g^{ma}t|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad{}=\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{00}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}+\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{even}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}$
$\displaystyle\qquad\quad{}+\sum_{\begin{subarray}{c}a|2(n^{\prime},k^{\prime})\\\
a\
\text{odd}\end{subarray}}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{01}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)}=\sum_{a|2(n^{\prime},k^{\prime})}\frac{1}{a}\operatorname{Tr}\big{(}(g^{m}t)^{a}|V^{\natural}_{\left(\frac{4n^{\prime}k^{\prime}}{a^{2}}\right)}\big{)},$

and the theorem is proven. ∎

We can now use Theorem 5.1 to reprove Höhn’s result

######  Theorem 5.2.

The McKay–Thompson series $T_{t,g}(z)$, for $g\in 2\cdot\mathbb{B}$, are the
$q$-expansions of Hauptmoduls.

###### Proof.

We know that the McKay–Thompson series $T_{t,g}$ are completely
$(2{+})$-replicable and consequently satisfy the recurrence relations from
Theorem 4.12. We know that $T^{[n\sqrt{2}]}_{t,g}=T_{1,tg^{n}}$ is a monstrous
function and therefore its coefficients are known once we know in what class
in the monster the element $tg$ is, for every $g\in 2\cdot\mathbb{B}$. This
can be done with GAP [17]. Hence, the first five coefficients of every
$T_{t,g}$ determine all the coefficients of the $T_{t,g}$ completely. From
Section 3 we also have some completely $(2{+})$-replicability results for some
Hauptmoduls and thus these Hauptmoduls satisfy the same recurrence relations
from Theorem 4.12. To prove that every McKay–Thompson series is a Hauptmodul
it would be enough to compare, for every $g\in 2\cdot\mathbb{B}$, the first
five coefficients of $T_{t,g},T_{1,tg},T_{t,g^{2}},T_{1,tg^{2}},\ldots$ with
those of $f,f^{[\sqrt{2}]},f^{[2]},f^{[2\sqrt{2}]},\ldots$, respectively, for
some Hauptmodul $f$ in Tables 1 and 2.

However, not all McKay–Thompson series correspond to Hauptmoduls listed in
Tables 1 and 2 and, because of that, this method for proving that the
McKay–Thompson series are Hauptmoduls works for all 247 classes in
$2\cdot\mathbb{B}$ with 13 exceptions. This happens because the Hauptmodul
associated to each of these 13 classes is neither a completely replicable
function nor a dash (see [14] for the definition of the dash operator) of a
completely replicable function and our Tables 1 and 2 only contain such
functions.

Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$ | Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$
---|---|---|---|---|---
$1a\>(1a)$ | $2A$ | $2A$ | $2a\>(1a)$ | $4\sim b$ | $1A$
$2b\>(1a)$ | $2a$ | $2A$ | $2c\>(1a)$ | $4A$ | $2B$
$2d\>(1a)$ | $2B$ | $2A$ | $2e\>(1a)$ | $4C$ | $2B$
$3a\>(3a)$ | $6A$ | $6A$ | $3b\>(3b)$ | $6D$ | $6D$
$4a\>(2a)$ | $8\sim b$ | $4B$ | $4b\>(2d)$ | $4a$ | $4A$
$4c\>(2d)$ | $4B$ | $4A$ | $4d\>(2d)$ | $4C$ | $4C$
$4e\>(2c)$ | $8a$ | $4B$ | $4f\>(2e)$ | $8A$ | $4C$
$4g\>(2e)$ | $8\sim d$ | $4A$ | $4h\>(2d)$ | $4D$ | $4C$
$4i\>(2c)$ | $8B$ | $4B$ | $4j\>(2e)$ | $8E$ | $4C$
$4k\>(2e)$ | $8D$ | $4D$ | $5a\>(5a)$ | $10A$ | $10A$
$5b\>(5b)$ | $10C$ | $10C$ | $6a\>(3a)$ | $12\sim d$ | $3A$
$6b\>(3b)$ | $12\sim f$ | $3B$ | $6c\>(3a)$ | $6a$ | $6A$
$6d\>(3a)$ | $6b$ | $6A$ | $6e\>(3a)$ | $12A$ | $6C$
$6f\>(3a)$ | $6C$ | $6A$ | $6g\>(3b)$ | $6c$ | $6D$
$6h\>(3a)$ | $12c$ | $6C$ | $6i\>(3b)$ | $12B$ | $6E$
$6j\>(3b)$ | $6E$ | $6D$ | $6k\>(3a)$ | $12E$ | $6C$
$6l\>(3b)$ | $12H$ | $6E$ | $6m\>(3b)$ | $12\sim h$ | $6B$
$6n\>(3b)$ | $12I$ | $6E$ | $7a\>(7a)$ | $14A$ | $14A$
$8a\>(4a)$ | $16\sim a$ | $8C$ | $8b\>(4d)$ | $8a$ | $8A$
$8c\>(4c)$ | $8b$ | $8B$ | $8d\>(4c)$ | $8c$ | $8B$
$8e\>(4d)$ | $8B$ | $8A$ | $8f\>(4c)$ | $8C$ | $8B$
$8g\>(4d)$ | $8D$ | $8E$ | $8h\>(4g)$ | $16\sim d$ | $8D$
$8i\>(4d)$ | $8E$ | $8E$ | $8j\>(4h)$ | $8F$ | $8D$
$8k\>(4f)$ | $16A$ | $8B$ | $8l\>(4j)$ | $16C$ | $8E$
$8m\>(4j)$ | $16\sim e$ | $8A$ | $8n\>(4i)$ | $16a$ | $8C$
$8o\>(4j)$ | $16B$ | $8E$ | $8p\>(4k)$ | $16d$ | $8F$
$9a\>(9a)$ | $18A$ | $18A$ | $9b\>(9b)$ | $18B$ | $18B$
$10a\>(5a)$ | $20\sim c$ | $5A$ | $10b\>(5b)$ | $20\sim d$ | $5B$
$10c\>(5a)$ | $10a$ | $10A$ | $10d\>(5a)$ | $20A$ | $10B$
$10e\>(5a)$ | $10B$ | $10A$ | $10f\>(5b)$ | $20C$ | $10E$
$10g\>(5b)$ | $10E$ | $10C$ | $10h\>(5a)$ | $20d$ | $10B$
$10i\>(5b)$ | $20F$ | $10E$ | $10j\>(5b)$ | $20\sim g$ | $10D$
$11a\>(11a)$ | $22A$ | $22A$ | $12a\>(6a)$ | $24\sim f$ | $12C$
$12b\>(6b)$ | $24\sim h$ | $12G$ | $12c\>(6f)$ | $12a$ | $12A$
$12d\>(6j)$ | $12G$ | $12B$ | $12e\>(6f)$ | $12b$ | $12A$
$12f\>(6f)$ | $12C$ | $12A$ | $12g\>(6e)$ | $24a$ | $12C$
$12h\>(6m)$ | $24\sim j$ | $12H$ | $12i\>(6m)$ | $24\sim k$ | $12B$
$12j\>(6f)$ | $12E$ | $12E$ | $12k\>(6e)$ | $24b$ | $12C$
$12l\>(6f)$ | $12d$ | $12E$ | $12m\>(6k)$ | $24B$ | $12E$
$12n\>(6k)$ | $24\sim m$ | $12A$ | $12o\>(6i)$ | $24c$ | $12G$
$12p\>(6e)$ | $24A$ | $12C$ | $12q\>(6j)$ | $12I$ | $12I$
$12r\>(6n)$ | $24C$ | $12I$ | $12s\>(6n)$ | $24\sim o$ | $12B$
$12t\>(6j)$ | $12F$ | $12H$ | $12u\>(6k)$ | $24h$ | $12E$
$12v\>(6m)$ | $24\sim q$ | $12I$ | $12w\>(6l)$ | $24H$ | $12F$
$12x\>(6n)$ | $24I$ | $12I$ | $12y\>(6n)$ | $24\sim r$ | $12H$
$13a\>(13a)$ | $26A$ | $26A$ | $14a\>(7a)$ | $28\sim c$ | $7A$
$14b\>(7a)$ | $14a$ | $14A$ | $14c\>(7a)$ | $14c$ | $14A$
$14d\>(7a)$ | $28B$ | $14B$ | $14e\>(7a)$ | $14B$ | $14A$
$14f\>(7a)$ | $28C$ | $14B$ | $15a\>(15a)$ | $30B$ | $30B$
$15b\>(15b)$ | $30F$ | $30F$ | $16a\>(8e)$ | $16b$ | $16A$
$16b\>(8e)$ | $16c$ | $16A$ | $16c\>(8i)$ | $16B$ | $16B$
$16d\>(8i)$ | $16A$ | $16C$ | $16e\>(8e)$ | $16a$ | $16A$
$16f\>(8i)$ | $16A$ | $16C$ | $16g\>(8l)$ | $32B$ | $16A$
$16h\>(8o)$ | $32A$ | $16B$ | $16i\>(8o)$ | $32\sim e$ | $16C$
$17a\>(17a)$ | $34A$ | $34A$ | $18a\>(9a)$ | $36\sim h$ | $9B$
$18b\>(9b)$ | $36\sim e$ | $9A$ | $18c\>(9b)$ | $18c$ | $18B$
$18d\>(9b)$ | $18c$ | $18B$ | $18e\>(9b)$ | $36A$ | $18C$

Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$ | Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$
---|---|---|---|---|---
$18f\>(9b)$ | $18C$ | $18B$ | $18g\>(9b)$ | $36f$ | $18C$
$18h\>(9a)$ | $36D$ | $18D$ | $18i\>(9a)$ | $36\sim q$ | $18E$
$19a\>(19a)$ | $38A$ | $38A$ | $20a\>(10a)$ | $40\sim c$ | $20B$
$20b\>(10e)$ | $20a$ | $20A$ | $20c\>(10g)$ | $20c$ | $20C$
$20d\>(10e)$ | $20b$ | $20A$ | $20e\>(10e)$ | $20B$ | $20A$
$20f\>(10d)$ | $40a$ | $20B$ | $20g\>(10d)$ | $40B$ | $20B$
$20h\>(10j)$ | $40\sim h$ | $20F$ | $20i\>(10j)$ | $40\sim i$ | $20C$
$20j\>(10g)$ | $20E$ | $20F$ | $20k\>(10i)$ | $40C$ | $20E$
$21a\>(21a)$ | $42A$ | $42A$ | $22a\>(11a)$ | $44\sim b$ | $11A$
$22b\>(11a)$ | $22a$ | $22A$ | $22c\>(11a)$ | $22a$ | $22A$
$22d\>(11a)$ | $44A$ | $22B$ | $22e\>(11a)$ | $22B$ | $22A$
$23a\>(23a)$ | $46C$ | $46C$ | $23b\>(23b)$ | $46C$ | $46C$
$24a\>(12b)$ | $48\sim c$ | $24G$ | $24b\>(12f)$ | $24d$ | $24A$
$24c\>(12f)$ | $24e$ | $24A$ | $24d\>(12f)$ | $24g$ | $24A$
$24e\>(12f)$ | $24f$ | $24A$ | $24f\>(12j)$ | $24b$ | $24B$
$24g\>(12q)$ | $24c$ | $24C$ | $24h\>(12j)$ | $24A$ | $24B$
$24i\>(12i)$ | $48\sim h$ | $24H$ | $24j\>(12n)$ | $48\sim i$ | $24D$
$24k\>(12m)$ | $48A$ | $24A$ | $24l\>(12q)$ | $24H$ | $24I$
$24m\>(12v)$ | $48\sim j$ | $24I$ | $24n\>(12v)$ | $48\sim k$ | $24C$
$24o\>(12t)$ | $24F$ | $24H$ | $24p\>(12w)$ | $48h$ | $24F$
$25a\>(25a)$ | $50A$ | $50A$ | $26a\>(13a)$ | $52\sim c$ | $13A$
$26b\>(13a)$ | $26a$ | $26A$ | $27a\>(27a)$ | $54A$ | $54A$
$28a\>(14a)$ | $56\sim d$ | $28A$ | $28b\>(14e)$ | $28A$ | $28B$
$28c\>(14e)$ | $28C$ | $28C$ | $28d\>(14e)$ | $28a$ | $28B$
$28e\>(14d)$ | $56a$ | $28A$ | $28f\>(14f)$ | $56A$ | $28C$
$28g\>(14f)$ | $56\sim g$ | $28B$ | $30a\>(15a)$ | $60\sim c$ | $15A$
$30b\>(15b)$ | $60\sim l$ | $15C$ | $30c\>(15a)$ | $30a$ | $30B$
$30d\>(15a)$ | $30d$ | $30B$ | $30e\>(15a)$ | $60B$ | $30C$
$30f\>(15a)$ | $30C$ | $30B$ | $30g\>(15a)$ | $60a$ | $30C$
$30h\>(15b)$ | $60D$ | $30G$ | $30i\>(15b)$ | $60\sim m$ | $30A$
$30j\>(15b)$ | $60C$ | $30G$ | $30k\>(15b)$ | $30G$ | $30F$
$30l\>(15b)$ | $60C$ | $30G$ | $30m\>(15b)$ | $30G$ | $30F$
$31a\>(31a)$ | $62A$ | $62A$ | $31b\>(31b)$ | $62A$ | $62A$
$32a\>(16c)$ | $32B$ | $32A$ | $32b\>(16c)$ | $32B$ | $32A$
$32c\>(16d)$ | $32b$ | $32B$ | $32d\>(16d)$ | $32b$ | $32B$
$33a\>(33a)$ | $66A$ | $66A$ | $34a\>(17a)$ | $68\sim b$ | $17A$
$34b\>(17a)$ | $34a$ | $34A$ | $34c\>(17a)$ | $34a$ | $34A$
$35a\>(35a)$ | $70A$ | $70A$ | $36a\>(18b)$ | $72\sim c$ | $36C$
$36b\>(18f)$ | $36C$ | $36A$ | $36c\>(18e)$ | $72a$ | $36C$
$36d\>(18i)$ | $72\sim p$ | $36D$ | $36e\>(18i)$ | $72\sim q$ | $36B$
$38a\>(19a)$ | $76\sim b$ | $19A$ | $38b\>(19a)$ | $38a$ | $38A$
$38c\>(19a)$ | $38a$ | $38A$ | $39a\>(39a)$ | $78A$ | $78A$
$40a\>(20a)$ | $80\sim a$ | $40A$ | $40b\>(20e)$ | $40b$ | $40B$
$40c\>(20e)$ | $40c$ | $40B$ | $40d\>(20e)$ | $40A$ | $40B$
$40e\>(20g)$ | $80a$ | $40A$ | $40f\>(20i)$ | $80\sim e$ | $40C$
$40g\>(20i)$ | $80\sim e$ | $40C$ | $42a\>(21a)$ | $84\sim e$ | $21A$
$42b\>(21a)$ | $42a$ | $42A$ | $42c\>(21a)$ | $42b$ | $42A$
$44a\>(22e)$ | $44c$ | $44A$ | $44b\>(22e)$ | $44c$ | $44A$
$46a\>(23a)$ | $92\sim b$ | $23A$ | $46b\>(23b)$ | $92\sim b$ | $23A$
$46c\>(23c)$ | $92A$ | $46A$ | $46d\>(23a)$ | $46A$ | $46C$
$46e\>(23b)$ | $92A$ | $46A$ | $46f\>(23b)$ | $46A$ | $46C$
$47a\>(47a)$ | $94A$ | $94A$ | $47b\>(47b)$ | $94A$ | $94A$
$48a\>(24h)$ | $48a$ | $48A$ | $48b\>(24h)$ | $48b$ | $48A$
$50a\>(25a)$ | $100\sim c$ | $25A$ | $52a\>(26a)$ | $104\sim c$ | $52A$
$54a\>(27a)$ | $108\sim g$ | $27A$ | $55a\>(55a)$ | $110A$ | $110A$
$56a\>(28c)$ | $56a$ | $56A$ | $56b\>(28c)$ | $56a$ | $56A$
$60a\>(30a)$ | $120\sim d$ | $60A$ | $60b\>(30f)$ | $60b$ | $60B$

Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$ | Class of $g$ (of $g^{2}$) | $T_{t,g}$ | $T^{[\sqrt{2}]}_{t,g}=T_{1,gt}$
---|---|---|---|---|---
$60c\>(30e)$ | $120a$ | $60A$ | $60d\>(30i)$ | $120\sim g$ | $60D$
$60e\>(30i)$ | $120\sim h$ | $60C$ | $62a\>(31a)$ | $124\sim b$ | $31A$
$62b\>(31b)$ | $124\sim b$ | $31A$ | $66a\>(33a)$ | $132\sim c$ | $33B$
$66b\>(33a)$ | $66a$ | $66A$ | $66c\>(33a)$ | $66a$ | $66A$
$68a\>(34a)$ | $136\sim b$ | $68A$ | $70a\>(35a)$ | $140\sim b$ | $35A$
$70b\>(35a)$ | $70a$ | $70A$ | $70c\>(35a)$ | $70a$ | $70A$
$78a\>(39a)$ | $156\sim d$ | $39A$ | $84a\>(42a)$ | $168\sim c$ | $84A$
$94a\>(47a)$ | $188\sim b$ | $47A$ | $94b\>(47a)$ | $188\sim b$ | $47A$
$104a\>(52a)$ | $208\sim a$ | $104A$ | $104b\>(52a)$ | $208\sim a$ | $104A$
$110a\>(55a)$ | $220\sim b$ | $55A$ |  |  |

Table 3: Classes in $2\cdot\mathbb{B}$ and corresponding McKay–Thompson series
together with their $\big{[}\sqrt{2}\big{]}$-replicates.

