* •

$(c,h)=\left(3,0\right)$:

$\displaystyle\begin{split}\chi_{0}=q^{-\tfrac{1}{8}}(1+&25q+51q^{2}+196q^{3}+297q^{4}+780q^{5}+1223q^{6}\\\
+&2551q^{7}+3798q^{8}+7201q^{9}+10582q^{10}+\mathcal{O}(q^{11})),\end{split}$
(247)

where since $h=0$ both the characters are identical. This is a single-
character solution which surprisingly is not “yet” any $(1,\ell)$ solution and
also does not possess a McKay-Thompson interpretation.

  * •

$(c,h)=\left(6,\frac{1}{4}\right)$:

$\displaystyle\chi_{0}=j_{2^{+}}^{\tfrac{1}{4}},$ (248)
$\displaystyle\chi_{\frac{1}{4}}=\text{unstable}$

  * •

$(c,h)=\left(7,\frac{1}{3}\right)$:

$\displaystyle\chi_{0}=j_{2^{+}}^{\tfrac{1}{4}}\otimes\chi_{0}^{{}^{(2)}\mathcal{W}_{1}}$
$\displaystyle\chi_{\frac{1}{3}}=j_{2^{+}}^{\tfrac{1}{4}}\otimes\chi_{\frac{1}{3}}^{{}^{(2)}\mathcal{W}_{1}}$
(249)

  * •

$(c,h)=\left(9,\frac{1}{2}\right)$:

$\displaystyle\chi_{0}=j_{2^{+}}^{\tfrac{1}{2}}\otimes\chi_{0}^{{}^{(2)}\mathcal{W}_{3}}$
$\displaystyle\chi_{\frac{1}{2}}=j_{2^{+}}^{\tfrac{1}{4}}\otimes\chi_{\frac{1}{2}}^{{}^{(2)}\mathcal{W}_{3}}$
(250)

  * •

$(c,h)=\left(11,\frac{2}{3}\right)$:

$\displaystyle\chi_{0}=j_{2^{+}}^{\tfrac{2}{3}}\otimes\chi_{0}^{{}^{(2)}\mathcal{W}_{5}}$
$\displaystyle\chi_{\frac{2}{3}}=j_{2^{+}}^{\tfrac{1}{4}}\otimes\chi_{\frac{2}{3}}^{{}^{(2)}\mathcal{W}_{5}}$
(251)

  * •

$(c,h)=\left(12,\frac{3}{4}\right)$:

$\displaystyle\begin{split}\chi_{0}=&j_{2^{+}}^{\tfrac{1}{2}}\\\
=&q^{-\tfrac{1}{2}}(1+52q+834q^{2}+4760q^{3}+24703q^{4}+94980q^{5}+343998q^{6}+1077496q^{7}\\\
&{}\ \ \ \ \ \ \ \ +3222915q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\frac{3}{4}}=&\text{unstable}.\end{split}$ (252)

The identity character in 248, along with 213 and 231 forms a set of central
charge $c=6$ admissible solutions that correspond to the $8C$ conjugacy class
of $\mathbb{M}$. Also, the identity character in 252 surprisingly turns out to
be the McKay-Thompson series of the $4B$ conjugacy class of $\mathbb{M}$ (OEIS
sequence A007247 A007247 ). Let ${}^{(2)}\mathcal{W}_{7}$,
${}^{(2)}\mathcal{W}_{9}$ and ${}^{(2)}\mathcal{W}_{11}$ denote the
$c\in\\{7,9,11\\}$ admissible character-like (2,2) solutions. Note that, here
also, we have some bilinear identities,

$\displaystyle
j_{2^{+}}^{\tfrac{3}{4}}=\chi_{0}^{{}^{(2)}\mathcal{W}_{7}}\chi_{0}^{{}^{(2)}\mathcal{W}_{11}}+2\,\chi_{\frac{1}{3}}^{{}^{(2)}\mathcal{W}_{7}}\chi_{\frac{2}{3}}^{{}^{(2)}\mathcal{W}_{11}},$
(253) $\displaystyle
j_{2^{+}}^{\tfrac{1}{4}}=\chi_{0}^{{}^{(2)}\mathcal{W}_{9}}\chi_{0}^{{}^{(2)}\mathcal{W}_{9}}+16\,\chi_{\frac{1}{2}}^{{}^{(2)}\mathcal{W}_{9}}\chi_{\frac{1}{2}}^{{}^{(2)}\mathcal{W}_{9}},$
(254)

where 254 is a self-dual relation. Note, 253 and 254 are nothing but
$j_{2^{+}}^{\tfrac{1}{2}}$ multiplied with 215 and 216 respectively. More
bilinear pairs can be constructed by multiplying $j_{2^{+}}^{\tfrac{1}{4}}$
with 215 and 216.
Note that, we can make use of the fact that $j_{2^{+}}^{x}$ with
$x\in\\{\frac{3}{4}\\}$ can also appear as a solution to the above $(2,2)$
MLDE and hence we get for this value of $x$, $\mu_{3}=0$. With $\mu_{3}=0$, we
get from the $i=1$ equation: $m_{1}=\frac{N(N-31)}{15-N}$ where
$N=-24\alpha_{0}$. This gives the allowed range of central charge as,
$15<c<32$. Scanning this space of $c$ we get the following admissible
solutions.

  * •

$(c,h)=\left(19,\frac{4}{3}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{19}{24}}(1+57q+2147q^{2}+31540q^{3}+260243q^{4}+1691798q^{5}+8887877q^{6}\\\
&{}\ \ \ \ \ \ \ \ \ +41091167q^{7}+168938614q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\frac{4}{3}}=&q^{\tfrac{13}{24}}(133+2717q+33839q^{2}+246468q^{3}+1506510q^{4}+7478629q^{5}+33354310q^{6}\\\
&{}\ \ \ \ \ \ \ \ \ \
+132591975q^{7}+489341675q^{8}+\mathcal{O}(q^{9})).\end{split}$ (255)

  * •

$(c,h)=\left(21,\frac{3}{2}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{7}{8}}(1+35q+2394q^{2}+40873q^{3}+405426q^{4}+2946132q^{5}+17381133q^{6}\\\
&{}\ \ \ \ \ \ \ \ +88016962q^{7}+395953299q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\frac{3}{2}}=&q^{\tfrac{5}{8}}(7+161q+2160q^{2}+17409q^{3}+115003q^{4}+619101q^{5}+2962183q^{6}\\\
&{}\ \ \ \ \ \ +12628973q^{7}+49681296q^{8}+\mathcal{O}(q^{9})).\end{split}$
(256)

  * •

$(c,h)=\left(23,\frac{5}{3}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{23}{24}}(1+23q+3335q^{2}+67068q^{3}+793776q^{4}+6461689q^{5}\\\
&{}\ \ \ \ \ \ \ \ \
+42601060q^{6}+236116965q^{7}+1157910689q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\frac{5}{3}}=&q^{\tfrac{17}{24}}(506+12903q+185725q^{2}+1639463q^{3}+11650696q^{4}+67590238q^{5}\\\
&{}\ \ \ \ \ \ \ \ \ \
+345549517q^{6}+1572505630q^{7}+6570405101q^{8}+\mathcal{O}(q^{9})).\end{split}$
(257)

For $\mu_{3}=-112$, we get from the $i=1$ equation:
$m_{1}=\frac{N(N-10)}{18-N}$ where $N=-24\alpha_{0}$. This gives the allowed
range of central charge as, $9<c<18$. Scanning this space of $c$ we find no
admissible solutions. Also, we note that we did not get any quasi-character
solutions.

####  V.2.5 Non-trivial bilinear pairs

We present some more bilinear pairs that are non-trivial in the sense that
they cannot be derived from 215 and 216. Let ${}^{(2)}\mathcal{W}_{19}$,
${}^{(2)}\mathcal{W}_{21}$ and ${}^{(2)}\mathcal{W}_{23}$ denote the
$c\in\\{19,21,23\\}$ admissible character-like (2,2) solutions (obtained with
$\mu_{3}=0$).

$\displaystyle
j_{2^{+}}-32=\chi_{0}^{{}^{(2)}\mathcal{W}_{5}}\chi_{0}^{{}^{(2)}\mathcal{W}_{19}}+2\,\chi_{\frac{2}{3}}^{{}^{(2)}\mathcal{W}_{5}}\chi_{\frac{4}{3}}^{{}^{(2)}\mathcal{W}_{19}},$
(258) $\displaystyle
j_{2^{+}}-64=\chi_{0}^{{}^{(2)}\mathcal{W}_{3}}\chi_{0}^{{}^{(2)}\mathcal{W}_{21}}+256\,\chi_{\frac{1}{2}}^{{}^{(2)}\mathcal{W}_{3}}\chi_{\frac{3}{2}}^{{}^{(2)}\mathcal{W}_{21}},$
(259) $\displaystyle
j_{2^{+}}-80=\chi_{0}^{{}^{(2)}\mathcal{W}_{1}}\chi_{0}^{{}^{(2)}\mathcal{W}_{23}}+2\,\chi_{\frac{1}{3}}^{{}^{(2)}\mathcal{W}_{1}}\chi_{\frac{5}{3}}^{{}^{(2)}\mathcal{W}_{23}},$
(260)

where the above bilinears are for the following relation:
$(2,0)\overset{c^{\mathcal{H}}=24}{\underset{n_{1}=2}{\longleftrightarrow}}(2,2)$.
Furthermore, we also have,

$\displaystyle j_{2^{+}}^{\tfrac{3}{4}}(j_{2^{+}}-102)=$
$\displaystyle\chi_{0}^{{}^{(2)}\mathcal{W}_{19}}\chi_{0}^{{}^{(2)}\mathcal{W}_{23}}+2\,\chi_{\frac{4}{3}}^{{}^{(2)}\mathcal{W}_{19}}\chi_{\frac{5}{3}}^{{}^{(2)}\mathcal{W}_{23}}$
$\displaystyle=$ $\displaystyle
q^{-\tfrac{7}{4}}\left(1+80q+6793q^{2}+472680q^{3}+18944362q^{4}+\mathcal{O}(q^{5})\right),$
(261) $\displaystyle j_{2^{+}}^{\tfrac{3}{4}}(j_{2^{+}}-112)=$
$\displaystyle\chi_{0}^{{}^{(2)}\mathcal{W}_{21}}\chi_{0}^{{}^{(2)}\mathcal{W}_{21}}+256\,\chi_{\frac{3}{2}}^{{}^{(2)}\mathcal{W}_{21}}\chi_{\frac{3}{2}}^{{}^{(2)}\mathcal{W}_{21}}$
(262) $\displaystyle=$ $\displaystyle
q^{-\tfrac{7}{4}}\left(1+70q+6013q^{2}+450030q^{3}+18635582q^{4}+\mathcal{O}(q^{5})\right),$
(263)

where the above bilinears are for the following relation:
$(2,2)\overset{c^{\mathcal{H}}=42}{\underset{n_{1}=3}{\longleftrightarrow}}(2,2)$.777A
similar analysis should be performed with the $\frac{k}{12}$ derivative
operator (the traditional one used in the case of $\text{SL}(2,\mathbb{Z})$).
In this case, we can tune the $\mu_{1}$ parameter to even make this MLDE
consistent with the Riemann-Roch. This is because in this case there exists a
weight $2$ meromorphic form contribution to the MLDE due to the allowance of
poles in the coefficient functions present in the MLDE. We reserve this for
future work..

####  V.2.6 $\ell=3$

The MLDE, in this case, reads888The generic form of the $(2,3)$ and $(2,1)$
MLDEs are identical with only the values of the parameters being different,

$\displaystyle\left[\mathcal{D}^{2}+\left(\mu_{1}\frac{\left(E_{4}^{(2+)}\right)^{2}}{E_{6}^{2+}}+\mu_{2}\frac{\Delta_{2}}{E_{6}^{2+}}\right)\mathcal{D}+\mu_{3}\frac{E_{10}^{(2+)}}{E_{6}^{(2+)}}\right]\chi=0.$
(264)

From above we obtain the following recursion relation,

$\displaystyle\sum\limits_{k=0}^{i}\left[E_{6,k}^{2+}(\alpha^{j}+i-k)(\alpha^{j}+i-k-1)+E_{6,k}^{2+}(\alpha^{j}+i-k)-\frac{E_{2,k}^{2+}E_{2,k}^{2+}}{4}(\alpha^{j}+i-k)\right.$
$\displaystyle\left.\mu_{1}(E_{4}^{(2+)})^{2}_{,k}(\alpha^{j}+i-k)+\mu_{2}\Delta_{2,k}(\alpha^{j}+i-k)+\mu_{3}E_{10,k}\right]a^{j}_{i-k}=0,$
(265)

which gives the following indicial equation,

$\displaystyle\alpha^{2}+\left(\mu_{1}-\frac{1}{4}\right)\alpha+\mu_{3}=0,$
(266)

From Riemann-Roch (sum of exponents $=-\frac{1}{2}$), we see that
$\mu_{1}=\frac{3}{4}$. Now, let us determine the value of $\mu_{2}$ by
transforming the $\tau$-MLDE in 264 into the corresponding $j_{2^{+}}$-MLDE
(about $\tau=\rho_{2}$),

$\displaystyle\left[(j_{2^{+}})^{2}\partial_{j_{2^{+}}}^{2}+\frac{j_{2^{+}}}{2}\partial_{j_{2^{+}}}-\frac{j_{2^{+}}}{j_{2^{+}}-256}\left((64+\mu_{2})\right)\partial_{j_{2^{+}}}+\frac{\mu_{3}j_{2^{+}}}{j_{2^{+}}-256}\right]\chi=0,$
(267)

where we have set $\mu_{1}=\frac{3}{4}$ from the $\ell=3$ indicial equation
(see 266). Now, note that about $\tau=\rho_{2}$, we have,
$\chi_{0}\sim(j_{2^{+}})^{x}$ and $j_{2^{+}}\to 0$, with $x=\frac{3}{4}$ for
$\ell=3$ (see single-character solutions). So, substituting
$\chi=(j_{2^{+}})^{x}$ about $j_{2^{+}}=0$, the $j_{2^{+}}$-indicial equation
(which is basically, coefficient of $(j_{2^{+}})^{x}$
$=\left.0\right|_{j_{2^{+}}=0}$) reads,

$\displaystyle x^{2}-\frac{x}{2}+\frac{1}{4}+\frac{\mu_{2}}{256}=0,$ (268)

which yields $\mu_{2}=-112$ for $x=\frac{3}{4}$. Similar analysis gives,
$(x,\mu_{2})=(\frac{1}{4},-48)$, $(x,\mu_{2})=(\frac{1}{2},-64)$ as solution
to 268.

For $\mu_{2}=-48$, we get from the $i=1$ equation:
$m_{1}=\frac{N(N+22)}{18-N}$ where $N=-24\alpha_{0}$. This gives the allowed
range of central charge as, $0<c<18$. Scanning this space of $c$ we get the
following admissible solutions:

  * •

$(c,h)=\left(6,0\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{4}}(1+14q+331q^{2}+4506q^{3}+112795q^{4}+3965662q^{5}\\\
&{}\ \ \ \ \ \ \ \
+175520124q^{6}+8718679572q^{7}+470448806341q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\text{non-id}}=&\text{unstable},\end{split}$ (269)

where the above is a single-character solution (not yet a $(1,\ell)$
solution).

For $\mu_{2}=-64$, we get from the $i=1$ equation:
$m_{1}=\frac{N(N+14)}{18-N}$ where $N=-24\alpha_{0}$. This gives the allowed
range of central charge as, $0<c<18$. Scanning this space of $c$ we get the
following admissible solutions:

  * •

$(c,h)=\left(6,0\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{4}}(1+10q+367q^{2}+6422q^{3}+169555q^{4}+6322026q^{5}\\\
&{}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+286760396q^{6}+14591718924q^{7}+802225180229q^{8}+\mathcal{O}(q^{9})),\\\
\chi_{\text{non-id}}=&\text{unstable},\end{split}$ (270)

where the above is a single-character solution (not yet a $(1,\ell)$
solution).

  * •

$(c,h)=\left(12,\frac{1}{2}\right)$:

$\displaystyle\begin{split}\chi_{0}=&j_{2^{+}}^{\tfrac{1}{2}},\\\
\chi_{\text{non-id}}=&1.\end{split}$ (271)

We note that although the identity characters in 269 and 270 possess an
unstable descendant, these do not have a nice McKay-Thompson interpretation.
This is similar to what we observed for the case of the identity-only solution
$(c,h)=(3,0)$ in V.2.4. The identity character with a trivial descendant in
271 along with 252 forms a set of central charge $c=12$ admissible solutions
that correspond to the $4B$ conjugacy class of $\mathbb{M}$.

###  V.3 Three-character solutions

We restrict our analysis to the $(n,\ell)=(3,0)$ MLDEs and reserve the other
cases for future work. This modular invariant ODE in this case reads

$\displaystyle\left[\mathcal{D}^{3}+\omega_{2}(\tau)\mathcal{D}^{2}+\omega_{4}(\tau)\mathcal{D}+\omega_{6}(\tau)\right]\chi(\tau)=0.$
(272)

With choices $\omega_{4}(\tau)=\mu_{1}E_{4}^{(2^{+})}(\tau)$ and
$\omega_{6}(\tau)=\mu_{2}E_{6}^{(2^{+})}(\tau)$ and in terms of derivatives
$\tilde{\partial}$, the MLDE reads

$\displaystyle\left[\tilde{\partial}^{3}-\frac{3}{4}\left(\tilde{\partial}E_{2}^{(2^{+})}(\tau)\right)\tilde{\partial}^{2}+\frac{1}{8}\left(E_{2}^{(2^{+})}\right)^{2}-\frac{1}{4}\left(\tilde{\partial}E_{2}^{(2^{+})}\right)\tilde{\partial}+\mu_{1}E_{4}^{(2^{+})}(\tau)\tilde{\partial}+\mu_{2}E_{6}^{(2^{+})}(\tau)\right]\chi(\tau)=0.$
(273)

We now make use of the Ramanujan identity in IV.2 for $E_{2}^{(2^{+})}(\tau)$
to rewrite the MLDE as follows

$\displaystyle\left[\tilde{\partial}^{3}-\frac{3}{4}\left(\tilde{\partial}E_{2}^{(2^{+})}(\tau)\right)\tilde{\partial}^{2}+\left(\frac{3}{4}\left(\tilde{\partial}E_{2}^{(2^{+})}\right)+\frac{1}{8}E_{4}^{(2^{+})}+\mu_{1}E_{4}^{(2^{+})}(\tau)\right)\tilde{\partial}+\mu_{2}E_{6}^{(2^{+})}(\tau)\right]\chi(\tau)=0.$
(274)

The recursion relation reads,

$\displaystyle\left[(\alpha^{j}+i)(\alpha^{j}+i-1)(\alpha^{j}+i-2)+3(\alpha^{j}+i)(\alpha^{j}+i-1)+(\alpha^{j}+i)\right]a^{j}_{i}$
$\displaystyle+\sum\limits_{k=0}^{i}\left[-\frac{3}{4}E_{2,k}^{(2+)}a^{j}_{i-k}(\alpha^{j}+i-k)(\alpha^{j}+i-k-1)-\frac{3}{4}E_{2,k}^{(2+)}a^{j}_{i-k}(\alpha^{j}+i-k)\right.$
$\displaystyle\left.+\frac{1}{32}E_{4,k}^{2+}a^{j}_{i-k}(\alpha^{j}+i-k)+\frac{3}{32}\left(E_{2}^{(2+)}\right)^{2}_{,k}a^{j}_{i-k}(\alpha^{j}+i-k)+\mu_{1}E_{4,k}^{(2+)}a^{j}_{i-k}(\alpha^{j}+i-k)\right.$
$\displaystyle\left.+\mu_{2}E_{6,k}^{(2+)}a^{j}_{i-k}\right]=0.$ (275)

