integral so ${\rm pr}_{x}$ is integral. Since $\mathfrak{X}(N)$ is an integral
scheme, we have inclusions

$A\hookrightarrow{\cal O}_{\mathfrak{X}(N),{\rm pr}(x)}\hookrightarrow{\rm
Frac}(A).$

Therefore $K={\rm Frac}(A)$. Similarly, we conclude $L={\rm Frac}(B).$ Both
$\mathfrak{X}(N)$ and $\mathfrak{X}$ are normal schemes so ${\cal
O}_{\mathfrak{X}(N),{\rm pr}(x)}$ (resp. ${\cal O}_{\mathfrak{X},x}$) is
integrally closed in $K$ (resp. $L$). Having established

${\rm pr}_{x}:{\cal O}_{\mathfrak{X}(N),{\rm pr}(x)}\rightarrow{\cal
O}_{\mathfrak{X},x}$

is integral, the integral closure of ${\cal O}_{\mathfrak{X}(N),{\rm pr}(x)}$
in $L$ is precisely ${\cal O}_{\mathfrak{X},x}$.

Since ${\rm pr}$ is a finite map between two integral schemes, the extension
of function fields

$K(\mathfrak{X}(N))\rightarrow K(\mathfrak{X})$

is a finite extension of characteristic zero fields, and hence separable. Note
that $K(\mathfrak{X}(N))=K$ and $K(\mathfrak{X})=L$ so $L/K$ is a finite
separable extension. By [Ser79, I, §IV, Prop. 8], we can conclude ${\rm
pr}_{x}$ is finite. ∎

Since $\mathfrak{X}$ is integral, it’s only associated point is its unique
generic point. In Proposition 4.11, we deduced the map on function fields
$K(\mathfrak{X})\rightarrow K(\mathfrak{X}(N))$ is finite separable so ${\rm
pr}$ is étale at the generic point of $\mathfrak{X}$. By Proposition 4.10, we
have

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal O}_{\mathfrak{X}}([{\cal
R}])={\cal O}_{\mathfrak{X}}\left(\sum_{\begin{subarray}{c}x\in\mathfrak{X}\\\
\dim{\cal O}_{\mathfrak{X},x}=1\end{subarray}}{\rm mult}_{x}({\cal
R})\cdot\overline{\left\\{x\right\\}}\right)$ (17)

Since ${\cal O}_{\mathfrak{X},x}$ is a DVR in this case, ${\rm mult}_{x}({\cal
R})$ is equal to the valuation of ${\cal R}_{x}=\mathfrak{D}_{{\rm pr},x}$ in
${\cal O}_{\mathfrak{X},x}$. By Corollary 4.9 and Proposition 4.11,
$\mathfrak{D}_{{\rm pr},x}=\mathfrak{D}_{{\cal O}_{\mathfrak{X},x}/{\cal
O}_{\mathfrak{X}(N),{\rm pr}(x)}}$. Also, by Proposition 4.6(b), we can
identify $\mathfrak{D}_{{\rm pr},x}$ with

${\rm Ann}_{{\cal O}_{\mathfrak{X},x}}(\Omega_{{\cal O}_{\mathfrak{X},x}/{\cal
O}_{\mathfrak{X}(N),{\rm pr}(x)}}^{1}).$

Let $d_{x}$ denote the valuation of the different $\mathfrak{D}_{{\rm pr},x}$
in the DVR ${\cal O}_{\mathfrak{X},x}$. Recall by [Liu02, 8.3.4] the
codimension 1 points of $\mathfrak{X}$ are precisely the closed points of the
generic fiber, and the generic points of the special fiber. We will compute
$d_{x}$ when $x$ is a closed point of the generic fiber. If $x$ is a generic
point of the special fiber, then we show all $d_{x}$ contributions are the
same which taken together contribute nothing to the different of ${\rm pr}$.
However, one can compute the $d_{x}$ explicitly using strict Henselizations.

###  4.3 Computing $d_{x}$ for the closed points of the generic fiber

The generic fiber of $\mathfrak{X}$ is open so it suffices to compute $d_{x}$
over $\mathbb{Q}_{p}(\zeta_{Np^{r}})$ i.e. for the map

${\rm pr}_{x}:{\cal O}_{\mathfrak{X}(N)_{/\mathbb{Q}_{p}(\zeta_{Np^{r}})},{\rm
pr}(x)}\rightarrow{\cal O}_{\mathfrak{X},x}.$

First we show the value of $d_{x}$ does not change after base changing
$\mathbb{Q}_{p}(\zeta_{Np^{r}})$. In particular, we can compute the value of
$d_{x}$ over $\mathbb{C}$ and use the classical theory of modular curves as
compact Riemann surfaces.