For those $247-13=234$ classes covered by Tables 1 and 2 we use the
decomposition of the first five head characters given in [18]:

  * 1)

$H_{1}=\chi_{1}+\chi_{2}$,

  * 2)

$H_{2}=\chi_{185}$,

  * 3)

$H_{3}=2\chi_{1}+\chi_{2}+\chi_{3}+\chi_{4}$,

  * 4)

$H_{4}=2\chi_{185}+\chi_{186}$,

  * 5)

$H_{5}=3\chi_{1}+3\chi_{2}+2\chi_{3}+\chi_{4}+\chi_{6}+\chi_{7}$

to make the comparison and we obtain a proof of Höhn’s results which is
analogous to Borcherds proof of the original moonshine conjectures.

For the remaining 13 classes (their names are, using GAP notation: $12h$,
$12i$, $20h$, $20i$, $24i$, $24m$, $24n$, $36d$, $36e$, $40f$, $40g$, $60d$,
$60e$) we use again the recurrence relations from Theorem 4.12 to find the
first 23 coefficients of their McKay–Thompson series. Since we know that a
replicable function (we recall from Remark 2.13 that a $(2{+})$-replicable
function is replicable) is completely determined by its first 23 coefficients,
a simple check among the power series expansions of the 616 Hauptmoduls for
genus zero groups with rational integer coefficients allows us to match every
such class with some Hauptmodul (see references [9, 14, 24] for details of
these functions and the corresponding groups). This is analogous to Höhn’s
proof but the recurrence relations from Theorem 4.12 simplify the computations
greatly. ∎

A list of $2\cdot\mathbb{B}$ classes with their corresponding McKay–Thompson
series and $\left[\sqrt{2}\right]$-replicates is given in Table 3 (see [24]
for the labelling of the Hauptmoduls).

##  6 Comments and questions

The results of the previous sections raise some natural questions. For
example, to what extent can these results be extended to other groups arising
as centralizers of elements in classes $p+$ or $p-$, for a prime $p$, in the
monster?

For groups of type $p+$ there is an obvious generalization of
$(2{+})$-replication to $(p+)$-replication. Computationally we have found the
$(p+)$-replicates of all of the rational replicable functions in Norton’s list
of 616 such functions [24]. In all the cases we have checked, this replication
structure respects the power map structure in the group arising as a
centralizer of the element of class $p+$ in the monster. However, even if we
could generalize all the results of the current paper, this would not be
sufficient to give a separate proof akin to Borcherds’ monstrous proof for
each of these groups. In the first place, an explicit construction of each
VOA-module would be required. However, more importantly, for $p>2$ there are
Hauptmodul which have some irrational integer coefficients. There are two such
Hauptmodul for $3\cdot Fi_{24}$. Although it is possible to write down the
identities which are required in these cases, we have not found a natural way
to do so. The crux of the problem is that for the monster and
$2\cdot\mathbb{B}$ a $p$-th replicate contains information about the $p$th
power of an element $g$ even when $p$ divides the order of $g$ and this
information is contained in the essentially combinatorial definition of the
$p$-replicate. No information about $g$ or the underlying action of $g$ on the
relevant VOA-module is needed. But when the corresponding Hauptmodul has
irrational coefficients we have not found a way to extract this information
(other than to look at the underlying VOA-module structure, of course). A
solution to this problem for $3\cdot Fi_{24}$ would be very interesting. The
classes in question are $18m$ and $18p$ and the Hauptmoduls are $54\sim c$ and
$54\sim b$ [24].

For groups of type $p-$ the situation is more complicated and we have found no
natural way to choose an operator in the Hecke algebra that applied to the
Hauptmodul for $p-$ yields the Faber polynomial while at the same time
respecting the power map structure in corresponding group. This seems to be
due to the fact that $p-$, unlike $p+$, has more than one cusp. A solution to
this problem in the simplest case $2-$ would be very interesting.

We should also mention that complete replicability plays a key role in the
work of Martin [21] on modularity of “$j$-final” functions and also in the
work of Kozlov [19] and Cummins and Gannon [10] on modularity from formal
modular equations. The aim of these papers was to see if the identities
implied by complete replicability imply the modularity of Thompson–McKay
series thus avoiding the part of Borcherds’ proof which shows modularity by
recurrence relations and a comparison of initial terms. This approach was used
by Carnahan in his work on generalized moonshine. We believe it would be of
interest to extend these results to complete $(2{+})$-replicability, but we
have not done so here.

### Acknowledgements

The authors would like to thank the referees for their valuable comments and
suggestions to improve the quality of the paper. The second author
acknowledges financial assistance from the Center for Mathematics of the
University of Coimbra—UID/MAT/00324/2013, funded by the Portuguese Government
through FCT/MCTES and co-funded by the European Regional Development Fund
through the Partnership Agreement PT2020.

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