From which the indicial equation reads,

$\displaystyle\alpha^{3}-\frac{3}{4}\alpha^{2}+\left(\mu_{1}+\frac{1}{8}\right)\alpha+\mu_{2}=0,$
(276)

where we have defined $\tilde{\mu}_{1}\equiv\mu_{1}+\tfrac{1}{8}$. Using the
equation we obtain at $k=1$, we find the following relations among the roots
of the indicial equation, $\alpha_{i},i=0,1,2$, and the free parameters

$\displaystyle\begin{split}\tilde{\mu}_{1}=&\alpha_{0}\alpha_{1}+\alpha_{1}\alpha_{2}+\alpha_{0}\alpha_{2},\\\
\mu_{2}=&-\alpha_{0}\alpha_{1}\alpha_{2}.\end{split}$ (277)

We can now fix $\mu_{2}$ from the indicial equation and express $\mu_{1}$ in
terms of the roots $\alpha_{i}$ and the ratio $m_{1}$ obtain ed via the
equation at $k=1$. At $k=2$, we obtain the following equation

$\displaystyle-24m_{1}^{2}+9m_{1}m_{2}-1004m_{1}\alpha_{0}-184m_{1}^{2}\alpha_{0}+1064m_{2}\alpha_{0}+12m_{1}m_{2}\alpha_{0}+22560\alpha_{0}^{2}-6192m_{1}\alpha_{0}^{2}$
$\displaystyle-672m_{1}^{2}\alpha_{0}^{2}+2016m_{2}\alpha_{0}^{2}-141696\alpha_{0}^{3}-18304m_{1}\alpha_{0}^{3}-512m_{1}^{2}\alpha_{0}^{3}+1024m_{2}\alpha_{0}^{3}+150528\alpha_{0}^{4}=0.$
(278)

We observe that if we define the variable

$\displaystyle N=-1176\alpha_{0},$ (279)

then the polynomial equation simplifies to (after multiplying $7^{6}\times
3^{3}\times 2^{2}$ with Eq.(278) and using the above identification)

$\displaystyle-304946208m_{1}^{2}+114354828m_{1}m_{2}+10847718m_{1}N+1988028m_{1}^{2}N-11495988m_{2}N$
$\displaystyle-129654m_{1}m_{2}N+207270N^{2}-56889m_{1}N^{2}-6174m_{1}^{2}N^{2}+18522m_{2}N^{2}+1107N^{3}$
$\displaystyle+143m_{1}N^{3}+4m_{1}^{2}N^{3}-8m_{2}N^{3}+N^{4}=0.$ (280)

The integer root theorem, $N=-1176\alpha_{0}$ above should be an integer and
hence, $N=49c\in\mathbb{Z}$. this tells us that the central charge has to be a
rational number whose denominator is always $49$. Since $\alpha_{1}$ and
$\alpha_{2}$ are to be rational, this implies that the discriminant (which is
an integer, say $k^{2}$ with $k\in\mathbb{Z}$) has to be the perfect square to
be able to result in rational roots Thus, we have,

$\displaystyle
46694888100m_{1}^{2}-5006200248m_{1}N-190591380m_{1}^{2}N+77533092N^{2}$
$\displaystyle+17114328m_{1}N^{2}+194481m_{1}^{2}N^{2}-22932N^{3}-15582m_{1}N^{3}-143N^{4}=k^{2}.$
(281)

From this, we find the three roots of the indicial equation to read

$\displaystyle\alpha_{0}=-\frac{N}{1176},$ (282)
$\displaystyle\alpha_{1}=\frac{129654m_{1}-11466N+147m_{1}N-13N^{2}-k}{2352(147m_{1}-13N)},$
(283)
$\displaystyle\alpha_{2}=\frac{129654m_{1}-11466N+147m_{1}N-13N^{2}+k}{2352(147m_{1}-13N)}.$
(284)

This equation where all the unknown variables $N$, $m_{1}$, and $k$ are
positive integers is known as a Diophantine equation.

Now, solving 280 and 281, for the range $1\leq N\leq 2058$, yields,

  * •

$(c,h_{1},h_{2})=\left(\frac{6}{7},\frac{5}{7},\frac{1}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{28}}(1+q^{2}+q^{3}+2q^{4}+q^{5}+3q^{6}+2q^{7}+4q^{8}+3q^{9}+6q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{7}}$
$\displaystyle=q^{\tfrac{19}{28}}(1+q^{2}+q^{4}+q^{5}+2q^{6}+q^{7}+3q^{8}+2q^{9}+4q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{7}}$
$\displaystyle=q^{\tfrac{3}{28}}(1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+3q^{7}+5q^{8}+5q^{9}+7q^{10}+\mathcal{O}(q^{11})).$
(285)

where the identity character above is similar to that of a Virasoro minimal
model CFT by which we mean that $m_{1}=0$ for such models since they do not
possess any Kac-Moody currents. From here on we shall use the notation minimal
model to denote any admissible solution whose identity character has
$m_{1}=0$.

  * •

$(c,h_{1},h_{2})=\left(1,\frac{13}{24},\frac{1}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{24}}(1+q+2q^{2}+q^{3}+3q^{4}+3q^{5}+5q^{6}+5q^{7}+8q^{8}+8q^{9}+12q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{13}{24}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{3}}$
$\displaystyle=q^{\tfrac{7}{24}}(1+q^{2}+q^{3}+2q^{4}+q^{5}+3q^{6}+2q^{7}+5q^{8}+4q^{9}+7q^{10}+\mathcal{O}(q^{11})),$
(286)

where above is a two-character solution paired with an unstable character.
This is the $(2,0)$ two-character solution found in 210 that makes a
reappearance here.

  * •

$(c,h_{1},h_{2})=\left(2,\frac{2}{3},\frac{1}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{12}}(1+2q+5q^{2}+6q^{3}+12q^{4}+16q^{5}+29q^{6}+38q^{7}+61q^{8}+80q^{9}$
(287) $\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+121q^{10}+\mathcal{O}(q^{11})),$ $\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{\tfrac{7}{12}}(1+2q^{2}+2q^{3}+5q^{4}+4q^{5}+11q^{6}+10q^{7}+22q^{8}+22q^{9}+41q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{3}}$
$\displaystyle=q^{\tfrac{1}{4}}(1+q+3q^{2}+3q^{3}+8q^{4}+9q^{5}+17q^{6}+20q^{7}+36q^{8}+43q^{9}+70q^{10}+\mathcal{O}(q^{11})).$
(288)

We note here that the identity character can be expressed in terms of modular
forms of $\Gamma_{0}(7)$, akin to what we saw in the two-character case at low
central charge $c=1$ in 214, as follows

$\displaystyle\chi_{0}^{(c=2)}=\left(\frac{\left(j_{2}\Delta_{2}^{\infty}\right)(3\tau)}{\left(j_{2}\Delta_{2}^{\infty}\right)^{\tfrac{1}{3}}(\tau)}\right)^{\tfrac{1}{4}}\left(\frac{1}{\Delta_{2}(\tau)}\right)^{\tfrac{1}{12}}.$
(289)

  * •

$(c,h_{1},h_{2})=\left(3,\frac{5}{8},\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{8}}(1+5q+11q^{2}+20q^{3}+41q^{4}+76q^{5}+127q^{6}+211q^{7}+342q^{8}+541q^{9}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ +838q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{8}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{3}{8}}(1+q+5q^{2}+6q^{3}+16q^{4}+21q^{5}+46q^{6}+61q^{7}+117q^{8}+157q^{9}$
(290) $\displaystyle\ \ \ \ \ \ \ \ \ \ +273q^{10}+\mathcal{O}(q^{11})).$

This is the $(2,0)$ two-character solution we found in 211 that makes a
reappearance here.

  * •

$(c,h_{1},h_{2})=\left(3,0,\frac{9}{8}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{8}}(1+25q+51q^{2}+196q^{3}+297q^{4}+780q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+1223q^{6}+2551q^{7}+3798q^{8}+7201q^{9}+10582q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{8}}(1+25q+51q^{2}+196q^{3}+297q^{4}+780q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+1223q^{6}+2551q^{7}+3798q^{8}+7201q^{9}+10582q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{9}{8}}$ $\displaystyle=\text{unstable}.$ (291)

This is the single character identified in 247 that makes a reappearance here.

  * •

$(c,h_{1},h_{2})=\left(\frac{30}{7},\frac{5}{7},\frac{4}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{28}}(1+10q+25q^{2}+70q^{3}+145q^{4}+330q^{5}+610q^{6}+1200q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+2095q^{8}+3800q^{9}+6336q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{7}}$
$\displaystyle=q^{15/28}(4+5q+30q^{2}+45q^{3}+130q^{4}+204q^{5}+480q^{6}+745q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \ \
+1510q^{8}+2330q^{9}+4294q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{4}{7}}$
$\displaystyle=q^{11/28}(5+10q+45q^{2}+74q^{3}+205q^{4}+350q^{5}+770q^{6}+1260q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \ \
+2470q^{8}+3950q^{9}+7115q^{10}+\mathcal{O}(q^{11})).$ (292)

  * •

$(c,h_{1},h_{2})=\left(5,\frac{17}{24},\frac{2}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{24}}(1+15q+40q^{2}+135q^{3}+290q^{4}+726q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+1415q^{6}+3000q^{7}+5485q^{8}+10565q^{9}+18407q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{17}{24}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{\tfrac{11}{24}}(5+11q+55q^{2}+100q^{3}+290q^{4}+535q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+1235q^{6}+2160q^{7}+4400q^{8}+7465q^{9}+13945q^{10}+\mathcal{O}(q^{11})).$
(293)

This is the $(2,0)$ two-character solution we found in 212 that makes a
reappearance here.

  * •

$(c,h_{1},h_{2})=\left(6,1,\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{4}}(1+m_{1}q+m_{2}q^{2}+\mathcal{O}(q^{3})),$
$\displaystyle\chi_{1}$
$\displaystyle=q^{\tfrac{3}{4}}(1+2q+11q^{2}+22q^{3}+69q^{4}+134q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+330q^{6}+616q^{7}+1324q^{8}+2382q^{9}+4675q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{1}{4}}(1+6q+21q^{2}+62q^{3}+162q^{4}+384q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+855q^{6}+1806q^{7}+3648q^{8}+7110q^{9}+13434q^{10}+\mathcal{O}(q^{11}))$
(294)

where for the above identity character we have,
$m_{1}\in\mathbb{N}\cup\\{0\\}$, $m_{2}=27+2m_{1}$ and $k=2247336-86436m_{1}$
for $m_{1}<26$ and $k=-2247336+86436m_{1}$ for $m_{1}\geq 27$.999This kind of
infinite one-parameter solutions (here the parameter being $m_{1}$) were first
observed in Das:2021uvd  where it was explained using an argument using
Inhomogenous MLDE. We have checked that the same argument applies here too
with the following replacement
$\eta(\tau)^{2k}\leftrightarrow(\eta(\tau)\eta(2\tau))^{k}$. For $m_{1}=26$,
the id. character is just $j_{2^{+}}^{\frac{1}{4}}$

  * •

$(c,h_{1},h_{2})=\left(7,\frac{31}{24},\frac{1}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{7}{24}}(1+27q+107q^{2}+458q^{3}+1268q^{4}+3673q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+8722q^{6}+21061q^{7}+45251q^{8}+97657q^{9}+195561q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{31}{24}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{3}}$
$\displaystyle=q^{\tfrac{1}{24}}(1+26q+80q^{2}+353q^{3}+862q^{4}+2564q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+5728q^{6}+13956q^{7}+28861q^{8}+62621q^{9}+122250q^{10}+\mathcal{O}(q^{11}))$
(295)

This is the $(2,2)$ two-character solution we found in • ‣ V.2.4 that makes a
reappearance here.

  * •

$(c,h_{1},h_{2})=\left(\frac{54}{7},\frac{9}{7},\frac{3}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{9}{28}}(1+27q+135q^{2}+591q^{3}+1836q^{4}+5481q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+14037q^{6}+35046q^{7}+79866q^{8}+178083q^{9}+374328q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{9}{7}}$
$\displaystyle=q^{\tfrac{27}{28}}(26+81q+405q^{2}+1047q^{3}+3375q^{4}+7884q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+20352q^{6}+43983q^{7}+99927q^{8}+202994q^{9}+422901q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{3}{7}}$
$\displaystyle=q^{\tfrac{3}{28}}(3+54q+189q^{2}+840q^{3}+2268q^{4}+6858q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+16397q^{6}+40986q^{7}+89505q^{8}+199410q^{9}+407457q^{10}+\mathcal{O}(q^{11})).$
(296)

  * •

$(c,h_{1},h_{2})=\left(\frac{54}{7},\frac{10}{7},\frac{2}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{9}{28}}(1+30q+135q^{2}+594q^{3}+1836q^{4}+5484q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+14040q^{6}+35052q^{7}+79869q^{8}+178092q^{9}+374334q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{10}{7}}$
$\displaystyle=q^{\tfrac{31}{28}}(51+186q+837q^{2}+2262q^{3}+6852q^{4}+16388q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+40977q^{6}+89490q^{7}+199395q^{8}+407436q^{9}+838032q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{2}{7}}$
$\displaystyle=q^{-\tfrac{1}{28}}(3+26q+84q^{2}+408q^{3}+1053q^{4}+3378q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+7893q^{6}+20358q^{7}+43995q^{8}+99936q^{9}+203012q^{10}+\mathcal{O}(q^{11})).$
(297)

  * •

$(c,h_{1},h_{2})=\left(9,\frac{11}{8},\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{8}}(1+31q+220q^{2}+1027q^{3}+3815q^{4}+12189q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+35157q^{6}+93733q^{7}+234357q^{8}+556019q^{9}+1262556q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{11}{8}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{1}{8}}(1+27q+110q^{2}+541q^{3}+1648q^{4}+5402q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+14153q^{6}+37936q^{7}+89648q^{8}+212550q^{9}+465321q^{10}+\mathcal{O}(q^{11})).$
(298)

  * •

$(c,h_{1},h_{2})=\left(10,\frac{4}{3},\frac{2}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{12}}(1+30q+305q^{2}+1470q^{3}+6230q^{4}+20952q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+66035q^{6}+184830q^{7}+494290q^{8}+1228810q^{9}+2950340q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{4}{3}}$
$\displaystyle=q^{\tfrac{11}{12}}(25+110q+671q^{2}+2210q^{3}+8125q^{4}+22730q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+66020q^{6}+165620q^{7}+418470q^{8}+966350q^{9}+2222205q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{\tfrac{1}{4}}(5+86q+420q^{2}+2040q^{3}+6925q^{4}+23130q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+65371q^{6}+180730q^{7}+453375q^{8}+1111940q^{9}+2560355q^{10}+\mathcal{O}(q^{11})).$
(299)

  * •

$(c,h_{1},h_{2})=\left(10,\frac{5}{3},\frac{1}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{12}}(1+40q+305q^{2}+1490q^{3}+6250q^{4}+21002q^{5}+66075q^{6}+184940q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+494390q^{8}+1229030q^{9}+2950560q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{3}}$
$\displaystyle=q^{\tfrac{5}{4}}(1+5q+25q^{2}+85q^{3}+285q^{4}+806q^{5}+2230q^{6}+5595q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+13725q^{8}+31605q^{9}+71258q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{3}}$
$\displaystyle=q^{-\tfrac{1}{12}}(5+60q+245q^{2}+1372q^{3}+4480q^{4}+16330q^{5}+45605q^{6}+132230q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+331545q^{8}+837340q^{9}+1933305q^{10}+\mathcal{O}(q^{11})).$ (300)

  * •

$(c,h_{1},h_{2})=\left(11,\frac{35}{24},\frac{2}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{11}{24}}(1+41q+509q^{2}+2686q^{3}+12605q^{4}+45402q^{5}+153461q^{6}+455033q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+1288031q^{8}+3367910q^{9}+8491985q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{35}{24}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{\tfrac{5}{24}}(5+141q+736q^{2}+4029q^{3}+14596q^{4}+52740q^{5}+157646q^{6}+462885q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+1221334q^{8}+3151415q^{9}+7594605q^{10}+\mathcal{O}(q^{11}))$ (301)

  * •

$(c,h_{1},h_{2})=\left(\frac{78}{7},\frac{9}{7},\frac{6}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{39}{84}}(1+26q+403q^{2}+2002q^{3}+9906q^{4}+35204q^{5}+122161q^{6}+361062q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+1038219q^{8}+2718820q^{9}+6930937q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{9}{7}}$
$\displaystyle=q^{\tfrac{23}{28}}(52+299q+2002q^{2}+7501q^{3}+29406q^{4}+90506q^{5}+278408q^{6}+751907q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+2002066q^{8}+4919317q^{9}+11872484q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{6}{7}}$
$\displaystyle=q^{\tfrac{11}{28}}(26+352q+2054q^{2}+9880q^{3}+37180q^{4}+127764q^{5}+388765q^{6}+1115322q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+2969551q^{8}+7572838q^{9}+18351476q^{10}+\mathcal{O}(q^{11})).$ (302)

  * •

$(c,h_{1},h_{2})=\left(12,\frac{5}{4},1\right)$: 2-char

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{2}}(1+m_{1}q+m_{2}q^{2}+\mathcal{O}(q^{3})),$
$\displaystyle\chi_{\frac{5}{4}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{1}$
$\displaystyle=q^{\tfrac{1}{4}}(1+12q+78q^{2}+376q^{3}+1509q^{4}+5316q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+16966q^{6}+50088q^{7}+138738q^{8}+364284q^{9}+913824q^{10}+\mathcal{O}(q^{11}))$
(303)

where for the above identity character we have,
$m_{1}\in\mathbb{N}\cup\\{0\\}$, $m_{2}=210+12m_{1}$ and
$k=2247336-43218m_{1}$ for $m_{1}<52$ and $k=-2247336+43218m_{1}$ for
$m_{1}\geq 53$. For $m_{1}=52$, the id. character is just
$j_{2^{+}}^{\frac{1}{2}}$.

  * •

$(c,h_{1},h_{2})=\left(14,\frac{1}{3},\frac{13}{6}\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{3}}$
$\displaystyle=q^{-\tfrac{1}{4}}(1+26q+79q^{2}+326q^{3}+755q^{4}+2106q^{5}+4460q^{6}+10284q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+20165q^{8}+41640q^{9}+77352q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{13}{6}}$ $\displaystyle=\text{unstable}.$ (304)

This is the $c=6$ single-character theory we found in 213 that makes a
reappearance as a descendant of a higher central charge three-character theory
with unstable identity and second descendant. We can actually re-interpret the
central charge (using unitary presentation kind of arguments of Harvey:2018rdc
) as $c_{\text{eff}}=c-24h_{1}=14-24\times\frac{1}{3}=6$. Hence, when the
$\frac{1}{3}$ descendant is re-interpreted as a $c=6$ theory with central
charge equal to $c_{\text{eff}}=6$, it is indeed $j_{2^{+}}^{\frac{1}{4}}$.

  * •

$(c,h_{1},h_{2})=\left(18,\frac{1}{2},\frac{5}{2}\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{\frac{1}{2}}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{\frac{5}{2}}$
$\displaystyle=q^{\tfrac{7}{4}}(1+10q+71q^{2}+390q^{3}+1831q^{4}+7602q^{5}+28712q^{6}+100292q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+328247q^{8}+1016012q^{9}+2996354q^{10}+\mathcal{O}(q^{11})).$ (305)

Above, the term Log unstable refers to some pathologies in the q-series
coefficients of these characters when we try to obtain their coefficients from
the recursion relation. At times, these happen when the corresponding
hypergeometric expressions for these characters, of ${}_{3}F_{2}$ types, when
expanded about $q=0$ result in a $\log[q]$ behaviour. However, in certain
cases, there is no such instability and it is just a shortcoming of the
recursion relation approach to obtain q-series coefficients of characters.
This happens when one tries to obtain the q-series coefficients of $D_{4,1}$
WZW model in the $\text{SL}(2,\mathbb{Z})$ case. To remedy this, one uses the
theta series representations for the characters of $D_{4,1}$ to find the
q-series of its characters. However, for the congruence subgroups, we are not
aware whether such theta series representations exist for the characters and
hence we have not been able to conclude whether these are really Log
singularities in the q-series or just an artefact of the recursion relation.
In any case, we call them Log unstable here and shall return in the future to
understand this feature better in the realm of congruence subgroups.
101010Note that, we have only reported the Log unstable solutions for the
three-character case and we postpone the analysis of such instabilities in the
two-character case to future work. Indeed, these log instabilities do appear
even in the two-character case whenever one of the conformal dimensions is a
non-zero positive integer. For the $\text{SL}(2,\mathbb{Z})$ case, whenever a
descendant had a conformal dimension $\in\mathbb{N}$, the theory is called a
logarithmic CFT. Here, we do not know how to interpret such things just as
yet, and thus, we reserve this for future work too.