######  Lemma 4.12.

Let $\pi:X\rightarrow Y$ be a finite type morphism of normal curves over a
field $K$ and let $L$ be a field extension of $K$. Let $p:X_{L}\rightarrow X$
denote the usual projection morphism from base change and let $x\in X_{L}$.
Then we have $d_{x}=d_{p(x)}$.

###### Proof.

We can equate $d_{x}$ with the valuation of the annihilator ideal of
$\Omega_{X_{L}/Y_{L},x}^{1}$ in ${\cal O}_{X_{L},x}$. Since Kahler
differentials are compatible with base change, we have

$\Omega_{X_{L}/Y_{L},x}^{1}\simeq(p^{*}\Omega_{X/Y}^{1})_{x}\simeq\Omega_{X/Y,p(x)}^{1}\otimes_{{\cal
O}_{X,p(x)}}{\cal O}_{X_{L},x}.$

Since $K\rightarrow L$ is flat, the map ${\cal O}_{X,p(x)}\rightarrow{\cal
O}_{X_{L},x}$ is flat. Furthermore, $\Omega_{X/Y,p(x)}^{1}$ is finite over
${\cal O}_{X,p(x)}$ since $\pi$ is finite type. Thus, by Lemma 2.5, we have

$\displaystyle{\rm Ann}_{{\cal O}_{X_{L}},x}(\Omega_{X_{L}/Y_{L},x}^{1})$
$\displaystyle={\rm Ann}_{{\cal
O}_{X_{L}},x}(\Omega_{X/Y,p(x)}^{1}\otimes_{{\cal O}_{X,p(x)}}{\cal
O}_{X_{L},x})$ $\displaystyle={\rm Ann}_{{\cal
O}_{X},p(x)}(\Omega_{X/Y,x}^{1}){\cal O}_{X_{L},x}.$

Hence $d_{x}=d_{p(x)}$. ∎

Thus we can compute $d_{x}$ in the situation that our modular curves are over
$\mathbb{C}$. Let $X(M)$ denote the modular curve over $\mathbb{C}$ of level
$\Gamma(M)$. In this situation, ${\cal O}_{X(Np^{r}),x}$ is tamely ramified
over ${\cal O}_{X(N),{\rm pr}(x)}$ so by Proposition 4.6(a), we have
$d_{x}=e_{x}-1$ where $e_{x}$ is the ramification index of $x$. We will
investigate the ramification index of all points in $X(Np^{r})$ under the
usual projection map ${\rm pr}:X(Np^{r})\rightarrow X(N)$.

Let $\Gamma\leq{\rm SL}_{2}(\mathbb{Z})$ be a congruence subgroup. We also let
${\cal H}$ denote the (complex) upper half plane and ${\cal H}^{*}={\cal
H}\cup\mathbb{P}^{1}(\mathbb{Q})$. As we will see shortly, investigating the
ramification of a point $z\in{\cal H}^{*}$ amounts to understanding the
stabilizer group

$\Gamma_{z}=\left\\{\gamma\in\Gamma:\gamma\cdot z=z\right\\}.$

The points of $z\in{\cal H}^{*}$ can be classified depending on which elements
of $\Gamma$ fix $z$. Note that

$\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\cdot z=z\iff
cz^{2}+(d-a)z-b=0$

for matrices not equal to $\pm I$. Hence the fixed points of a given
$\gamma\in\Gamma$ are either conjugate complex numbers, a single real number,
or two distinct real numbers. Based on this fact, the following definition
from [Miy89, §1.3, p.27].

######  Definition 4.13.

We say $z\in{\cal H}^{*}$ is a/an

  * •

elliptic point of $\Gamma$ if there exists $\gamma\in\Gamma_{z}$ such that
$\left|{\rm tr}(\gamma)\right|<2$, or equivalently $\gamma$ has two distinct
fixed points $z$ and $\bar{z}$.

  * •

cusp of $\Gamma$ if there exists $\gamma\in\Gamma_{z}$ such that $\left|{\rm
tr}(\gamma)\right|=2$, or equivalently $\gamma$ has a unique real fixed point
$z$.