  * •

$(c,h_{1},h_{2})=\left(22,\frac{2}{3},\frac{17}{6}\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{-\tfrac{1}{4}}(1+26q+79q^{2}+326q^{3}+755q^{4}+2106q^{5}+4460q^{6}+10284q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+20165q^{8}+41640q^{9}+77352q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{17}{6}}$ $\displaystyle=\text{unstable}.$ (306)

The first descendant is precisely equal to the first descendant in • ‣ V.3,
which is a reappearing single-character theory.

  * •

$(c,h_{1},h_{2})=\left(24,\frac{291}{292},\frac{201}{73}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-1}(1+16091q+4372q^{2}+96256q^{3}+1240002q^{4}+10698752q^{5}+74428120q^{6}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+431529984q^{7}+2206741887q^{8}+10117578752q^{9}+42616961892q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{291}{292}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{201}{73}}$ $\displaystyle=\text{unstable}.$ (307)

The identity is an $\ell=4$, $c=24$ single-character theory $j_{2^{+}}+15987$.

  * •

$(c,h_{1},h_{2})=\left(26,\frac{2}{3},\frac{10}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{13}{12}}(1+286q+6799q^{2}+102466q^{3}+1374100q^{4}+12916124q^{5}+98031453q^{6}$
$\displaystyle{}\ \ \ \ \ \ \ \ \
+619853026q^{7}+3426275008q^{8}+16919200870q^{9}+76289142098q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{2}{3}}$
$\displaystyle=q^{-\tfrac{5}{12}}(65+2782q+26387q^{2}+366938q^{3}+3482752q^{4}+26387140q^{5}+174154539q^{6}$
$\displaystyle{}\ \ \ \ \ \ \ \ \
+989454102q^{7}+5051982756q^{8}+23348937742q^{9}+99790466750q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{10}{3}}$
$\displaystyle=q^{\tfrac{9}{4}}(13+198q+1872q^{2}+13728q^{3}+83655q^{4}+447174q^{5}+2143011q^{6}+9418266q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+38395656q^{8}+146938090q^{9}+531665433q^{10}+\mathcal{O}(q^{11})).$ (308)

  * •

$(c,h_{1},h_{2})=\left(27,\frac{1}{8},4\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{\frac{1}{8}}$
$\displaystyle=q^{-1}(69+9224q+301668q^{2}+6641664q^{3}+85560138q^{4}+738213888q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+5135540280q^{6}+29775568896q^{7}+152265190203q^{8}$ (309) $\displaystyle{}\ \
\ \ \ \ \ \ \ \ \ \ \
+698112933888q^{9}+2940570370548q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{4}$ $\displaystyle=\text{unstable}.$ (310)

  * •

$(c,h_{1},h_{2})=\left(30,\frac{1}{2},4\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{-\frac{3}{4}}(25+2110q+51953q^{2}+1015950q^{3}+15480440q^{4}+163643540q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+1372184025q^{6}+9575867298q^{7}+58118855135q^{8}$ (311) $\displaystyle{}\ \ \
\ \ \ \ \ \ \ \ \ \
+314673021160q^{9}+1550715669375q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{4}$
$\displaystyle=q^{\frac{11}{4}}(15+286q+3085q^{2}+25370q^{3}+170775q^{4}+1000690q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+5228922q^{6}+24959680q^{7}+110204890q^{8}$ $\displaystyle{}\ \ \ \ \ \ \ \ \
\ \ \ +455662830q^{9}+1777830395q^{10}+\mathcal{O}(q^{11})).$ (312)

  * •

$(c,h_{1},h_{2})=\left(30,\frac{5}{6},\frac{11}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{4}}(1+670q+21195q^{2}+261130q^{3}+4298745q^{4}+52456626q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+486596870q^{6}+3700866400q^{7}+23959194930q^{8}$ (313) $\displaystyle{}\ \ \
\ \ \ \ \ \ \ \ \ +137023596950q^{9}+706045556129q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{6}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{11}{3}}$ $\displaystyle=\text{unstable}.$ (314)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(30,\frac{37}{38},\frac{67}{19}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{4}}(1+3740q+101015q^{2}+503660q^{3}+5299565q^{4}+54774476q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+493062290q^{6}+3714558600q^{7}+23990766810q^{8}$ (315) $\displaystyle{}\ \ \
\ \ \ \ \ \ \ \ \ +137085503500q^{9}+706173390929q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{37}{38}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{67}{19}}$ $\displaystyle=\text{unstable}.$ (316)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(33,\frac{7}{8},4\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{unstable},$ (317)
$\displaystyle\chi_{\frac{7}{8}}$
$\displaystyle=q^{-\tfrac{1}{2}}(1+52q+834q^{2}+4760q^{3}+24703q^{4}+94980q^{5}+343998q^{6}+1077496q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+3222915q^{8}+8844712q^{9}+23381058q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{4}$ $\displaystyle=\text{unstable}.$ (318)

The single-character theory found in 187 makes a reappearance as a descendant
of a higher central charge three-character theory with unstable identity and
second descendant.

  * •

$(c,h_{1},h_{2})=\left(36,\frac{1}{4},5\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$ (319)
$\displaystyle\chi_{\frac{1}{4}}$
$\displaystyle=q^{-\tfrac{5}{4}}(5+906q+42431q^{2}+1112574q^{3}+20696981q^{4}+260437910q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+2427837286q^{6}+18493431760q^{7}+119770840554q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +685068701490q^{9}+3530126012485q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{5}$ $\displaystyle=\text{unstable}.$ (320)

  * •

$(c,h_{1},h_{2})=\left(36,\frac{1}{2},\frac{19}{4}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{2}}(1+412q+23926q^{2}+628600q^{3}+11629865q^{4}+185255140q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+2265641766q^{6}+22044942216q^{7}+179314657672q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1260100453460q^{9}+7865573552644q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{-1}(13+1608\,q+56836q^{2}+1251328q^{3}+16120026q^{4}+139083776q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+967565560q^{6}+5609889792q^{7}+28687644531q^{8}$ $\displaystyle{}\ \ \ \ \ \
\ \ \ \ \ \ \ +131528523776q^{9}+554020504596q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{19}{4}}$ $\displaystyle=\text{unstable}.$ (321)

The descendant character, $\chi_{\frac{1}{2}}$, is $13j_{2^{+}}+256$ which can
be thought of as twelve copies of $j_{2^{+}}$ single characters and one copy
of the single character $j_{2^{+}}+256$.

  * •

$(c,h_{1},h_{2})=\left(36,\frac{23}{28},\frac{31}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{2}}(1+940q+51382q^{2}+1068952q^{3}+14143145q^{4}+198298324q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+2315791206q^{6}+22226573160q^{7}+179883575560q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1261802152580q^{9}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{23}{28}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{31}{7}}$ $\displaystyle=\text{unstable}.$ (322)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(36,\frac{11}{12},\frac{13}{3}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-3/2}(1+1884q+100470q^{2}+1856248q^{3}+18636585q^{4}+221617956q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+2405452326q^{6}+22551307272q^{7}+180900731784q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ \ +1264844584340q^{9}+7878592968708q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{11}{12}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{13}{3}}$ $\displaystyle=\text{unstable}.$ (323)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(36,\frac{151}{156},\frac{167}{39}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{2}}(1+4719q+247890q^{2}+4220638q^{3}+32131185q^{4}+291650961q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+2674720626q^{6}+23526541602q^{7}+183955432944q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1273981548365q^{9}+7903667727228q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{151}{156}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{167}{39}}$ $\displaystyle=\text{unstable}.$ (324)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(36,\frac{193}{196},\frac{209}{49}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{2}}(1+9760q+510022q^{2}+8424832q^{3}+56126345q^{4}+416178784q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+3153514806q^{6}+25260635520q^{7}+189387090280q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1290228262880q^{9}+7948253920420q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{193}{196}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{209}{49}}$ $\displaystyle=\text{unstable}$ (325)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(36,\frac{235}{236},\frac{251}{59}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{3}{2}}(1+34966q+1820734q^{2}+29446636q^{3}+176106905q^{4}+1038842602q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+5547580686q^{6}+33931449108q^{7}+216546454456q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1371465058370q^{9}+8171193731092q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{235}{236}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{251}{59}}$ $\displaystyle=\text{unstable}.$ (326)

The above is a single-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(39,\frac{5}{8},5\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{\frac{5}{8}}$
$\displaystyle=q^{-1}(117+14216q+511524q^{2}+11261952q^{3}+145080234q^{4}+1251753984q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+8708090040q^{6}+50489008128q^{7}+258188800779q^{8}$ $\displaystyle{}\ \ \ \ \
\ \ \ \ \ \ \ +1183756713984q^{9}+4986184541364q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{5}$ $\displaystyle=\text{unstable}.$ (327)

  * •

$(c,h_{1},h_{2})=\left(42,0,6\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{6}$
$\displaystyle=q^{\frac{17}{4}}(98+3196q+53669q^{2}+642838q^{3}+6133855q^{4}+49500906q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+350187964q^{6}+2224588380q^{7}+12909977905q^{8}$ $\displaystyle{}\ \ \ \ \ \
\ \ \ \ \ \ +69325524646q^{9}+347916765224q^{10}+\mathcal{O}(q^{11})).$ (328)

##  VI $\mathbf{\Gamma_{0}^{+}(3)}$

###  VI.1 Single character solutions

From Umasankar:2022kzs , we find the following form for the character in a
$n=1$ theory

$\displaystyle\begin{split}&\chi(\tau)=j_{3^{+}}^{w_{\rho}}(\tau),\\\
&c=24w_{\rho}=8\ell,\end{split}$ (329)

where $w_{\rho}\in\left\\{0,\tfrac{4}{3},2\right\\}$ corresponding to the
three characters at $\ell=0$, $\ell=4$, and $\ell=6$ respectively. Here, we
will present the admissible solutions at $\ell$ values all the way to the
bulk-pole.

####  VI.1.1 $\ell=1$

For $\ell=1$, the RHS of Valence Formula reads $\frac{1}{3}$ (since
$k=2\ell$). The MLDE takes the following form,

$\displaystyle\left[\mathcal{D}+\mu_{1}\frac{E_{4}^{(3+)}E_{6}^{(3+)}}{\left(E_{4}^{(3+)}\right)^{2}+\mu\Delta_{3^{+},8}}+\mu_{2}\frac{\Delta_{3^{+},10}}{(E_{4}^{(3+)})^{2}+\mu\Delta_{3^{+},8}}\right]\chi(\tau)=0.$
(330)

Now using the indicial equation we can set $\mu_{1}=\frac{1}{3}$. Now let us
take a guess, $\chi=j_{3^{+}}^{\tfrac{1}{3}}$. One can readily check that this
is an admissible central charge $c=8$ solution with $\mu_{2}=\frac{1}{3}$. The
$q$-series expansion of this character reads

$\displaystyle\begin{split}\chi=q^{-\tfrac{1}{3}}(1+&14q+65q^{2}+156q^{3}+456q^{4}+1066q^{5}+2250q^{6}+4720q^{7}\\\
+&9426q^{8}+17590q^{9}+\mathcal{O}(q^{10})),\end{split}$ (331)

which turns out to be the McKay-Thompson series of the $9a$ conjugacy class of
$\mathbb{M}$ (OEIS sequence A058092 A058092 ).

####  VI.1.2 $\ell=2$

The MLDE takes the following form,

$\displaystyle\left[\mathcal{D}+\mu_{1}\frac{E_{6}^{(3+)}}{E_{4}^{(3+)}}\right]\chi(\tau)=0,$
(332)
$\displaystyle\left[E_{4}^{3+}\mathcal{D}+\mu_{1}E_{6}^{3+}\right]\chi(\tau)=0.$
(333)

The indicial equation ($\alpha_{0}+\mu_{1}=0$) dictates $\mu_{1}=\frac{2}{3}$.
Consider the ansatz, $\chi=j_{3^{+}}^{\tfrac{2}{3}}$. Then,
$\tilde{\partial}_{\tau}j_{3^{+}}^{\tfrac{2}{3}}=-\frac{2}{3}j_{3^{+}}^{\tfrac{2}{3}}\frac{E_{6}^{(3+)}}{E_{4}^{(3+)}}$.
Note, that this $\chi$ indeed satisfies 332 with $\mu_{1}=\frac{2}{3}$.

####  VI.1.3 $\ell=3$

From Riemann-Roch, we notice that this is the bulk-pole MLDE which takes the
following form

$\displaystyle\left[\mathcal{D}+\mu_{1}\frac{\left(E_{4}^{(3+)}\right)^{2}}{E_{6}^{(3+)}}+\mu_{2}\frac{\Delta_{3^{+},8}}{E_{6}^{(3+)}}\right]\chi(\tau)=0,$
(334)
$\displaystyle\left[E_{6}^{(3+)}\mathcal{D}+\mu_{1}\left(E_{4}^{(3+)}\right)^{2}+\mu_{2}\Delta_{3^{+},8}\right]\chi(\tau)=0.$
(335)

Note that in above $\mathcal{D}=\theta_{q}$. Also, the indicial equation
($\alpha_{0}+\mu_{1}=0$) dictates $\mu_{1}=1$.
Consider now the ansatz, $\chi=j_{3^{+}}+\mathcal{N}$. Substituting this in
ODE 335 yields

$\displaystyle\left(E_{6}^{(3+)}\mathcal{D}+\mu_{1}\left(E_{4}^{(3+)}\right)^{2}+\mu_{2}\Delta_{3^{+},8}\right)(j_{3^{+}}+\mathcal{N})=\Delta_{3^{+},8}\left[(108+b+\mathcal{N})j_{3^{+}}+\mathcal{N}b\right]$
(336)

So, we see that $j_{3^{+}}+\mathcal{N}$ is a solution to the above
differential equation if and only if,

$\displaystyle
b=0\,\,\,\text{and}\,\,\,\mathcal{N}=-108\longrightarrow\text{soln.:}\,\,j_{3^{+}}-108,$
(337) $\displaystyle
b=-108\,\,\,\text{and}\,\,\,\mathcal{N}=0\longrightarrow\text{soln.:}\,\,j_{3^{+}},$
(338)

Note that, $j_{3^{+}}-108$ has $m_{1}<0$ and hence is not an admissible
solution. So, the only admissible solution is: $j_{3^{+}}$ which happens when
$b=-108$ and $\mathcal{N}=0$. Thus, unlike the $\text{SL}(2,\mathbb{Z})$ or
the Fricke level $2$ case, here we do not get $j_{3^{+}}+\mathcal{N}$ as a
solution to the bulk-pole MLDE. This probably happens because of the following
reason. In the bulk-pole MLDE of both $\text{SL}(2,\mathbb{Z})$ and
$\Gamma^{+}_{0}(2)$ which appears at $\ell=6$ and $\ell=4$ respectively, the
bulk-pole shows up in the denominator ($D^{r}=E_{4}^{3}+\mu\Delta$ and
$D^{r}=\left(E_{4}^{(2+)}\right)^{2}+\mu\Delta_{2}$ respectively) while in
$\Gamma^{+}_{0}(3)$ the bulk-pole appears in the numerator
($N^{r}=\left(E_{4}^{(3+)}\right)^{2}+\mu\Delta_{3^{+},8}$).

###  VI.2 Two-character solutions

From the dimension of the space of modular forms in 94, we obtain the
following expression for the number of free parameters

$\displaystyle\begin{split}\\#(\mu)=\begin{cases}\left\lfloor\left.\frac{\ell}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+1}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+2}{3}\right\rfloor\right.+2,\
&2\ell\not\equiv 2,6\ (\text{mod}\ 12),\ 2\ell>2\\\
\left\lfloor\left.\frac{\ell}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+1}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+2}{3}\right\rfloor\right.-1,\
&2\ell\equiv 2,6\ (\text{mod}\ 12),\ 2\ell>2.\end{cases}\end{split}$ (339)

On the other hand, for the case of $k\equiv 2,6\ (\text{mod}\ 12),k\in
2\mathbb{Z}$, we are to use the following expression

$\displaystyle\\#(\mu)=\left\lfloor\left.\frac{\ell}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+1}{3}\right\rfloor\right.+\left\lfloor\left.\frac{\ell+2}{3}\right\rfloor\right.-1,\
2\ell\equiv 2,6\ (\text{mod}\ 12).$ (340)

####  VI.2.1 $\ell=0$

The space $\mathcal{M}_{2}(\Gamma_{0}^{+}(3))$ is zero-dimensional and there
are no modular forms of weight $2$ in $\Gamma_{0}^{+}(3)$. Hence, we set
$\phi_{1}(\tau)=0$. The space $\mathcal{M}_{4}(\Gamma_{0}^{+}(2))$ is one-
dimensional and hence, we set $\phi_{0}(\tau)=\mu E_{4}^{(3^{+})}(\tau)$. For
these choices, the second-order MLDE takes the following form

$\displaystyle\left[\mathcal{D}^{2}+\mu_{1}E_{4}^{(3^{+})}(\tau)\right]\chi(\tau)=0,$
(341)

where $\mu_{1}$ is an independent parameter. Now, since the covariant
derivative transforms a weight $r$ modular form into one of weight $r+2$, the
double covariant derivative of a weight $0$ form is

$\displaystyle\begin{split}\mathcal{D}^{2}=\mathcal{D}_{(2)}\mathcal{D}_{(0)}=&\left(\frac{1}{2\pi
i}\frac{d}{d\tau}-\frac{1}{3}E^{(3^{+})}_{2}(\tau)\right)\frac{1}{2\pi
i}\frac{d}{d\tau}\\\
=&\tilde{\partial}^{2}-\frac{1}{3}E^{(2^{+})}_{2}(\tau)\tilde{\partial}.\end{split}$
(342)

The MLDE 341 now reads

$\displaystyle\left[\tilde{\partial}^{2}-\frac{1}{3}E^{(3^{+})}_{2}(\tau)\tilde{\partial}+\mu_{1}E_{4}^{(3^{+})}(\tau)\right]\chi(\tau)=0.$
(343)

This equation can be solved by making the following mode expansion
substitution for the character $\chi(\tau)$ and the other modular forms,

$\displaystyle\chi(\tau)=$ $\displaystyle
q^{\alpha}\sum\limits_{n=0}^{\infty}\chi_{n}q^{n},$ (344) $\displaystyle
E_{2}^{(3^{+})}(\tau)=$
$\displaystyle\sum\limits_{n=0}^{\infty}E_{2,n}^{(3^{+})}q^{n}$
$\displaystyle=$ $\displaystyle 1-6q-18q^{2}-42q^{3}-42q^{4}+\ldots,$
$\displaystyle E_{4}^{(2^{+})}(\tau)=$
$\displaystyle\sum\limits_{n=0}^{\infty}E_{4,n}^{(3^{+})}q^{n}$ (345)
$\displaystyle=$ $\displaystyle 1+24q+216q^{2}+888q^{3}+1752q^{4}+\ldots$
(346)

Substituting these expansions in the MLDE, we obtain Substituting these
expansions in the MLDE, we obtain

$\displaystyle
q^{\alpha}\sum\limits_{n=0}^{\infty}\chi_{n}q^{n}\left[(n+\alpha)^{2}-(n+\alpha)\sum\limits_{m=0}^{\infty}\frac{1}{3}E^{(3^{+})}_{2,m}q^{m}+\mu\sum\limits_{m=0}^{\infty}E_{4,m}^{(3^{+})}q^{m}\right]=0.$
(347)

When $n=0,m=0$, with $E^{(3^{+})}_{2,0}=E^{(3^{+})}_{4,0}=1$, we obtain the
following indicial equation

$\displaystyle\alpha^{2}-\frac{1}{3}\alpha+\mu=0.$ (348)

Solving this indicial equation, we have

$\displaystyle\begin{split}\alpha_{0}=&\frac{1}{6}\left(1-\sqrt{1-36\mu}\right)\equiv\frac{1}{6}(1-x),\\\
\alpha_{1}=&\frac{1}{6}\left(1+\sqrt{1-36\mu}\right)\equiv\frac{1}{6}(1+x),\end{split}$
(349)

where we have set $x=\sqrt{1-36\mu}$. The smaller solution
$\alpha_{0}=\tfrac{1}{5}(1-x)$ corresponds to the identity character which
behaves as $\chi(\tau)\sim q^{\tfrac{1-x}{6}}\left(1+\mathcal{O}(q)\right)$.
We know that the identity character $\chi_{0}$, associated with a primary of
weight $h_{0}=0$ behaves as $\chi_{0}(\tau)\sim
q^{-\tfrac{c}{24}}\left(1+\mathcal{O}(q)\right)$. Comparing the two
behaviours, we obtain the following expression for the central charge

$\displaystyle c=4(x-1).$ (350)

To find the conformal dimension $h$, we compare the behaviours with the larger
solution for $\alpha$, i.e. $\chi(\tau)\sim
q^{\tfrac{1+x}{6}}\left(1+\mathcal{O}(q)\right)$ and $\chi(\tau)\sim
q^{-\tfrac{c}{24}+h}\left(1+\mathcal{O}(q)\right)$. This gives us

$\displaystyle h=\frac{x}{3}.$ (351)

Using the Cauchy product formulae, we obtain the following recurrence relation

$\displaystyle\chi_{n}=\left((n+\alpha)^{2}-\frac{1}{3}(n+\alpha)+\mu\right)^{-1}\sum\limits_{m=1}^{n}\left(\frac{(n+\alpha-m)}{3}E^{(3^{+})}_{2,m}-\mu
E^{(3^{+})}_{4,m}\right)\chi_{n-m}.$ (352)

When $n=1$, we solve for the ratio $\tfrac{\chi_{1}}{\chi_{0}}$ in terms of
$\alpha$ with coefficients $E^{(3^{+})}_{2,1}=-6$ and $E_{4,1}^{(3^{+})}=24$
to obtain

$\displaystyle m_{1}^{(i)}+14\alpha_{0}+3m_{1}\alpha_{0}-36\alpha_{0}^{2}=0.$
(353)

for $i=0,1$ corresponding to ratios of $m^{(0)}_{1}$ and $m^{(1)}_{1}$ taken
with respect to values $\alpha_{0}$ and $\alpha_{1}$ respectively. Restricting
to $i=0$ and assuming that $\alpha_{0}=-\tfrac{c}{24}$, we can recast this
equation by identifying $N=-12\alpha_{0}=\tfrac{c}{2}$,

$\displaystyle-4m_{1}+5N+m_{1}N+N^{2}=0,$ (354)

$a_{1}$ can now be expressed in terms of $N$ as,

$\displaystyle m_{1}=\frac{N(N+5)}{4-N},$ (355)

which sets an upper bound on the central charge $c<8$. Furthermore, applying
the integer root theorem to 354, we see that $c=2N$ and the central charge, in
this case, is restricted to be an even integer, i.e.

$\displaystyle m_{1}=\frac{c(10+c)}{8-c}.$ (356)

Dropping the subscript associated with the index $i$, we see that for
$m_{1}\geq 0$, we require $c<8$. Rewriting this, we have

$\displaystyle c^{2}+c(m_{1}+10)=8m_{1}.$ (357)

This tells us that for $(n,\ell)=(2,0)$ in $\Gamma_{0}^{+}(3)$, we have
$c\in\mathbb{Z}$ and $c<8$. For $c$ to be rational, we demand the determinant,
$\sqrt{25+26m_{1}+m_{1}^{2}}$ , to be rational. We further demand that the
square root is an integer and this gives us

$\displaystyle(m_{1}+13)^{2}-144=k^{2},$ (358)

where $k^{2}\in\mathbb{Z}$. We now set $p$ to be

$\displaystyle p=13+m_{1}-k,$ (359)

and we recast 358 as follows

$\displaystyle m_{1}+13=\frac{144+p^{2}}{2p}=\frac{72}{p}+\frac{p}{2}.$ (360)

Restricting $k$ to be positive, we see that $p<13$. We conclude that all
possible values of $m_{1}$ are found by those values of $p$ below $13$ that
divide $72$ and are even. The list of these values reads $p=\\{2,4,6,8,12\\}$.
We ignore the case corresponding to $p=12$ since $m_{1}=-1$. Table 6 contains
CFT data.

$p$ | 2 | 4 | 6 | 8
---|---|---|---|---
$\mu_{1}$ | $-\tfrac{7}{48}$ | $-\tfrac{1}{12}$ | $-\tfrac{5}{144}$ | $0$
$m_{1}$ | 24 | 7 | 2 | 0
$c$ | 6 | 4 | 2 | 0
$h$ | $\tfrac{5}{6}$ | $\tfrac{2}{3}$ | $\tfrac{1}{2}$ | $\tfrac{1}{3}$
Table 6: $c$ and $\Delta$ data corresponding to the Fricke group
$\Gamma_{0}^{+}(3)$ for the choice $\phi_{0}=\mu_{1}E^{(3^{+})}_{4}(\tau)$
with $\ell=0$.

Using recursion relation 347, when $n=2$, we obtain

$\displaystyle
m_{2}^{(i)}\equiv\frac{\chi_{2}^{(i)}}{\chi_{0}^{(i)}}=\frac{9\alpha_{i}(36\alpha_{i}-13)}{6\alpha_{i}+5}+\frac{3m_{1}^{(i)}(-1+\alpha_{i}(-5+12\alpha_{i}))}{6\alpha_{i}+5}.$
(361)

We discard the case $p=8$ since further computations show that all
coefficients $m_{i}$ are null valued thus making this a trivial solution. The
values of $m_{2}$ fr $i=0$ is tabulated in table 7.

$p$ | 2 | 4 | 6
---|---|---|---
$c$ | 6 | 4 | 2
$m_{2}$ | $\tfrac{243}{7}$ | 8 | 2
Table 7: Values of $m_{2}$ for corresponding to the Fricke group
$\Gamma_{0}^{+}(3)$ for the choice $\phi_{0}=\mu_{1}E^{(3^{+})}_{4}(\tau)$
with $\ell=0$.

In a similar fashion, computing coefficients for $i=0,1$, we obtain the
coefficients of the two characters. Checking to higher-order (up to
$\mathcal{O}(q^{5000})$) for both characters, for $0<c<8$ (even $c$), we find
the admissible character shown below.

  * •

$(c,h)=\left(2,\frac{1}{2}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{12}}(1+2q+2q^{2}+4q^{3}+5q^{4}+6q^{5}+11q^{6}+14q^{7}+17q^{8}+24q^{9}+31q^{10}+\mathcal{O}(q^{11})),\\\
\chi_{\frac{1}{2}}=&q^{\tfrac{5}{12}}(1+q^{2}+2q^{3}+2q^{4}+2q^{5}+5q^{6}+4q^{7}+7q^{8}+10q^{9}+11q^{10}+\mathcal{O}(q^{11})).\end{split}$
(362)

We also find a identity only admissible solution, which only has a nice
identity character but an unstable non-trivial character. This is found at
$(c,h)=\left(4,\frac{2}{3}\right)$.

$\displaystyle\begin{split}\chi_{0}=q^{-\tfrac{1}{6}}(1+&7q+8q^{2}+22q^{3}+42q^{4}+63q^{5}+106q^{6}+190q^{7}+267q^{8}\\\
+&428q^{9}+652q^{10}+\mathcal{O}(q^{11}))\\\
=&j^{\tfrac{1}{6}}_{3^{+}}.\end{split}$ (363)

At central charge $c=6$, both characters are unstable. We here note that the
identity character in 362 can be expressed in terms of (semi-)modular forms,
which we shall denote by $\mathcal{S}^{0}_{k}(\Gamma)$, of the Hecke group
$\Gamma_{0}(3)$ as follows

$\displaystyle\begin{split}\chi_{0}^{(c=2)}=\left(\frac{\Delta_{3}^{0}(2\tau)}{\left(\Delta_{3}^{0}\right)^{\tfrac{1}{2}}(\tau)}\right)^{\tfrac{1}{3}}\left(\frac{1}{\Delta_{3}(\tau)}\right)^{\tfrac{1}{12}},\
\ \
\Delta_{3}^{0}(\tau)=\frac{\eta^{9}(\tau)}{\eta^{3}(3\tau)}\in\mathcal{S}^{0}_{3}(\Gamma_{0}(3)),\end{split}$
(364)

where $\Delta_{3}^{0}(\tau)$ is the $2^{\text{nd}}$ semi-modular form of
weight $2$ in $\Gamma_{0}(3)$ Junichi .

####  VI.2.2 $\ell=2$

The (2,2) MLDE takes the following form

$\displaystyle\left[\mathcal{D}^{2}+\mu_{1}\frac{E_{6}^{(3+)}}{E_{4}^{(3+)}}\mathcal{D}+\mu_{2}E_{4}^{(3+)}+\mu_{3}\frac{\Delta_{3^{+},8}}{E_{4}^{(3+)}}\right]\chi(\tau)=0,$
(365)

where $\chi$ is a weight 0 modular function for $\Gamma^{+}_{0}(3)$.
Substituting for the Fourier expansions of the relevant terms in 365, we
obtain the following recursion relation

$\displaystyle\sum\limits_{k=0}^{i}\left[E_{4,k}^{(3+)}(\alpha^{j}+i-k)(\alpha^{j}+i-k-1)+E_{4,k}^{(3+)}(\alpha^{j}+i-k)-\frac{E_{2,k}^{(3+)}E_{4,k}^{(3+)}}{3}(\alpha^{j}+i-k)\right.$
$\displaystyle\left.\mu_{1}E_{6,k}^{(3+)}(\alpha^{j}+i-k)+\mu_{2}(E_{4}^{(3+)})^{2}_{,k}+\mu_{3}\Delta_{3^{+},8,k}\right]a^{j}_{i-k}=0,$
(366)

where $a_{i}$s are Fourier coefficients of the character $q$-series expansion.
Setting $i=0$ and $k=0$ in 366, we obtain the following indicial equation

$\displaystyle\alpha^{2}+\left(\mu_{1}-\frac{1}{3}\right)\alpha+\mu_{2}=0.$
(367)

Setting $\alpha=\alpha_{0}$ in 367 and observing Riemann-Roch we get,

$\displaystyle\mu_{1}=\frac{2}{3},$ (368)
$\displaystyle\mu_{2}=-\left(\alpha_{0}^{2}-\frac{\alpha_{0}}{3}\right).$
(369)

Tthe value of $\mu_{3}$ can now be determined by transforming the $\tau$-MLDE
in 365 into the corresponding $j_{3^{+}}$-MLDE (about $\tau=\rho_{3}$),

$\displaystyle\left[(j_{3^{+}})^{2}\partial_{j_{3^{+}}}^{2}+\frac{j_{3^{+}}}{6}\partial_{j_{3^{+}}}+\frac{(j_{3^{+}})^{2}}{2(j_{3^{+}}-108)}\partial_{j_{3^{+}}}+\frac{\mu_{2}j_{3^{+}}}{j_{3^{+}}-108}+\frac{\mu_{3}}{j_{3^{+}}-108}\right]\chi=0,$
(370)

where we have set $\mu_{1}=\frac{2}{3}$ from the $\ell=2$ indicial equation
(see 367). Now, note that about $\tau=\rho_{3}$, we have, $\chi_{0}\sim
j_{3^{+}}^{x}$ and $j_{3^{+}}\to 0$, with $x=\frac{2}{3}$ for $\ell=2$ (see
single-character solutions). Hence, substituting $\chi=j_{3^{+}}^{x}$ about
$j_{3^{+}}=0$, the $j_{3^{+}}$-indicial equation (which is basically,
coefficient of $j_{3^{+}}^{x}$ $=\left.0\right|_{j_{3^{+}}=0}$) reads,

$\displaystyle x^{2}-\frac{5}{6}x-\frac{\mu_{3}}{108}=0,$ (371)

which yields $\mu_{2}=-12$ for $x=\frac{2}{3}$. Similar analysis gives,
$(x,\mu_{2})=(\frac{1}{3},-18)$ as solution to 371. For $\mu_{3}=-12$, we get
from the $i=1$ equation: $m_{1}=9-N$ where $N=-12\alpha_{0}$. This gives the
allowed range of central charge as, $0<c<10$ with $c\in 2\mathbb{Z}$. Scanning
this space of $c$ we get the following admissible solutions.

  * •

$(c,h)=\left(4,0\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{6}}(1+7q+8q^{2}+22q^{3}+42q^{4}+63q^{5}+106q^{6}+190q^{7}\\\
&{}\ \ \ \ \ \ \ \ +267q^{8}+428q^{9}+652q^{10}+\mathcal{O}(q^{11}))\\\
=&j_{3^{+}}^{\tfrac{1}{6}},\\\ \chi_{\text{non-id}}=&\chi_{0}\end{split}$
(372)

  * •

$(c,h)=\left(6,\frac{1}{6}\right)$:

$\displaystyle\begin{split}\chi_{0}&=q^{-\tfrac{1}{4}}(1+6q+8q^{3}+18q^{4}+17q^{6}+54q^{7}+\mathcal{O}(q^{9})),\\\
\chi_{\frac{1}{6}}&=q^{-\tfrac{1}{12}}(1+8q+17q^{2}+46q^{3}+98q^{4}+198q^{5}+371q^{6}+692q^{7}+1205q^{8}+\mathcal{O}(q^{9})).\end{split}$
(373)

For $\mu_{3}=-18$, we get from the $i=1$ equation:
$m_{1}=\frac{N^{2}-17N+108}{8-N}$ where $N=-12\alpha_{0}$. This gives the
allowed range of central charge as, $0<c<8$ with $c\in 2\mathbb{Z}$. Scanning
this space of $c$ we get the following admissible solutions.

  * •

$(c,h)=\left(4,0\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{6}}(1+13q+50q^{2}+76q^{3}+222q^{4}+405q^{5}+664q^{6}+1222q^{7}\\\
&{}\ \ \ \ \ \ \ \ +2121q^{8}+3146q^{9}++\mathcal{O}(q^{10})),\\\
\chi_{\text{non-id}}=&\chi_{0}\end{split}$ (374)

  * •

$(c,h)=\left(8,\frac{1}{3}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{3}}(1+14q+65q^{2}+156q^{3}+456q^{4}+1066q^{5}+2250q^{6}+4720q^{7}\\\
&{}\ \ \ \ \ \ \ \ \ +9426q^{8}+17590q^{9}+\mathcal{O}(q^{10})),\\\
=&j_{3^{+}}^{\tfrac{1}{3}},\\\
\chi_{\frac{1}{3}}=&\text{unstable}.\end{split}$ (375)

  * •

$(c,h)=\left(10,\frac{1}{2}\right)$:

$\displaystyle\chi_{0}=j_{3^{+}}^{\tfrac{1}{3}}\otimes\chi_{0}^{{}^{(3)}\mathcal{W}_{2}}$
$\displaystyle\chi_{\frac{1}{2}}=j_{3^{+}}^{\tfrac{1}{3}}\otimes\chi_{\frac{1}{2}}^{{}^{(3)}\mathcal{W}_{2}}$
(376)

  * •

$(c,h)=\left(12,\frac{2}{3}\right)$:

$\displaystyle\begin{split}\chi_{0}=&q^{-\tfrac{1}{2}}(1+21q+171q^{2}+745q^{3}+2418q^{4}+7587q^{5}+20510q^{6}+51351q^{7}+\mathcal{O}(q^{8}))\\\
=&j_{3^{+}}^{\tfrac{1}{2}},\\\ \chi_{\frac{2}{3}}=&\text{unstable}\end{split}$
(377)

We note that the identity character in 372, and also 363, turns out to be the
McKay-Thompson series of the $18b$ conjugacy class of $\mathbb{M}$ (OEIS
sequence A058537 A058537 ), the identity character in 375 is the single
character solution we found in 331 which is nothing but the McKay-Thompson
series of the $9a$ conjugacy class of $\mathbb{M}$ (OEIS sequence A058092
A058092 ), and the identity character in 377 turns out to be the McKay-
Thompson series of the $6b$ conjugacy class of $\mathbb{M}$ (OEIS sequence
A007261 A007261 ).

###  VI.3 Three-character solutions

The $(3,0)$ MLDE reads,

$\displaystyle\left(\mathcal{D}^{3}+\mu_{1}E_{4}^{(3+)}\mathcal{D}+\mu_{2}E_{6}^{(3+)}\right)\chi=0.$
(378)

The recursion relation corresponding to the above $(3,0)$ MLDE reads

$\displaystyle\left[(\alpha^{j}+i)(\alpha^{j}+i-1)(\alpha^{j}+i-2)+3(\alpha^{j}+i)(\alpha^{j}+i-1)+(\alpha^{j}+i)\right]a^{j}_{i}$
$\displaystyle+\sum\limits_{k=0}^{i}\left[-E_{2,k}^{3+}a^{j}_{i-k}(\alpha^{j}+i-k)(\alpha^{j}+i-k-1)-E_{2,k}^{3+}a^{j}_{i-k}(\alpha^{j}+i-k)\right.$
$\displaystyle\left.+\frac{1}{18}E_{4,k}^{3+}a^{j}_{i-k}(\alpha^{j}+i-k)+\frac{1}{6}(E_{2}^{3+})^{2}_{,k}a^{j}_{i-k}(\alpha^{j}+i-k)+\mu_{1}E_{4,k}^{3+}a^{j}_{i-k}(\alpha^{j}+i-k)+\mu_{2}E_{6,k}^{3+}a^{j}_{i-k}\right]=0.$
(379)

From which the indicial equation reads,

$\displaystyle\alpha^{3}-\alpha^{2}+\left(\mu_{1}+\frac{2}{9}\right)\alpha+\mu_{2}=0.$
(380)

Also,

$\displaystyle
4m_{1}m_{2}+108m_{1}\alpha_{0}-18m_{1}^{2}\alpha_{0}+180m_{2}\alpha_{0}+6m_{1}m_{2}\alpha_{0}+3348\alpha_{0}^{2}-594m_{1}\alpha_{0}^{2}$
$\displaystyle-126m_{1}^{2}\alpha_{0}^{2}+360m_{2}\alpha_{0}^{2}-14472\alpha_{0}^{3}-2268m_{1}\alpha_{0}^{3}-108m_{1}^{2}\alpha_{0}^{3}+216m_{2}\alpha_{0}^{3}+10368\alpha_{0}^{4}=0.$
(381)

The Diophantine equation reads, (which is obtained by multiplying
$16^{3}\times 2$ with 381 and then identifying $N=-96\alpha_{0}$),

$\displaystyle 32768m_{1}m_{2}-9216m_{1}N+1536m_{1}^{2}N-15360m_{2}N$
$\displaystyle-512m_{1}m_{2}N+2976N^{2}-528m_{1}N^{2}-112m_{1}^{2}N^{2}+320m_{2}N^{2}+134N^{3}$
$\displaystyle+21m_{1}N^{3}+m_{1}^{2}N^{3}-2m_{2}N^{3}+N^{4}=0.$ (382)

By integer root theorem $N=-96\alpha_{0}$ above should be an integer and hence
$N=4c\in\mathbb{Z}$. Next, we have that the discriminant of the cubic indicial
equation should be a perfect square, say $k^{2}$ with $k\in\mathbb{Z}$. Thus
we have,

$\displaystyle 262144m_{1}^{2}-163840m_{1}N-16384m_{1}^{2}N+21504N^{2}$
$\displaystyle+8192m_{1}N^{2}+256m_{1}^{2}N^{2}-448N^{3}-96m_{1}N^{3}-7N^{4}=k^{2}.$
(383)

The three roots of the indicial equation read,

$\displaystyle\alpha_{0}=-\frac{N}{96},$ (384)
$\displaystyle\alpha_{1}=\frac{1536m_{1}-672N+16m_{1}N-7N^{2}-3k}{192(16m_{1}-7N)},$
(385)
$\displaystyle\alpha_{2}=\frac{1536m_{1}-672N+16m_{1}N-7N^{2}+3k}{192(16m_{1}-7N)}.$
(386)

Now solving 382 and 383, for the range $1\leq N\leq 96$, yields,

  * •

$(c,h_{1},h_{2})=\left(1,\frac{7}{8},\frac{1}{4}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{24}}(1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+q^{7}+2q^{8}+3q^{9}+3q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{7}{8}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{4}}$
$\displaystyle=q^{\tfrac{5}{24}}(1+q+q^{3}+q^{4}+q^{5}+2q^{6}+2q^{7}+2q^{8}+3q^{9}+3q^{10}+\mathcal{O}(q^{11})).$
(387)

The $q$-series expansions of the identity and the second descendant of this
theory can be expressed in terms of ratios of Ramanujan theta series following
the sequences in A097242  (OEIS sequence A097242) and A328796  (OEIS sequence
A328796) respectively.

  * •

$(c,h_{1},h_{2})=\left(2,\frac{3}{4},\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{12}}(1+2q+2q^{2}+4q^{3}+5q^{4}+6q^{5}+11q^{6}+14q^{7}+17q^{8}+24q^{9}+31q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{3}{4}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{5}{12}}(1+q^{2}+2q^{3}+2q^{4}+2q^{5}+5q^{6}+4q^{7}+7q^{8}+10q^{9}+11q^{10}+\mathcal{O}(q^{11})).$
(388)

This is a $(2,0)$ solution found in 362 that makes a reappearance here.

  * •

$(c,h_{1},h_{2})=\left(4,1,\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{6}}(1+m_{1}q+m_{2}q^{2}+\mathcal{O}(q^{3})),$
$\displaystyle\chi_{1}$
$\displaystyle=q^{\tfrac{5}{6}}(1+2q^{2}+4q^{3}+5q^{4}+8q^{5}+18q^{6}+20q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+36q^{8}+56q^{9}+76q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{1}{3}}(1+2q+3q^{2}+8q^{3}+13q^{4}+20q^{5}+37q^{6}+56q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+83q^{8}+134q^{9}+196q^{10}+\mathcal{O}(q^{11})),$ (389)

where for the above identity character we have,
$m_{1}\in\mathbb{N}\cup\\{0\\}$, $m_{2}=8$ and $k=1792-256m_{1}$ for $m_{1}<7$
and $k=-1792+256m_{1}$ for $m_{1}\geq 8$.

  * •

$(c,h_{1},h_{2})=\left(6,\frac{5}{4},\frac{1}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{4}}(1+9q+24q^{2}+56q^{3}+135q^{4}+264q^{5}+497q^{6}+945q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+1656q^{8}+2830q^{9}+4815q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{5}{4}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{\tfrac{1}{4}}(1+7q+9q^{2}+31q^{3}+66q^{4}+117q^{5}+227q^{6}+436q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \
+702q^{8}+1241q^{9}+2072q^{10}+\mathcal{O}(q^{11})).$ (390)

This is a new two-character solution that we did not find in our analysis with
$(2,\ell)$ MLDEs.

  * •

$(c,h_{1},h_{2})=\left(7,\frac{9}{8},\frac{3}{4}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{7}{24}}(1+7q+35q^{2}+70q^{3}+189q^{4}+420q^{5}+833q^{6}+1631q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \
+3143q^{8}+5530q^{9}+9863q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{9}{8}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{3}{4}}$
$\displaystyle=q^{\tfrac{11}{20}}(7+22q+56q^{2}+161q^{3}+343q^{4}+686q^{5}+1421q^{6}+2653q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+4782q^{8}+8638q^{9}+14847q^{10}+\mathcal{O}(q^{11})).$ (391)

This is a two-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(7,\frac{7}{4},\frac{1}{8}\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{7}{4}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{1}{8}}$
$\displaystyle=q^{-\tfrac{1}{6}}(1+7q+8q^{2}+22q^{3}+42q^{4}+63q^{5}+106q^{6}+190q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+267q^{8}+428q^{9}+652q^{10}+\mathcal{O}(q^{11})).$ (392)

The stable descendant with $h=\tfrac{1}{8}$ is a $c=4$ single-character theory
found in 363 that also reappears as a two-character theory with an unstable
descendant in 372.

  * •

$(c,h_{1},h_{2})=\left(8,1,1\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{1}{3}}(1+m_{1}q+m_{2}q^{2}+\mathcal{O}(q^{3})),$ ,
$\displaystyle\chi_{1}$
$\displaystyle=q^{-\tfrac{2}{3}}(1+2q+8q^{2}+22q^{3}+47q^{4}+102q^{5}+224q^{6}+422q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+815q^{8}+1516q^{9}+2688q^{10}+\mathcal{O}(q^{11})),$ $\displaystyle\chi_{1}$
$\displaystyle=q^{-\tfrac{2}{3}}(1+2q+8q^{2}+22q^{3}+47q^{4}+102q^{5}+224q^{6}+422q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+815q^{8}+1516q^{9}+2688q^{10}+\mathcal{O}(q^{11})).$ (393)

where for the above identity character we have,
$m_{1}\in\mathbb{N}\cup\\{0\\}$, $m_{2}=37+2m_{1}$ and $k=0$. Also, note that,
in the above case the two descendants are equal. This is exactly similar to
how $D_{4,1}$ (and its GHM dual (Gaberdiel:2016zke ) with central charge 20)
appear as 3-character solutions to the $(3,0)$ MLDE in the
$\text{SL}(2,\mathbb{Z})$ case (see Das:2021uvd ).

  * •

$(c,h_{1},h_{2})=\left(16,0,3\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$ (394)
$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{3}$
$\displaystyle=q^{\tfrac{7}{3}}(1+7q+35q^{2}+140q^{3}+490q^{4}+1547q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ +4480q^{6}+12192q^{7}+31402q^{8}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+77119q^{9}+182119q^{10}+\mathcal{O}(q^{11})).$ (395)

  * •

$(c,h_{1},h_{2})=\left(20,0,\frac{7}{2}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{6}}(1+47q+440q^{2}+5782q^{3}+83517q^{4}+239124q^{5}+1070594q^{6}+4192784q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+14782019q^{8}+47772551q^{9}+144263738q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{0}$
$\displaystyle=q^{-\tfrac{5}{6}}(1+47q+440q^{2}+5782q^{3}+83517q^{4}+239124q^{5}+1070594q^{6}+4192784q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+14782019q^{8}+47772551q^{9}+144263738q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{7}{2}}$ $\displaystyle=\text{unstable}.$ (396)

This is a two-character solution (not yet a $(2,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(20,\frac{1}{2},3\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$ (397)
$\displaystyle\chi_{\frac{1}{2}}$
$\displaystyle=q^{-\tfrac{1}{3}}(1+14q+65q^{2}+156q^{3}+456q^{4}+1066q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ +2250q^{6}+4720q^{7}+9426q^{8}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \
+17590q^{9}+32801q^{10}+\mathcal{O}(q^{11})),$ $\displaystyle\chi_{3}$
$\displaystyle=\text{unstable}.$ (398)

  * •

$(c,h_{1},h_{2})=\left(24,\frac{6}{7},\frac{22}{7}\right)$:

$\displaystyle\chi_{0}$
$\displaystyle=q^{-1}(1+238q+783q^{2}+8672q^{3}+65367q^{4}+371520q^{5}+1741655q^{6}+7161696q^{7}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \
+26567946q^{8}+90521472q^{9}+288078201q^{10}+\mathcal{O}(q^{11})),$
$\displaystyle\chi_{\frac{6}{7}}$ $\displaystyle=\text{unstable},$
$\displaystyle\chi_{\frac{22}{7}}$ $\displaystyle=\text{unstable}.$ (399)

This is a one-character solution (not yet a $(1,\ell)$ solution).

  * •

$(c,h_{1},h_{2})=\left(24,0,4\right)$:

$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$ (400)
$\displaystyle\chi_{0}$ $\displaystyle=\text{Log unstable},$
$\displaystyle\chi_{4}$
$\displaystyle=q^{3}(3185+34398q+231231q^{2}+1201824q^{3}+5321511q^{4}+20914816q^{5}$
$\displaystyle{}\ \ \ \ \ \ \ \ \ \ \ \ \ \
+74740679q^{6}+247766688q^{7}+770976635q^{8}$ $\displaystyle{}\ \ \ \ \ \ \ \
\ \ \ \ \ \ +2272634846q^{9}+6394256424q^{10}+\mathcal{O}(q^{11})).$ (401)

##  VII Lattice relations

The $\Theta$-function of a $d$-dimensional lattice $\Lambda$ is a weight
$\tfrac{d}{2}$ modular form defined as follows

$\displaystyle\Theta_{\Lambda}(\tau)=\sum\limits_{x\in\Lambda}N(m)q^{m}.$
(402)

Here, the sum runs over all the vectors $x$ in the lattice $\Lambda$ whose
length is $m=x\cdot x$, $N(m)$ denotes the number of vectors of norm $m$, and
$q\equiv e^{2\pi i\tau}$. The lattice partition function $\mathcal{Z}$ is
defined as follows

$\displaystyle\mathcal{Z}(\tau)\equiv\frac{\Theta_{\Lambda}(\tau)}{\eta^{d}(\tau)}.$
(403)

It is always possible to define a holomorphic modular-covariant CFT with
central charge $c$ with a partition function given by 403 corresponding to an
even self-dual lattice of integer dimension $d$. The packing radius $\rho$,
the kissing number $\mathscr{K}$ can be read off from the $q$-series expansion
of the $\Theta$-function of a lattice $\Lambda$ as follows

$\displaystyle\Theta_{\Lambda}(\tau)=1+\mathscr{K}q^{4\rho^{2}}+\ldots$ (404)

Let $\Theta_{2^{+}}(\tau)=1+\sum_{m=1}^{\infty}N_{2^{+}}(m)q^{m}$ denote the
lattice corresponding to the single character solution of central charge
$c=24$ found in $\Gamma_{0}^{+}(2)$ with partition function
$\mathcal{Z}(\tau)=j_{2^{+}}(\tau)$. This $\Theta$-series was found to be
related to the $\Theta$-series of odd Leech lattice $O_{24}$ as follows
Umasankar:2022kzs

$\displaystyle\Theta_{O_{24}}(\tau)=1+\sum\limits_{n=1}^{\infty}m^{(23A)}_{n}q^{n}-\Theta_{2^{+}}\left(\frac{\tau}{2}\right).$
(405)

where $m^{(23A)}_{i}$ indicated the coefficients obtained by linear
combinations of character degrees of the McKay-Thompson series of class $23A$
for $\mathbb{M}$. From V.1.4, we see that the partition function,
$\mathcal{Z}(\tau)=j_{2^{+}}(\tau)+\mathcal{N}$, represents a large set of
admissible solutions that includes the single character solution with
$\mathcal{N}=0$ at $\ell=4$, the non-trivial bilinear pairs in 258, 259, and
260 with $\mathcal{N}=-32,-64,-80$ respectively, and possibly more which we
haven’t found yet in the current analysis. With this partition function, we
obtain the following $q$-series expansion of the $\Theta$-function

$\displaystyle\eta^{24}(\tau)\left(j_{2^{+}}(\tau)-80+\tilde{a}\right)=1+\tilde{a}q+(4048-24\tilde{a})q^{2}+(-4096+252\tilde{a})q^{3}+\ldots,$
(406)

where $\tilde{a}=\mathcal{N}a$. Now, upon setting $\tilde{a}=0$, we find the
kissing number and lattice radius to be $\mathscr{K}=4048$ and
$\rho=\tfrac{1}{\sqrt{2}}$ respectively. Thus, a wide set of $c=24$ single-
character solutions and concocted bilinear pairs all possess the same
$\Theta$-series that is related to the odd Leech lattice in $d=24$ via the
relation shown in 405.

The space of modular forms of weight $4$ is one-dimensional and it turns out
that the modular form is related to a lattice theta function as follows

$\displaystyle\begin{split}\omega^{(2^{+})}_{4}(\tau)=&E_{4}^{(2^{+})}(\tau)\\\
=&\tfrac{1}{4}\left(\theta_{3}^{4}(\tau)+\theta_{4}^{4}(\tau)\right)^{2}\\\
=&\Theta_{\mathbf{D}_{4}\oplus\mathbf{D}_{4}}(\tau).\end{split}$ (407)

The Fricke weight $4$ Eisenstein series of level $2$ is the theta function of
two copies of the $\mathbf{D}_{4}$ lattice (OEIS sequence A008658 A008658 ).
Such lattice relations are not observed for $E_{k}^{(2^{+}}(\tau)$ with $k\geq
4$. Moving over to $\Gamma_{0}^{+}(3)$, we the following connections between
the modular forms lattice theta functions

$\displaystyle\begin{split}\omega^{(3^{+})}_{4}(\tau)=&E_{4}^{(3^{+})}(\tau)=\Theta_{4\cdot\mathbf{H}}(\tau),\\\
E_{2,3^{{}^{\prime}}}(\tau)=&\Theta_{2\cdot\mathbf{H}}(\tau)\end{split}$ (408)

The Fricke weight $4$ Eisenstein series of level $3$ is the theta function of
six copies of the Hexagonal lattice $\mathbf{H}$ (OEIS sequence A008655
A008655 ) and the modular form
$E_{2,3^{{}^{\prime}}}(\tau)\in\mathcal{M}_{3}(\Gamma_{0}(3))$ is the theta
function of two copies of $\mathbf{H}$ (OEIS sequence A008653 A008653 . Such
lattice relations are not observed for $E_{k}^{(p^{+}}(\tau)$ with $k\geq 4$
and $p=3,5,7$. However, in $\Gamma_{0}^{+}(7)$, we find that the modular form
$E_{2,7^{{}^{\prime}}}(\tau)$ (see appendix A for definition) is the theta
series of the square of Kleinian lattice
$\mathbb{Z}\left[\tfrac{-1+\sqrt{-7}}{2}\right]$ (OEIS sequence A028594
A028594 ). Similar relations can be found for other modular forms of Hecke and
Fricke groups.

##  VIII Discussion

###  VIII.1 Fricke Monsters lurk in the Borcherds product

From MLDE analysis for groups $\Gamma_{0}^{+}(2)$ and $\Gamma_{0}^{+}(3)$, we
found that the identity characters with unstable or trivial descendants
possess coefficients equal to a McKay-Thompson series of a certain conjugacy
class of $\mathbb{M}$. There exists a Borcherds product interpretation for
this observation as pointed out in Duncan:2022afh  (see table $10$). We
summarize this observation here. Consider the following infinite product
expansion of the $\text{SL}(2,\mathbb{Z})$ Haputmodul,

$\displaystyle
j(\tau)-j(z)=q_{\tau}^{-1}\prod\limits_{\begin{subarray}{c}m>0\\\
n\in\mathbb{Z}\end{subarray}}\left(1-q_{\tau}^{m}q_{z}^{n}\right)^{c(m,n)},$
(409)

where $q_{a}=e^{2\pi ia}$, and $c(r)$ is the r${}^{\text{th}}$ Fourier
coefficient of $J(\tau)=j(\tau)-744$. When $m=1,n=1$, we find the RHS to read
$q_{\tau}^{-1}-q_{z}^{-1}$ which indeed matches the antisymmetric LHS although
it may not look it at first sight. This $j$-product formula also called the
Koike-Norton-Zagier infinite product identity, was independently discovered by
Borcherds, Koike, Norton, and Zagier. This is a generalization of the Weyl
denominator formula for the finite-dimensional Lie algebra which reads

$\displaystyle\sum\limits_{w\in\mathcal{W}}\varepsilon(w)w(P)=q_{\rho}^{-1}\prod\limits_{\alpha>0}\left(1-q_{\alpha}\right)^{M(\alpha)},$
(410)

where $\mathcal{W}$ denotes the Weyl group, $\varepsilon(w)=(-1)^{\ell}(w)$ is
the sign function with $\ell(w)$ denoting the length of the Weyl group
element, $w(P)=q_{\rho}^{-1}\sum_{k}\varepsilon(k)q_{k}$ where the index $k$
runs over all the sums of pairwise orthogonal simple roots, $\rho$ is the Weyl
vector of a group $G$, $\alpha$ are the roots of the Lie algebra, and
$M(\alpha)$, which is always unity, is the multiplicity. Comparing 410 and
409, we see that the infinite sums over all positive vectors correspond to the
infinite sum over all positive roots, the $q_{\tau}^{-1}$ is matched by the
$q_{\rho}^{-1}$, and expanding the difference of the Hauptmodules, we observe
that we obtain an alternating sum that is matched by the alternating sum in
the Weyl formula. We direct the reader to section $4$ of
Carnahan2012GeneralizedMI  for a more detailed explanation of the same. The
generalized Borcherds product is obtained by taking a product sum of the
$j$-product over binary quadratic forms $Q=[a,b,c]=ax^{2}+bxy+cy^{2}$, where
$a,b,c\in\mathbb{Z}$, belonging to the orbit space
$\mathcal{Q}_{D}^{(N)}/\Gamma_{0}^{+}(N)$ (we note that a similar construction
is also applicable to Hecke groups), where $D=b^{2}-4ac$ is the discriminant
of $Q$ and $\mathcal{Q}_{D}$ denotes the space of binary quadratic forms with
$D<0$ and $a>0$. For a unique choice of the discriminant, this takes the
following form

$\displaystyle\prod\limits_{Q\in\mathcal{Q}_{D_{0}}^{(N)}/\Gamma_{0}^{+}(N)}\left(j_{N^{+}}(\tau)-j_{N^{+}}(\alpha_{Q})\right)^{\frac{1}{|s_{Q}|}}=\left(j_{N^{+}}(\tau)+\mathcal{N}\right)^{x},$
(411)

where the root $\alpha_{Q}$ is the complex multiplication point of $Q$,
$s_{Q}$ is the stabilizer in $\Gamma_{0}^{+}(N)/\mathbb{Z}_{2}$, and
$\mathcal{N},x\in\mathbb{Z}$. The values of $j(\alpha_{Q})$ solely depends on
the $\Gamma_{0}^{+}(N)$-equivalence class of $Q$. For the case of $N=2$, we
notice that the set
$(\mathcal{N},x)=\left\\{\left(0,\tfrac{1}{4}\right),\left(0,\tfrac{1}{2}\right),(104,1)\right\\}$
corresponding to solutions with central charges $c=6$ (213, 231, 248), $c=12$
(252, 271), and to the single character solution with central charge $c=24$
(V.1.4) respectively. Next, when $N=3$, we find that the set
$(\mathcal{N},x)=\left\\{\left(0,\tfrac{1}{6}\right),\left(0,\tfrac{1}{3}\right),\left(0,\tfrac{1}{2}\right)\right\\}$
that correspond to solutions with central charges $c=4$ (363, 372), $c=8$
(375), and $c=12$ (377).

From Duncan:2022afh , we can also predict the admissible solutions with
unstable descendants for higher-level Fricke groups. This, along with the
results thus far for Fricke levels $p=2,3$ are summarized in table 8. Of the
$14$ prime divisor levels, only $10$ turn out to possess admissible identity
solutions with an unstable descendant. At this point, we do not have enough
information to state if these solutions correspond to new characters not found
by the single character analysis, for this we would need to resort to a
complete analysis.

Level $p$ | Conjugacy class | $c$ | Hauptmodul construction
---|---|---|---
2 | 8C | 6 | $j_{2^{+}}^{\tfrac{1}{4}}$
4B | 12 | $j_{2^{+}}^{\tfrac{1}{2}}$
3 | 18b | 4 | $j_{3^{+}}^{\tfrac{1}{6}}$
9a | 8 | $j_{3^{+}}^{\tfrac{1}{3}}$
6b | 12 | $j_{3^{+}}^{\tfrac{1}{2}}$
5 | 10a | 12 | $j_{5^{+}}^{\tfrac{1}{2}}$
7 | 21C | 8 | $\left(j_{7^{+}}+13\right)^{\tfrac{1}{3}}$
13 | 39B | 8 | $\left(j_{13^{+}}+5\right)^{\tfrac{1}{3}}$
Table 8: This table lists all the admissible solutions to $n=2$ MLDEs, for
Fricke groups of prime divisor levels of $\mathbb{M}$, that possess an
unstable descendant. The Hauptmodules are constructed for higher-level groups
by using 411 as a guide and playing around with the exponent to obtain the
correct sequence. The central charges are found by reading off the value in
the exponent of the identity characters (Hauptmodul constructions here), i.e.
$j_{p^{+}}^{x}=q^{-\tfrac{c}{24}}(1+\ldots)$.

###  VIII.2 General formula for identity character with c-chronology

For the case of $\Gamma_{0}^{+}(2)$, we found in 214 and 289 that the identity
characters with ascending central charges in two- and three-character theories
can be expressed in terms of modular forms of the Hecke group $\Gamma_{0}(2)$.
We find the following general formula for all the identities one might expect
to find with ascending central charges in higher-character theories

$\displaystyle\begin{split}\chi^{(2^{+})}_{0}(\tau)=&\left(\mathfrak{p}_{2^{+}}\Delta_{2}^{-\tfrac{1}{24}}\right)^{c},\\\
\mathfrak{p}_{2^{+}}(\tau)\equiv&\left(\frac{\left(j_{2}\Delta_{2}^{\infty}\right)(3\tau)}{\left(j_{2}\Delta_{2}^{\infty}\right)^{\tfrac{1}{3}}(\tau)}\right)^{\tfrac{1}{8}}.\end{split}$
(412)

We can write down a similar expression for $c$-chronology in
$\Gamma_{0}^{+}(3)$ as follows

$\displaystyle\begin{split}\chi^{(3^{+})}_{0}(\tau)=&\left(\mathfrak{p}_{3^{+}}\Delta_{3}^{-\tfrac{1}{6}}\right)^{c},\\\
\mathfrak{p}_{3^{+}}(\tau)\equiv&\left(\frac{\Delta_{3}^{0}(2\tau)}{\left(\Delta_{3}^{0}\right)^{\tfrac{1}{2}}(\tau)}\right)^{\tfrac{1}{3}}.\end{split}$
(413)

Analysis of higher-character theories of $\Gamma_{0}^{+}(2)$ and
$\Gamma_{0}^{+}(3)$ would reveal more such solutions. The reason why only the
identity corresponding to a theory with a low central charge possesses this
nice closed-form expressed would be interesting to study further. On the other
hand, MLDE analysis of $\Gamma_{0}(2)$ and $\Gamma_{0}(3)$ could reveal some
of these solutions since the Fricke groups are super groups of Hecke groups.

###  VIII.3 From Fricke solutions to Schellekens’ list

Schellekens classified all $c=24$ single-character theories and found that
there is a total of $71$ CFTs, all of which possess a partition function of
the form Schellekens:1992db

$\displaystyle\begin{split}&Z(\tau)=J(\tau)+\mathcal{N},\ \ \ \ \
\mathcal{N}=12m,\\\
&m\in\left\\{0,2,3,4,5,6,7,8,9,10,12,13,14,16,18,20,22,24,25,26,28,30,32,34,38,46,52,62,94\right\\}\end{split}$
(414)

For $\Gamma_{0}^{+}(2)$, the partition function corresponding to the general
solution to $c=24$ single-character theories we have established in V.1.4 is
$Z_{2^{+}}(\tau)=j_{2^{+}}(\tau)+\mathcal{N}$, where $\mathcal{N}\geq-104$.
The only solutions we have come across are 258, 259, 260, • ‣ V.3, and the
$\ell=4$ solution with $\mathcal{N}=0$ found in Umasankar:2022kzs . This gives
us the list $\mathcal{N}\in\\{-80,-64,-32,0,15987\\}$. We understand the
relationship between Klein’s function, i.e. the Hauptmodul of the modular
group, and the Hauptmodul of the Fricke group of level $2$ which reads

$\displaystyle
j(\tau)=\frac{1}{2}\left[\left(j_{2^{+}}-81\right)\sqrt{j_{2^{+}}(256-j_{2^{+}})}+\left(j^{2}_{2^{+}}-207j_{2^{+}}+3456\right)\right](\tau).$
(415)

This helps us map $\mathcal{Z}_{2^{+}}(\tau)$ to $Z(\tau)$ for
$\mathcal{N}_{2^{+}}=0$ but does not act as a general map. It would be
interesting to complete the list and find a general map between the single-
character partition functions. Additionally, since we have the following
relations

$\displaystyle\begin{split}j_{2^{+}}(\tau)=&\left(j_{2}+4096j_{2}^{-1}+128\right)(\tau),\\\
j(\tau)=&\left(\frac{256+j_{2}}{j_{2}^{2}}\right)(\tau),\end{split}$ (416)

we can utilize these relationships to better understand the maps among the
partition functions post an exhaustive analysis of the Hecke group of level
$2$.

###  VIII.4 Hecke images

We obtain the following Hecke relations between some of our single-character
solutions using the Hecke operators constructed in Harvey:2018rdc . For
single-character theories, the conductor $N$ is computed by observing the
smallest positive integer $N$ such that $\left(e^{-\frac{i\pi
c}{12}}\right)^{N}=1$. Now, if,

$\displaystyle\chi=\sum\limits_{n\in\mathbb{Z}}a(n)q^{\frac{n}{N}},$ (417)

Then, acting the above with a Hecke operator $T_{p}$ where
$\text{gcd}(p,N)=1$, we get,

$\displaystyle
T_{p}(\chi)=\sum\limits_{n\in\mathbb{Z}}a^{(p)}(n)q^{\frac{n}{N}},$ (418)

where,

$\displaystyle
a^{(p)}(n)=\begin{cases}&pa(pn)\,\,\,\,\,\text{if}\,\,p\,\not|\hskip
2.0pt\,n\\\ &pa(pn)+a(\frac{n}{p})\,\,\,\,\,\text{if}\,\,p\,|\,n\end{cases}$
(419)

We report a few Hecke images of single-character solutions we found in Fricke
groups of levels $2$ and $3$ here.
$\mathbf{\Gamma^{+}_{0}(2)}$:
The conductor is $N=4$ for $j_{2^{+}}^{\frac{1}{4}}$.

$\displaystyle
T_{3}\left(j_{2^{+}}^{\frac{1}{4}}\right)=j_{2^{+}}^{\frac{3}{4}},$ (420)
$\displaystyle
T_{5}\left(j_{2^{+}}^{\frac{1}{4}}\right)=j_{2^{+}}^{\frac{1}{4}}\left(j_{2^{+}}-130\right),$
(421) $\displaystyle
T_{7}\left(j_{2^{+}}^{\frac{1}{4}}\right)=j_{2^{+}}^{\frac{3}{4}}\left(j_{2^{+}}-182\right).$
(422)

$\mathbf{\Gamma^{+}_{0}(3)}$:
The conductor is $N=3$ for $j_{3+}^{\frac{1}{3}}$.

$\displaystyle T_{2}\left(j_{3+}^{\frac{1}{3}}\right)=j_{3+}^{\frac{2}{3}},$
(423) $\displaystyle
T_{5}\left(j_{3+}^{\frac{1}{3}}\right)=j_{3+}^{\frac{2}{3}}\left(j_{3+}-70\right),$
(424) $\displaystyle
T_{7}\left(j_{3+}^{\frac{1}{3}}\right)=j_{3+}^{\frac{1}{3}}\left(j_{3+}^{2}-98j_{3+}+917\right).$
(425)

##  IX Future directions

There are various interesting future directions we reserve for upcoming
work(s) which we present briefly below while also summarizing the results of
this paper.

Are there new theories out in the wild?
We have obtained many single-, two-, and three-character solutions
corresponding to $\Gamma^{+}_{0}(2)$ and $\Gamma^{+}_{0}(3)$ congruence
subgroups of $\text{SL}(2,\mathbb{Z})$. Many of the single-character solutions
have nice interpretations with regard to the conjugacy classes of
$\mathbb{M}$. We noticed the reappearance of certain single-character
solutions aided by unstable descendants as solutions to $(2,\ell)$ and $(3,0)$
MLDEs which possess a neat description in terms of the Borcherds product.
However, these solutions themselves do not describe any CFTs “yet”. It would
be nice to see if these solutions, by themselves, could be related to some
known theories which have some restricted conformal symmetry such that their
corresponding partition functions are only modular (up to a phase if required)
with respect to a congruence subgroup rather than the full modular group. For
instance, we notice that many of our two-character solutions form nice
bilinear pairs with many other two-character solutions to give single-
character solutions. This phenomenon was very useful in identifying coset
pairs in the $\text{SL}(2,\mathbb{Z})$ case (see Gaberdiel:2016zke ). This
perhaps hints at the fact that there could be some nice coset relations even
for these solutions belonging to the congruence subgroups. In the CFT
literature, it is known that there exist some theories belonging to the Hecke
subgroup at level 2, for example, fermionic RCFTs (Bae:2021mej ), SCFTs and
parafermion theories Anderson:1987ge . So, it would be natural to expect that
there might be theories out in the wild for Fricke subgroups too.

Looking at the bulk and beyond for $\mathbf{\Gamma_{0}^{+}(2)}$ and
$\mathbf{\Gamma_{0}^{+}(3)}$.
For two-character and three-character analysis, we have analyzed the MLDEs and
their solutions for $\mathbf{\Gamma_{0}^{+}(2)}$ and
$\mathbf{\Gamma_{0}^{+}(3)}$ up to the bulk point. There exists a complete
classification (see Chandra:2018pjq ) of admissible two-character solutions in
the $\text{SL}(2,\mathbb{Z})$ case for $\ell\geq 6$ (note, $\ell=6$ is the
bulk point for $\text{SL}(2,\mathbb{Z})$). The way this classification works
is that any solution above the bulk point can be written down in terms of the
solutions below the bulk point. Performing an MLDE analysis beyond the bulk
point for these congruence subgroups might enable us to get a complete
classification of admissible solutions using the solutions obtained above
since these solutions are below the bulk point solutions.

Higher-character MLDEs for $\mathbf{\Gamma_{0}^{+}(2)}$ and
$\mathbf{\Gamma_{0}^{+}(3)}$?
It is natural to extend our analysis to four- and five-character MLDEs and
probe the landscape, at least at $\ell=0$, for admissible solutions
(reappearances of other lower-character solutions, minimal models, and of
genuine solutions). Additionally, its worth mentioning that in the case of
$\Gamma_{0}^{+}(2)$, all the minimal models possess central charges
$c\in\tfrac{6}{7}\times\\{1,5,9,13\\}$, where
$\tfrac{6}{7}=\tfrac{4}{7}\times\tfrac{3}{2}=c_{\mathcal{M}(7,2)}\cdot\mu_{0}^{+}$,
where $\mathcal{M}(7,2)$ is a three-character minimal model with
$(c,h_{1},h_{2})=\left(\tfrac{4}{7},\tfrac{5}{7},\tfrac{1}{7}\right)$. It
would be interesting to explore this relationship further and also to see if
such relations show up for minimal models at higher-character theories. The
equation would be simple to work out thanks to the ease of computation with
modular forms of Fricke levels $2$ and $3$. Additionally, it would be nice to
provide a general prescription for modular re-parameterization to
$n$-character theories for these levels (which should work out nicely since
the re-parameterization setup for these groups mimics the prescription done in
the case of $\text{SL}(2,\mathbb{Z})$).

What’s up with $\mathbf{\Gamma_{0}^{+}(5)}$ and $\mathbf{\Gamma_{0}^{+}(7)}$?
Although we found that $\Gamma_{0}^{+}(7)$ is quite hard to re-parameterize,
we should be able to find three-character solutions to it and for Fricke level
$5$. It would be interesting to see if analysis at $(3,0)$ could reveal some
interesting solutions. Firstly, we could check if the identity of a particular
three-character solution is expressible in terms of Hecke modular forms of
level $5$ and $7$ thereby enabling us to identify the $c$-chronology and
devise a general expression akin to 412 and 413. Secondly, we can explore if
we continue to find minimal models at these levels as we did at levels
$p=2,3$. Lastly, it would be interesting to see if genuine three-character
solutions exist at these levels.

What about other prime divisors levels of $\mathbf{\Gamma_{0}^{+}(p)}$?
the most approachable higher Fricke level would be $p=11$ since the space of
modular forms is well studied Junichi . Thus, An exhaustive analysis at Fricke
level $11$ for single-, two-, and three-character solutions should be possible
and we can ask the same questions of finding minimal models and the existence
of genuine three-character solutions here too. The space of modular forms
becomes complicated at levels $p\geq 13$ which unfortunately limits the ease
of calculations. Nevertheless, at least an exhaustive single-character and a
$(n,\ell)=(2,0)$ analysis for the remaining prime divisor levels could help us
establish some common themes among the solutions in $\Gamma_{0}^{+}(p)$.

What about Hecke groups?
The natural direction to follow post-Fricke would be to investigate the Hecke
groups. Taking indications from $c$-chronology, where the identity character
could be expressed in terms of Hecke group modular forms, it would be
interesting to check if a set of solutions found in our analysis for
$\Gamma_{0}^{+}(p)$ reappears in $\Gamma_{0}(p)$. this is natural to expect
since Fricke groups are supergroups of Hecke groups. From section III, we see
that we can associate multiple definitions of the Serre-Ramanujan covariant
derivative to a particular Hecke group. This could yield a plethora of two-
and three-character solutions at the same Wronskian index. It would be nice to
check if there exist certain single character solutions found with $(2,\ell)$
that possess unstable descendants. If this is the case, then there should also
exist a way to read off the (unstable) theories from a Borcherds product with
the Hecke Hauptmodul. Lastly, it would be especially interesting to see if we
can relate the admissible solutions to existing minimal models, coset
theories, or other unitary (S)CFTs.

Getting lucky with other congruent subgroups.
Why leave out congruent subgroups $\Gamma_{1}(N)$, the conjugate groups
$\Gamma^{0}(N)$ and $\Gamma^{1}(N)$, and the groups $\Gamma_{0}^{0}(N)$? We
could fully comprehend how admissible solutions are categorized in these
congruence subgroups with the aid of MLDE analysis, but more importantly, this
could provide us with new insights into the relationships of character
solutions among the subgroups and between the subgroup and modular group
characters. For example, the $S$-transform of the $\Gamma_{0}(7)$ Hauptmodul,
$j_{7}(\tau)$, is found to be
$j_{\Gamma_{0}(7)}(S(\tau))=49\left(\tfrac{\eta(\tau)}{\eta\left(\tfrac{\tau}{7}\right)}\right)^{4}=49j_{\Gamma^{0}(7)}^{-1}(\tau)$.
Suppose $j_{\Gamma_{0}(7)}(\tau)$ and $j_{\Gamma^{0}(7)}(\tau)$ are admissible
single-character solutions, then we see that analyzing just the Hecke
solutions can help us find the related conjugate solutions by mere taking the
$S$-transform. Also, Hecke images of $\Gamma_{0}(N)$ characters can be related
to those of $\Gamma^{0}(N)$ characters. There also exists other Atkin-Lehner
groups $\Gamma_{0}^{*}(N)$ studied in Junichi  that are desirable candidates
to pick for MLDE analysis.

A modular tower of theories.
To better comprehend the categorization of CFTs, a more systematic
investigation of the two- and three-character MLDE for the Fricke groups and
related modular towers, $\Gamma_{0}^{+}(p^{n+2})$ would be interesting. The
natural choice of candidates for first consideration would be the towers
$\Gamma_{0}^{+}(2)\to\Gamma_{0}^{+}(4)\to\Gamma_{0}^{+}(8)\to\ldots$ (and the
corresponding Hecke tower), that follows from the modular tower
$X_{0}(2^{n})$, and
$\Gamma_{0}^{+}(3)\to\Gamma_{0}^{+}(6)\to\Gamma_{0}^{+}(12)\to\ldots$ (and the
corresponding Hecke tower), that follows from the modular tower
$X^{+}_{0}(3\cdot 2^{n})$. In Umasankar:2022kzs , an admissible single-
character solution to the first levels of the modular tower
$\Gamma_{0}^{+}(7^{n})$ was found in terms of the parameterization in the
affine equation of the elliptic modular curve $X_{0}(49)$. Since
$\Gamma_{0}^{+}(25)$ is also a Conway-Norton ghost similar to
$\Gamma_{0}^{+}(49)$, it would be interesting to examine what occurs for level
$5$ Fricke groups.

Finding new lattices.
It was reported in McKay2000FuchsianGA  that the Schwarz derivative of the
Hauptmodules of genus-zero Hecke groups and their conjugates with no elliptic
elements can be expressed in terms of Eisenstein series or certain lattice
$A_{4}$ and $D_{4}$-lattice theta functions. The Schwarzian of the Fricke
level $2$ Hauptmodule is found to possess the following $q$-series expansion

$\displaystyle\begin{split}-\\{j_{2^{+}}(\tau),\tau\\}=&-\left(\frac{1}{\left(j_{2^{+}}^{\prime}\right)^{2}}\left(2j^{\prime}_{2^{+}}j^{\prime\prime\prime}_{2^{+}}-3\left(j^{\prime\prime}_{2^{+}}\right)^{2}\right)\right)(\tau)\\\
=&1+44040q^{2}+3792000q^{3}+536995320q^{4}+57168341760q^{5}+\mathcal{O}(q^{6}),\end{split}$
(426)

where all the coefficients were found to possess positive signs. A first step
to try reconstruction of this series would be to use the Jacobi theta-series
combination
$\tfrac{1}{256}\left(\theta_{3}(q^{4})+\theta_{4}(q^{4})\right)^{8}$ which has
a null-valued coefficient at order $\mathcal{O}(q)$. This ansatz, however,
does not turn out to yield the required series. Thus, it could be that the
series we are after is some combination of the level $2$ Eisenstein series or
it could be constructed out of other integer lattices. Finding such
Schwarzian-lattice relations for all Fricke levels would form an interesting
study. Additionally, one can also construct $\Theta$-series for the modular
invariant partition functions 217, 218, and 219 and explore if it can be
expressed as a combination of a McKay-Thompson series and a lattice
$\Theta$-series akin to what was done with $\Theta_{2^{+}}(\tau)$.

Exhaustive Hecke Relations.
We observed a few Hecke relations (Harvey:2018rdc ) among some of our single-
character solutions. The analysis done here in terms of Hecke images is far
from complete. One natural extension to the analysis provided here is to
compute Hecke images for two-character and three-characters solutions obtained
above. Since taking Hecke images of higher characters involves the modular $S$
and $T$ matrices, we can in principle ask the opposite question for our case
considered above. Say, we construct Hecke images of our two-character and
three-character solutions. We would end up with some characters from this
procedure. Now, using the knowledge of how Hecke operators act on characters
in the $\text{SL}(2,\mathbb{Z})$ case, perhaps we can determine the restricted
modular $S$ and $T$ matrices which would govern the transformations of these
congruence subgroups characters. We aim to return to this kind of Hecke image
analysis for other Fricke and Hecke subgroups in the future.

MTCs for congruence subgroups.
It is known that every RCFT, whose partition function by definition is modular
(up to a phase) with respect to the full modular group, belongs to some
Modular Tensor Category (MTC) rowell2009classification . Say by following the
procedure outlined in the above point, one can find the restricted modular $S$
and $T$ matrices for a given character-like solution. Then, this begs the
question of which MTC this modular data would belong to. If such a thing would
exist then we could find, in principle, MTCs to which our two- and three-
character solutions belong. This idea is also motivated by the fact that we
observe nice bilinear relations among many of our character-like solutions. In
the $\text{SL}(2,\mathbb{Z})$ case, whenever two RCFTs are related by a
bilinear relation, it meant that if one belonged to an MTC then the other, in
the pair, belonged to the conjugate MTC. Such things might also exist in the
case of congruence subgroups.

Kaneko-Zagier Equations.
It was noted in Chandra:2018pjq  that the following Kaneko-Zagier equation,

$\displaystyle\left(\mathcal{D}^{2}_{(k)}-\frac{k(k+2)}{144}E_{4}(\tau)\right)f_{(k)}(\tau)=0,$
(427)

can be recast into the usual $(2,0)$ MMS equation,

$\displaystyle\left(\mathcal{D}^{2}_{(0)}-\frac{k(k+2)}{144}E_{4}(\tau)\right)f_{(0)}(\tau)=0,$
(428)

by doing the following substitution,

$\displaystyle f_{(k)}(\tau)=\eta(\tau)^{2k}f_{(0)}(\tau),\qquad c=2k.$ (429)

We figured out that a similar Kaneko-Zagier type equation can be obtained in
the $\Gamma^{+}_{0}(2)$ case. We start with the following Kaneko-Zagier type
equation,

$\displaystyle\left(\mathcal{D}^{2}_{(k)}-\frac{k(k+2)}{64}E^{(2+)}_{4}(\tau)\right)f_{(k)}(\tau)=0.$
(430)

Now we observe that
$\theta_{q}(\eta(\tau)\eta(2\tau))=\tfrac{1}{8}E_{2}^{(2+)}(\tau)(\eta(\tau)\eta(2\tau))$
which follows from $\theta_{q}\Delta_{2}=E_{2}^{(2+)}\Delta_{2}$ (see 67). So,
consider the substitution,

$\displaystyle f_{(k)}=(\eta(\tau)\eta(2\tau))^{k}f_{(0)},\qquad c=3k.$ (431)

This then leads to (from 430),

$\displaystyle\left(\mathcal{D}^{2}_{(0)}-\frac{k(k+2)}{64}E_{4}(\tau)\right)f_{(0)}(\tau)=0,$
(432)

In Chandra:2018pjq , an equation of the type 427 was studied to classify
quasi-characters. Motivated by this, we are interested to study the equation
of the form 430 to classify quasi-characters that appear as solutions in the
$\Gamma^{+}_{0}(2)$ case. Furthermore, from a mathematical standpoint,
equations like 427 find applications in the study of Supersingular
j-invariants, hypergeometric series, and Atkin’s orthogonal polynomials (see
KZ ). It would be interesting to investigate a similar equation (obtained
above in 430) for the $\Gamma^{+}_{0}(2)$ case with regards to applications
similar to those done in KZ . One can in principle also set up Kaneko-Zagier
type equations for other Fricke groups.

###### Acknowledgements.

We thank Sunil Mukhi for the helpful discussions. Additionally, AD would like
to thank Chethan N. Gowdigere, Jagannath Santara and Jishu Das for insightful
discussions pertaining to MLDEs. AD would also like to thank Nabil Iqbal for
useful discussions on CFTs. AD would also like to express his gratitude to
Sigma Samhita for help in LaTeX-formatting. NU would like to thank Mikhail
Isachenkov for multiple discussions, and useful suggestions, and Miranda Cheng
and Erik Verlinde for early discussions.

##  Appendix A Modular forms for levels 5 & 7

$\mathbf{\Gamma_{0}^{+}(5)}:$

The Fricke group of level $5$ is generated by
$\Gamma_{0}^{+}(5)=\langle\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right),W_{5},\left(\begin{smallmatrix}&1\\\
5&2\end{smallmatrix}\right)\rangle$. The elliptic points in the fundamental
domain are at $\tfrac{i}{\sqrt{5}}$ (also known as the Fricke involution
point), $\rho_{5,1}=-\tfrac{1}{2}+\tfrac{i}{2\sqrt{5}}$, and
$\rho_{5,2}=\tfrac{-2+i}{5}$ and this group has a cusp at $\tau=i\infty$. A
non-zero $f\in\mathcal{M}_{k}^{!}(\Gamma_{0}^{+}(5))$ satisfies the following
valence formula

$\displaystyle\nu_{\infty}(f)+\frac{1}{2}\nu_{\tfrac{i}{\sqrt{5}}}(f)+\frac{1}{2}\nu_{\rho_{5,1}}(f)+\frac{1}{2}\nu_{\rho_{5,2}}(f)+\sum\limits_{\begin{subarray}{c}p\in\Gamma_{0}^{+}(5)\backslash\mathbb{H}^{2}\\\
p\neq\tfrac{i}{\sqrt{5}},\rho_{5,1},\rho_{5,2}\end{subarray}}\nu_{p}(f)=\frac{k}{4}.$
(433)

The Hauptmodul and modular forms of $\Gamma_{0}^{+}(5)$ read

$\displaystyle\begin{split}j_{5^{+}}(\tau)=&\left(\frac{\left(E_{2,5^{{}^{\prime}}}\right)^{2}}{\Delta_{5}}\right)(\tau)=\left(\frac{E_{4}^{(5^{+})}}{\Delta_{5}}\right)(\tau)-\frac{36}{13}=\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^{6}+5^{3}\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}+22,\\\
\Delta_{5}(\tau)=&\left(\eta(\tau)\eta(5\tau)\right)^{4}\in\mathcal{S}_{4}(\Gamma_{0}^{+}(5)),\\\
E_{2,5^{{}^{\prime}}}(\tau)=&\frac{5E_{2,5^{{}^{\prime}}}(5\tau)-E_{2}(\tau)}{4}\in\mathcal{M}_{2}(\Gamma_{0}(5)),\\\
E_{k}^{(5^{+})}(\tau)=&\frac{5^{\tfrac{k}{2}}E_{k}(5\tau)+E_{k}(\tau)}{5^{\tfrac{k}{2}}+1}\in\mathcal{M}_{k}(\Gamma_{0}^{+}(5)),\
\text{for}\ k\geq 4\ \&\ k\in 2\mathbb{Z}.\end{split}$ (434)

$\mathbf{\Gamma_{0}^{+}(7)}:$

The Fricke group of level $7$ is generated by
$\Gamma_{0}^{+}(7)=\langle\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right),W_{7},\left(\begin{smallmatrix}3&1\\\
7&2\end{smallmatrix}\right)\rangle$. The elliptic points in the fundamental
domain are at $\tfrac{i}{\sqrt{7}}$ (also known as the Fricke involution
point), $\rho_{7,1}=-\tfrac{1}{2}+\tfrac{i\sqrt{7}}{10}$, and
$\rho_{7,2}=\tfrac{-5+i\sqrt{3}}{14}$ and this group has a cusp at
$\tau=i\infty$. A non-zero $f\in\mathcal{M}_{k}^{!}(\Gamma_{0}^{+}(7))$
satisfies the following valence formula

$\nu_{\infty}(f)+\frac{1}{2}\nu_{\rho_{F}}(f)+\frac{1}{2}\nu_{k_{2}}(f)+\frac{1}{3}\nu_{\rho_{2}}(f)+\sum\limits_{\begin{subarray}{c}p\in\
X_{0}^{+}(7)\\\
p\neq\tfrac{i}{\sqrt{7}},\rho_{7,1},\rho_{7,2}\end{subarray}}\nu_{p}(f)=\frac{k}{3}.$
(435)

The Hauptmodul and some semi-modular and cusp forms in the Hecke group
$\Gamma_{0}(7)$ are defined as follows Umasankar:2022kzs

$\displaystyle\begin{split}j_{7}(\tau)=&\left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^{4},\\\
\theta_{7}(\tau)=&\sqrt{E_{2,7^{{}^{\prime}}}(\tau)}=\left(\frac{7E_{2}(7\tau)-E_{2}(\tau)}{6}\right)^{\frac{1}{2}},\\\
\mathbf{k}(\tau)=&\frac{\eta^{7}(\tau)}{\eta(7\tau)},\ \ \
\mathbf{t}(\tau)=\frac{1}{j_{7}(\tau)},\\\
\Delta_{7}(\tau)=&\left(\mathbf{k}\mathbf{t}\right)(\tau)\in\mathcal{S}_{3}(\Gamma_{0}(7))\end{split}$
(436)

We now express the Hauptmodul and some semi-modular and cusp forms in
$\Gamma_{0}^{+}(7)$ in terms of the definitions in 436

$\displaystyle\begin{split}j_{7^{+}}(\tau)=&\left(\frac{E_{1,7^{{}^{\prime}}}^{3}}{\Delta_{7}}\right)(\tau)=\left(\frac{E_{4}^{(7^{+})}}{\Delta_{7^{+},4}}\right)-\frac{16}{5}=\left(j_{7}+\frac{49}{j_{7}}+13\right)(\tau),\\\
E_{1,7^{{}^{\prime}}}(\tau)=&\sqrt{E_{2,7^{{}^{\prime}}}(\tau)}=\theta_{7}(\tau),\\\
\Delta_{7^{+},4}(\tau)=&\left(\theta_{7}\mathbf{t}\mathbf{k}\right)(\tau)=\left(\sqrt{E_{2,7^{{}^{\prime}}}}\Delta_{7}\right)(\tau)\in\mathcal{S}_{4}(\Gamma_{0}^{+}(7)),\\\
\Delta_{7^{+},10}(\tau)=&\frac{559}{690}\left(\frac{41065}{137592}\left(E^{(7^{+})}_{4}(\tau)E^{(7^{+})}_{6}(\tau)-E^{(7^{+})}_{10}(\tau)\right)-E^{(7^{+})}_{6}(\tau)\Delta_{7^{+},4}(\tau)\right)\\\
=&\left(\Delta_{7^{+},4}^{2}\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)(\tau)\in\mathcal{S}_{10}(\Gamma_{0}^{+}(7)),\\\
\Delta_{7^{+},6}(\tau)=&\left(\frac{\Delta_{7^{+},10}}{\Delta_{7^{+},4}}\right)(\tau)=\left(\Delta_{7^{+},4}\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)(\tau)\in\mathcal{S}_{6}(\Gamma_{0}^{+}(7)),\end{split}$
(437)

where

$\displaystyle\begin{split}E_{k,7^{{}^{\prime}}}(\tau)\equiv&\frac{7^{\tfrac{k}{2}}E_{2}(7\tau)-E_{k}(\tau)}{7^{\tfrac{k}{2}}-1},\
\text{for}\ k\in 2\mathbb{Z},\\\
E_{k}^{(7^{+})}(\tau)\equiv&\frac{7^{\tfrac{k}{2}}E_{k}(7\tau)+E_{k}(\tau)}{7^{\tfrac{k}{2}}+1},\
\text{for}\ k\geq 2\ \&\ k\in 2\mathbb{Z}.\end{split}$ (438)

##  Appendix B Modular re-parameterization for $\Gamma_{0}^{+}(5)$

From 159, we find

$\displaystyle
E_{4}^{(5^{+})}(\tau)=\left(1-\frac{36}{13}\frac{1}{j_{5^{+}}(\tau)}\right)E_{2,5^{{}^{\prime}}}^{2}(\tau).$
(439)

We now make the following definitions

$\displaystyle
A_{5^{+}}(\tau)\equiv\left(\frac{E_{6}^{(5^{+})}}{E_{2,5^{{}^{\prime}}}^{2}}\right)(\tau),\
\ \ P_{5^{+}}(\tau)\equiv\frac{36}{13}\frac{1}{j_{5^{+}}(\tau)},$ (440)

where we see that the definition of $A_{5^{+}}$ naturally follows from
observing the form of the derivative of the Hauptmodul. The derivative of
$P_{5^{+}}$ is found to be

$\displaystyle
q\frac{d}{dq}P_{5^{+}}(\tau)=\left(P_{5^{+}}A_{5^{+}}\right)(\tau).$ (441)

With this we find the following expressions for $\theta_{q}$ and
$\mathcal{D}^{2}$,

$\displaystyle\begin{split}\theta_{q}=&A_{5^{+}}\theta_{P_{5^{+}}},\\\
\mathcal{D}^{2}=&\left(\theta_{q}-\frac{1}{2}E_{2}^{(5^{+})}\right)\theta_{q}\\\
=&A_{5^{+}}^{2}\theta_{P_{5^{+}}}^{2}+\left[E_{2,5^{{}^{\prime}}}^{2}\left(\frac{1250}{81}P_{5^{+}}^{2}-\frac{116}{9}(1-P_{5^{+}})P_{5^{+}}-\frac{3}{2}(1-P_{5^{+}})^{2}\right)+\frac{A_{5^{+}}^{2}}{1-P_{5^{+}}}\right]\theta_{P_{5^{+}}}.\end{split}$
(442)

The modular forms $E_{6}^{(5^{+})}(\tau)$ and
$E_{2,5^{{}^{\prime}}}^{2}(\tau)$ can be expressed in terms of
$K_{5^{+}}\equiv\tfrac{22+10\sqrt{5}}{j_{5^{+}}}$, but the relationship is
more complicated unlike what we observed previously at levels $2$ and $3$.
These modular forms are related to the inverse Hauptmodul $K_{5^{+}}$ via Heun
functions as shown below

$\displaystyle
E_{6}^{(5^{+})}(\tau)=\sqrt{1-\frac{143}{9}P_{5^{+}}-\frac{169}{81}P^{2}_{5^{+}}}\
H\ell_{5}^{6}(\tau),\ \ \
E_{2,5^{{}^{\prime}}}^{2}(\tau)=H\ell_{5}^{4}(\tau).$ (443)

Substituting this in the definition 440, we set

$\displaystyle A_{5^{+}}(\tau)\equiv$
$\displaystyle\frac{1}{9}\sqrt{81-1287P_{5^{+}}-169P_{5^{+}}^{2}}\
H\ell_{5}^{2}(\tau)$ (444) $\displaystyle=$ $\displaystyle
f(P_{5^{+}})H\ell_{5}^{2}(\tau).$ (445)

With
$\omega_{4}(\tau)=\mu_{1}E_{2,5^{{}^{\prime}}}^{2}(\tau)+\mu_{2}\Delta_{5}(\tau)$,
the re-parameterized MLDE reads

$\displaystyle\begin{split}\left[\theta_{P_{5^{+}}}^{2}+\left(\frac{4345P_{5^{+}}-1602P_{5^{+}}-243}{162f(P_{5^{+}})}+\frac{1}{1-P_{5^{+}}}\right)\theta_{P_{5^{+}}}+\frac{36\mu_{1}-13\mu_{2}P_{5^{+}}}{36f(P_{5^{+}})}\right]f(\tau)=0.\end{split}$
(446)

##  Appendix C Useful relations in level 7 groups

Some useful expressions of the $q$-derivatives of forms in $\Gamma_{0}(7)$ and
$\Gamma_{0}^{+}(7)$ are listed below

$\displaystyle\begin{split}&q\frac{d}{dq}j_{7}(\tau)=-\left(j_{7}\theta_{7}^{2}\right)(\tau),\\\
&q\frac{d}{dq}\mathbf{t}(\tau)=\left(\mathbf{t}\theta_{7}^{2}\right)(\tau),\\\
&q\frac{d}{dq}\mathbf{k}(\tau)=\frac{7}{24}\mathbf{k}(\tau)\left(E_{2}(\tau)-E_{2}(7\tau)\right)\\\
&{}\ \ \ \ \ \ \ \ \ \ \
=\mathbf{k}(\tau)\left[E_{2}^{(7^{+})}-\left(\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)-E_{2,7^{{}^{\prime}}}\frac{13\mathbf{t}+98\mathbf{t}^{2}}{1+13\mathbf{t}+49\mathbf{t}^{2}}\right](\tau),\\\
&q\frac{d}{dq}\Delta_{7}(\tau)=\left(\Delta_{7}E_{2}^{(7^{+})}\right)(\tau)\\\
&q\frac{d}{dq}E_{2,7^{{}^{\prime}}}(\tau)=\frac{2}{3}\left(E_{2,7^{{}^{\prime}}}E_{2}^{(7^{+})}-E_{4,7^{{}^{\prime}}}\right)(\tau),\\\
&q\frac{d}{dq}E_{4,7^{{}^{\prime}}}(\tau)=\frac{19}{8}\left(\frac{7^{3}E_{2}(7\tau)E_{4}(7\tau)-E_{2}(\tau)E_{4}(\tau)}{342}-E_{6,7^{{}^{\prime}}}(\tau)\right),\\\
&q\frac{d}{dq}\Delta_{7^{+},4}(\tau)=\frac{4}{3}\Delta_{7^{+},4}(\tau)\left(E_{2}^{(7^{+})}-\frac{1}{4}\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)(\tau),\\\
&q\frac{d}{dq}\Delta_{7^{+},10}(\tau)=\Delta_{10,7^{{}^{\prime}}}(\tau)\left[2\left(E_{2}^{(7^{+})}-\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)(\tau)+q\frac{d}{dq}\log{E_{4,7^{\prime}}}(\tau)\right]\\\
&{}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=\Delta_{10,7^{{}^{\prime}}}(\tau)\left[2\left(E_{2}^{(7^{+})}-\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)(\tau)+\frac{57}{8E_{4,7^{{}^{\prime}}}(\tau)}q\frac{d}{dq}\left(7^{3}E_{4}(7\tau)-E_{4}(\tau)\right)\right],\end{split}$
(447)

where we used
$q\tfrac{d}{dq}E_{4}(\tau)=\tfrac{1}{3}\left(E_{2}E_{4}-E_{6}\right)(\tau)$ to
obtain the last line.

##  Appendix D Heun function relations in $\Gamma_{0}^{+}(7)$ re-
parameterization

We suspect that it should be possible to simplify the covariant derivative
$\mathcal{D}^{2}$ by expressing modular forms $E_{k,7^{{}^{\prime}}}(\tau)$ in
terms of those belonging to $\Gamma_{0}(7)$ since we know that the
$\text{SL}(2,\mathbb{Z})$ Eisenstein series can be expressed in terms of
$\mathbf{k}(\tau)$ and $\mathbf{t}(\tau)$ as follows Liu

$\displaystyle\begin{split}E_{4}(\tau)=&\mathbf{k}^{\tfrac{4}{3}}\left(1+245\mathbf{t}+2401\mathbf{t}^{2}\right)\left(1+13\mathbf{t}+49\mathbf{t}^{2}\right)^{\tfrac{1}{3}}(\tau),\\\
E_{4}(7\tau)=&\mathbf{k}^{\tfrac{4}{3}}\left(1+5\mathbf{t}+\mathbf{t}^{2}\right)\left(1+13\mathbf{t}+49\mathbf{t}^{2}\right)^{\tfrac{1}{3}}(\tau),\\\
E_{6}(\tau)=&\mathbf{k}^{2}\left(1-7(5+2\sqrt{7}\mathbf{t}-7^{3}(21+8\sqrt{7}))\mathbf{t}^{2}\right)\left(1-7(5-2\sqrt{7}\mathbf{t}-7^{3}(21-8\sqrt{7}))\mathbf{t}^{2}\right)(\tau),\\\
E_{6}(7\tau)=&\mathbf{k}^{2}\left(1+(7+2\sqrt{7})\mathbf{t}+(21+8\sqrt{7})\mathbf{t}^{2}\right)\left(1+(7-2\sqrt{7})\mathbf{t}+(21-8\sqrt{7})\mathbf{t}^{2}\right)(\tau).\end{split}$
(448)

From this, we see that the modular forms
$E_{2,7^{{}^{\prime}}}^{(7^{+})}(\tau)$, $E_{4}^{(7^{+})}(\tau)$,
$E_{4,7^{{}^{\prime}}}(\tau)$, and it’s logarithmic derivative can be
rewritten using 448 as follows

$\displaystyle\begin{split}E_{2,7^{{}^{\prime}}}(\tau)=&\mathbf{k}^{\tfrac{2}{3}}\left(1+13\mathbf{t}+49\mathbf{t}^{2}\right)^{\tfrac{2}{3}}(\tau),\\\
E_{4}^{(7^{+})}(\tau)=&\frac{1}{5}\mathbf{k}^{\tfrac{4}{3}}\left(5+49\mathbf{t}+245\mathbf{t}^{2}\right)\left(1+13\mathbf{t}+49\mathbf{t}^{2}\right)^{\tfrac{1}{3}}(\tau),\\\
E_{4,7^{{}^{\prime}}}(\tau)=&\mathbf{k}^{\tfrac{4}{3}}\left(1-49\mathbf{t}^{2}\right)\left(1+13\mathbf{t}+49\mathbf{t}^{2}\right)^{\tfrac{1}{3}}(\tau)\\\
q\frac{d}{dq}\log{E_{4,7^{{}^{\prime}}}}(\tau)=&\frac{4}{3}\left[E_{2}^{(7^{+})}-\left(\frac{E_{4,7^{{}^{\prime}}}}{E_{2,7^{{}^{\prime}}}}\right)-\frac{13\mathbf{t}+98\mathbf{t}^{2}}{1+13\mathbf{t}+49\mathbf{t}^{2}}E_{2,7^{{}^{\prime}}}\right](\tau)\\\
&-\left[\frac{\mathbf{t}\left(19208\mathbf{t}^{3}+4459\mathbf{t}^{2}+196\mathbf{t}-13\right)}{(1-49\mathbf{t}^{2})(1+13\mathbf{t}+49\mathbf{t}^{2})}\frac{E_{2,7^{{}^{\prime}}}}{3}\right](\tau)\end{split}$
(449)

Using these expressions, the covariant derivative $\mathcal{D}^{2}$ now reads

$\displaystyle\mathcal{D}^{2}=A_{7^{+}}^{2}\theta_{K_{7^{+}}}^{2}+A_{7^{+}}\left(\frac{4}{3}E_{2}^{(7^{+})}-\frac{13+196\mathbf{t}+637\mathbf{t}^{2}}{(1-49\mathbf{t}^{2})(1+13\mathbf{t}+49\mathbf{t}^{2})}E_{2,7^{{}^{\prime}}}\right)\theta_{K_{7^{+}}}.$
(450)

Now, since $\text{dim}\ \mathcal{M}_{4}(\Gamma_{0}^{+}(7))=2$, we make the
choice
$\omega_{4}(\tau)=\mu_{1}E_{2,7^{{}^{\prime}}}^{2}+\mu_{2}\Delta_{7}E_{1,7^{{}^{\prime}}}$.
From Sakai2014TheAO , we have the following expressions for the Eisenstein
series and the cusp form in terms of Heun’s function,

$\displaystyle\begin{split}\Delta_{7}(\tau)=&\frac{1}{j_{7^{+}}}H\ell_{7}^{3}(\tau),\\\
E_{2,7^{{}^{\prime}}}^{2}(\tau)=&H\ell_{7}^{4}(\tau),\\\
E_{6}^{(7^{+})}(\tau)=&\sqrt{\left(1+\frac{1}{j_{7^{+}}}\right)\left(1-\frac{27}{j_{7^{+}}}\right)}\
H\ell_{7}^{6}(\tau),\\\
H\ell_{7}(\tau)\equiv&H\ell\left(-27,-2;\frac{1}{3},\frac{2}{3},1,\frac{1}{2};K_{7^{+}}(\tau)\right).\end{split}$
(451)

We now want to find similar Heun function relations for other modular forms.
Let us begin with the cusp form $\Delta_{7^{+},4}(\tau)$. From its definition
in 437, we have

$\displaystyle\begin{split}\Delta_{7^{+},4}(\tau)=&\left(\sqrt{E_{2,7^{{}^{\prime}}}}\Delta_{7}\right)(\tau)=\frac{1}{j_{7^{+}}}H\ell_{7}^{5}(\tau).\end{split}$
(452)

Next, consider the Eisenstein series $E_{4}^{(7^{+})}(\tau)$ which is found to
possess the following Heun function relation

$\displaystyle
E_{4}^{(7^{+})}(\tau)=\left(E_{2,7^{{}^{\prime}}}^{2}-\frac{16}{5}\Delta_{7^{+},4}\right)(\tau)=\left(H\ell_{7}^{4}-\frac{16}{5j_{7^{+}}}H\ell_{7}^{5}\right)(\tau).$
(453)

Consider now the cusp form $\Delta_{7^{+},10}(\tau)$. From its definition in
437, we see that the Eisenstein series $E_{10}^{(7^{+})}(\tau)$ is the only
modular form we do not possess a Heun function relation of. Consider now the
basis decomposition for the space of modular forms of weight $10$ reads

$\displaystyle\mathcal{M}_{10}(\Gamma_{0}^{+}(7))=\mathbb{C}E_{4}^{(7^{+})}E_{6}^{(7^{+})}\oplus\mathbb{C}E_{4}^{(7^{+})}\Delta_{7^{+},6}\oplus\mathbb{C}\Delta_{7^{+},10}.$
(454)

Considering an ansatz with three unique coefficients and comparing its
$q$-series expansion with that of $E_{10}^{(7^{+})}(\tau)$, we can find an
explicit expression for the latter to be

$\displaystyle\begin{split}E_{10}^{(7^{+})}(\tau)=&\left(E_{4}^{(7^{+})}E_{6}^{(7^{+})}-\frac{137592}{41065}\frac{\Delta_{7^{+},4}}{E_{2,7^{{}^{\prime}}}}E_{4}^{(7^{+})}E_{4,7^{{}^{\prime}}}-\frac{63504}{4775}\frac{\Delta_{7^{+},4}^{2}}{E_{2,7^{{}^{\prime}}}}E_{4,7^{{}^{\prime}}}\right)(\tau)\\\
=&\left(E_{4}^{(7^{+})}E_{6}^{(7^{+})}-\frac{137592}{41065}\frac{\Delta_{7^{+},4}}{E_{2,7^{{}^{\prime}}}}\left(E_{4}^{(7^{+})}\right)^{2}\left(\frac{5(1-49\mathbf{t}^{2})}{(5+49\mathbf{t}+245\mathbf{t}^{2})}\right)\right.\\\
&{}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\left.-\frac{63504}{4775}\frac{\Delta_{7^{+},4}^{2}}{E_{2,7^{{}^{\prime}}}}E_{4}^{7^{+}}\left(\frac{5(1-49\mathbf{t}^{2})}{(5+49\mathbf{t}+245\mathbf{t}^{2})}\right)\right)(\tau),\end{split}$
(455)

where we have used expressions in 449 to express $E_{4,7^{{}^{\prime}}}(\tau)$
in terms of $E_{4}^{(7^{+})}(\tau)$. Using this in the definition of
$\Delta_{7^{+},10}(\tau)$, we find

$\displaystyle\Delta_{7^{+},10}(\tau)=$
$\displaystyle\left[\frac{559}{690}\left(\frac{5(j_{7}-49)}{245+49j_{7}+5j_{7}^{2}}-\sqrt{\left(1-\frac{1}{j_{7^{+}}}\right)\left(1-\frac{27}{j_{7^{+}}}\right)}\right)\frac{1}{j_{7^{+}}}H\ell_{7}^{11}\right.$
(456) $\displaystyle\ \ \ \ \ \ \ \ \
\left.-\frac{5(j_{7}-49)}{245+49j_{7}+5j_{7}^{2}}\left(\frac{688}{345j_{7^{+}}^{3}}h\ell_{7}^{13}+\frac{3397}{1725j_{7^{+}}^{2}}H\ell_{7}^{12}\right)\right](\tau),$
(457)

where we have expressed all $\mathbf{t}(\tau)$ in terms of $j_{7}(\tau)$, the
Hauptmodul of $\Gamma_{0}(7)$. Using these relations, it should be possible to
express the re-parameterized MLDE in terms of the Hauptmodul
$j_{7^{+}}(\tau)$. From this, we find the following useful relations

$\displaystyle\begin{split}\left(\frac{\Delta_{7}^{4}}{E_{2,7^{{}^{\prime}}}^{6}}\right)(\tau)=&\mathfrak{a}=\frac{1}{j_{7^{+}}(\tau)},\\\
\left(\frac{\left(E_{6}^{(7^{+})}\right)^{2}}{\Delta_{7}^{2}}\right)(\tau)=&\mathfrak{b}=\left(j_{7^{+}}^{2}-26j_{7^{+}}-27\right)(\tau),\\\
\left(\frac{\left(E_{6}^{(7^{+})}\right)^{2}}{E^{6}_{2,7^{{}^{\prime}}}}\right)(\tau)=&\mathfrak{c}=\left(1-\frac{27}{j_{7^{+}}^{2}}-\frac{26}{j_{7^{+}}}\right)(\tau).\end{split}$
(458)

##  Appendix E Group data

$p$ | 2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 41 | 59 | 71
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\overline{\mu}_{0}^{+}$ | $\tfrac{3}{2}$ | 2 | 3 | 4 | 6 | 7 | 9 | 10 | 12 | 15 | 16 | 21 | 30 | 36
Table 9: Index data for Fricke groups of prime divisor levels of $\mathbb{M}$ with $\overline{\mu}_{0}^{+}=\left[\text{PSL}(2,\mathbb{Z}),\overline{\Gamma}_{0}^{+}(p)\right]$ $p$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 17 | 19 | 23 | 29 | 31 | 41 | 59 | 71
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\overline{\mu}_{0}$ | 3 | 4 | 6 | 6 | 12 | 8 | 12 | 12 | 18 | 12 | 24 | 14 | 18 | 20 | 24 | 30 | 32 | 42 | 60 | 72
Table 10: Index data for Fricke groups of prime divisor levels of $\mathbb{M}$
with
$\overline{\mu}_{0}=\left[\text{PSL}(2,\mathbb{Z}),\overline{\Gamma}_{0}(p)\right]$

##  Appendix F $(2,0)\ \Gamma_{0}^{+}(2)$ MLDE specifics

Substituting mode expansions in the MLDE, we obtain

$\displaystyle
q^{\alpha}\sum\limits_{n=0}^{\infty}f_{n}q^{n}\left[(n+\alpha)^{2}-(n+\alpha)\sum\limits_{m=0}^{\infty}\frac{1}{4}E^{(2^{+})}_{2,m}q^{m}+\mu\sum\limits_{m=0}^{\infty}E_{4,m}^{(2^{+})}q^{m}\right]=0.$
(459)

When $n=0,m=0$, with $E^{(2^{+})}_{2,0}=E^{(2^{+})}_{4,0}=1$, we obtain the
following indicial equation

$\displaystyle\alpha^{2}-\frac{1}{4}\alpha+\mu=0.$ (460)

The parameter $\mu$ can now be fixed using this equation. Let us denote the
roots of this equation, in increasing order, as $\alpha_{0}$ and $\alpha_{1}$.
Solving this indicial equation gives us

$\displaystyle\begin{split}\alpha_{0}=&\frac{1}{8}\left(1-\sqrt{1-64\mu}\right)\equiv\frac{1}{8}(1-x),\\\
\alpha_{1}=&\frac{1}{12}\left(1+\sqrt{1-64\mu}\right)\equiv\frac{1}{8}(1+x),\end{split}$
(461)

where we have set $x=\sqrt{1-64\mu}$. The smaller solution
$\alpha_{0}=\tfrac{1}{8}(1-x)$ corresponds to the identity character which
behaves as $f(\tau)\sim q^{\tfrac{1-x}{8}}\left(1+\mathcal{O}(q)\right)$. We
know that the identity character $\chi_{0}$, associated with a primary of
weight $h_{0}=0$ behaves as $\chi_{0}\sim
q^{-\tfrac{c}{24}}\left(1+\mathcal{O}(q)\right)$. Comparing the two
behaviours, we obtain the following expression for the central charge

$\displaystyle c=3(x-1).$ (462)

To find the conformal dimension $h$, we compare the behaviours with the larger
solution for $\alpha$, i.e. $f(\tau)\sim
q^{\tfrac{1+x}{8}}\left(1+\mathcal{O}(q)\right)$ and $\chi\sim
q^{-\tfrac{c}{24}+h}\left(1+\mathcal{O}(q)\right)$. This gives us

$\displaystyle h=\frac{x}{4}.$ (463)

Next, to obtain a recurrence relation among the coefficients $f_{n}$, we use
the Cauchy product of two infinite series which reads

$\displaystyle\left(\sum\limits_{i=0}^{\infty}\alpha_{i}\right)\cdot\left(\sum\limits_{j=0}^{\infty}\beta_{j}\right)=\sum\limits_{k=0}^{\infty}\gamma_{k},\
\gamma_{k}=\sum\limits_{p=0}^{k}\alpha_{p}\beta_{k-p}.$ (464)

This gives us

$\displaystyle\begin{split}\left(\sum\limits_{m=0}^{\infty}E^{(2^{+})}_{4,m}(\tau)q^{m}\right)\left(\sum\limits_{n=0}^{\infty}f_{n}q^{n}\right)=&\sum\limits_{k=0}^{\infty}\left(\sum\limits_{p=0}^{k}E^{(2^{+})}_{4,m}f_{k-p}\right)q^{k},\\\
\left(\sum\limits_{m=0}^{n}E^{(2^{+})}_{2,m}q^{m}\right)\left(\sum\limits_{n=0}^{\infty}f_{n}q^{n}(n+\alpha)\right)=&\sum\limits_{k=0}^{\infty}\left(\sum\limits_{p=0}^{k}E^{(2^{+})}_{2,p}(k+\alpha-p)f_{k-p}\right)q^{k}.\end{split}$
(465)

Substituting this back into the MLDE, we get

$\displaystyle
f_{n}=\left((n+\alpha)^{2}-\frac{1}{4}(n+\alpha)+\mu\right)^{-1}\sum\limits_{m=1}^{n}\left(\frac{(n+\alpha-m)}{4}E^{(2^{+})}_{2,m}-\mu
E^{(2^{+})}_{4,m}\right)f_{n-m}.$ (466)

When $n=1$, we solve for the ratio $\tfrac{f_{1}}{f_{0}}$ in terms of $\alpha$
with coefficients $E^{(2^{+})}_{2,1}=-8$ and $E_{4,1}^{(2^{+})}=48$ to obtain

$\displaystyle
m^{(i)}_{1}=\frac{f^{(i)}_{1}}{f^{(i)}_{0}}=\frac{8\alpha_{i}(-7+24\alpha_{i})}{3+8\alpha_{i}},$
(467)

for $i=0,1$ corresponding to ratios of $m^{(0)}_{1}$ and $m^{(1)}_{1}$ taken
with respect to values $\alpha_{0}$ and $\alpha_{1}$ respectively. Restricting
to $i=0$ and assuming that $\alpha_{i}=-\tfrac{c}{24}$, we find

$\displaystyle m_{1}^{(0)}=\frac{c(7+c)}{9-c}.$ (468)

Dropping the script associated with the index $i$, we see that for $m_{1}\geq
0$, we require $c<9$. Rewriting this, we have

$\displaystyle c^{2}+c(m_{1}+7)=9m_{1}.$ (469)

This tells us that $c$ is an integer. Hence, for two character theories with
$\ell=0$, we have $c\in\mathbb{Z}$ and the bound $c<9$. Solving the quadratic
equation, we notice that for $c$ to be rational, the following determinant has
to be rational

$\displaystyle\sqrt{49+50m_{1}+m_{1}^{2}}.$ (470)

Now, since we require $m_{1}\in\mathbb{Z}$, we must demand that the square
root is an integer and this gives us

$\displaystyle\left(m_{1}+25\right)^{2}-576=k^{2},$ (471)

where $k^{2}\in\mathbb{Z}$. We define $p$ such that we shift $k$ by an integer
amount that absorbs the first two terms on the left-hand side

$\displaystyle p=25+m_{1}-k,$ (472)

and we recast 471 as follows

$\displaystyle m_{1}+25=\frac{576+p^{2}}{2p}=\frac{288}{p}+\frac{p}{2}.$ (473)

This tells us that $p$ must be even and that it must divide $288$. Restricting
$k$ to be positive, we see that we have $k\geq m_{1}$ which implies that
$p<25$. We conclude that all possible values of $m_{1}$ are found by those
values of $p$ below $26$ that divide $288$ and are even. The list of these
values is $p=\\{2,4,6,8,12,16,18,24\\}$. We note that at $p=24$, we obtain
$(m_{1},c,h)=(-1,-3,0)$ which we ignore since we only want unitary theories.
Table 11 contains CFT data.

$p$ | $\mu$ | $m_{1}$ | $c$ | $h$
---|---|---|---|---
2 | $-\tfrac{7}{36}$ | 120 | 8 | $\tfrac{11}{12}$
4 | $-\tfrac{91}{576}$ | 49 | 7 | $\tfrac{5}{6}$
6 | $-\tfrac{1}{8}$ | 26 | 6 | $\tfrac{3}{4}$
8 | $-\tfrac{55}{576}$ | 15 | 5 | $\tfrac{2}{3}$
12 | $-\tfrac{3}{64}$ | 5 | 3 | $\tfrac{1}{2}$
16 | $-\tfrac{7}{576}$ | 1 | 1 | $\tfrac{1}{3}$
18 | $0$ | 0 | 0 | $\tfrac{1}{4}$
Table 11: $c$ and $h$ data corresponding to the Fricke group
$\Gamma_{0}^{+}(2)$ for the choice $\phi_{0}=\mu E^{(2^{+})}_{4}(\tau)$ with
$\ell=0$.

The ratio $m_{1}$ being non-negative is not sufficient proof to convince
ourselves that this data might indeed be related to a CFT. Hence, we compute
$m_{2}$ using recursion relation 466. If this turns out to be negative or
fractional, then we rule out the theory as a viable candidate. When $n=2$, we
get

$\displaystyle
m_{2}^{(i)}\equiv\frac{f_{2}^{(i)}}{f_{0}^{(i)}}=\frac{4\alpha_{i}(312\alpha_{i}-83)}{8\alpha_{i}+7}+\frac{4m_{1}^{(i)}(-1+\alpha_{i}(-7+24\alpha_{i}))}{8\alpha_{i}+7}.$
(474)

The values of $m_{2}$ for $i=0$ are tabulated in table 12.

$p$ | $c$ | $m_{2}$
---|---|---
2 | 8 | $\tfrac{6508}{13}$
4 | 7 | 173
6 | 6 | 79
8 | 5 | 40
12 | 3 | 11
16 | 1 | 2
18 | 0 | 0
Table 12: Values of $m_{2}$ for corresponding to the Fricke group
$\Gamma_{0}^{+}(2)$ for the choice $\phi_{0}=\mu E^{(2^{+})}_{4}(\tau)$ with
$\ell=0$.

We discard the case $p=2$ since the coefficient turns out to be fractional.
Furthermore, we also discard the case $p=18$ since further computation yields
all coefficients $m_{i}$ to be null valued which makes this a trivial
solution. We have now restricted to cases $p=\\{4,6,8,12,16\\}$. Next, we
compute $m_{3}$ using recursion relation 466 and this turns out to be

$\displaystyle
m_{3}^{(i)}\equiv\frac{f_{3}^{(i)}}{f_{0}^{(i)}}=\frac{32\alpha_{i}(168\alpha_{i}-43)}{24\alpha_{i}+33}+\frac{8m_{1}^{(i)}(-5+\alpha_{i}(-83+312\alpha_{i}))}{24\alpha_{i}+33}+\frac{8m_{2}^{(i)}(-2+\alpha_{i}(-7+24\alpha_{i}))}{24\alpha_{i}+33}.$
(475)

The integral values of $m_{3}$ for $i=0$ are shown in table 13.

$p$ | $c$ | $m_{2}$
---|---|---
6 | 6 | 326
8 | 5 | 135
12 | 3 | 20
16 | 1 | 1
Table 13: Values of $m_{3}$ for corresponding to the Fricke group
$\Gamma_{0}^{+}(2)$ for the choice $\phi_{0}=\mu E^{(2^{+})}_{4}(\tau)$ with
$\ell=0$.

In a similar fashion, computing coefficients for $i=0,1$, we obtain the
coefficients of the two characters.

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