  * •

hyperbolic point of $\Gamma$ if there exists $\gamma\in\Gamma_{z}$ such that
$\left|{\rm tr}(\gamma)\right|>2$, or equivalently $\gamma$ has two distinct
real fixed points.

  * •

ordinary point of $\Gamma$ if $\Gamma$ does not fix $z$, excluding $\pm I$.

Since the matrices we consider are in ${\rm SL}_{2}(\mathbb{Z})$, we will not
have any hyperbolic points appearing. We denote
$\bar{\Gamma}=\Gamma/\Gamma\cap\left\\{\pm I\right\\}$. The following is
[Miy89, Theorem 1.5.4] which describes the stabilizer groups $\Gamma_{z}$.

######  Theorem 4.14.

  1. a.

If $z\in{\cal H}$ is an elliptic point of $\Gamma$, then $\Gamma_{z}$ is a
finite cyclic group.

  2. b.

If $z\in\mathbb{Q}\cup\\{\infty\\}$ is a cusp of $\Gamma$, then
$\bar{\Gamma}_{z}\simeq\mathbb{Z}$.

The following proposition, which is [Shi94, Prop. 1.37], relates the
ramification index with the index of stabilizer groups.

######  Proposition 4.15.

Let $\Gamma^{\prime}\leq\Gamma$ be a finite index subgroup and consider the
projection $\pi:\Gamma^{\prime}\backslash{\cal
H}^{*}\rightarrow\Gamma\backslash{\cal H}^{*}$. The ramification index $e_{z}$
of a point $z\in\Gamma^{\prime}\backslash{\cal H}^{*}$ under $\pi$ is equal to

$e_{z}=\left[\bar{\Gamma}_{z}:\bar{\Gamma^{\prime}}_{z}\right].$

We are now ready to compute the ramification index under our map ${\rm pr}$.

######  Proposition 4.16.

Let ${\rm pr}:X(Np^{r})\rightarrow X(N)$ denote the usual projection map and
let $z\in X(Np^{r})$. We have

$e_{z}=\begin{cases}p^{r}&\mbox{if }z\mbox{ is a cusp}\\\
1&\mbox{otherwise}\end{cases}.$

###### Proof.

By [Shi94, Prop. 1.39], the congruence subgroup $\Gamma(M)$ has no elliptic
elements for any $M>1$. Consequently, the ordinary points are precisely the
points of $\Gamma(M)\backslash{\cal H}$. We will now apply Proposition 4.15 in
the case $\Gamma^{\prime}=\Gamma(Np^{r})$ and $\Gamma=\Gamma(N)$.

Note that the image of a cusp under ${\rm pr}$ is again a cusp and similarly
for ordinary points. If $z\in X(Np^{r})$ is ordinary, then its stabilizer
group is trivial so $e_{z}=1$. If $z$ is a cusp, then there exists
$\gamma\in{\rm SL}_{2}(\mathbb{Z})$ such that $\gamma\cdot z=\infty$.
Therefore

$\overline{\Gamma(N)}_{z}/\overline{\Gamma(Np^{r})}_{z}\simeq\gamma\overline{\Gamma(N)}_{z}\gamma^{-1}/\gamma\overline{\Gamma(Np^{r})}_{z}\gamma^{-1}=\overline{\Gamma(N)}_{\infty}/\overline{\Gamma(Np^{r})}_{\infty}$

so it suffices to compute $e_{\infty}$.

Let $A(d)=\left(\begin{array}[]{cc}1&d\\\ 0&1\end{array}\right).$ According to
[Shi94, bottom p. 22], we have for $M>1$ that
$\overline{\Gamma(M)}_{\infty}=\left\langle A(d)\right\rangle.$ Furthermore,
$-I\in\Gamma(M)$ if and only if $-1\equiv 1$ modulo $M$, or equivalently
$M=2$. Since $M\geq 3$ in our situation, we always have
$\overline{\Gamma(M)}_{\infty}=\Gamma(M)_{\infty}$. Note that for any $m\geq
0$, we have $A(d)^{m}=A(md)$. Therefore the order of $A(d)$ in
$\overline{\Gamma(N)}_{\infty}/\overline{\Gamma(Np^{r})}_{\infty}$ is equal to
$p^{r}$ so

$e_{\infty}=[\overline{\Gamma(N)}_{\infty}:\overline{\Gamma(Np^{r})}_{\infty}]=p^{r}.\qed$

###  4.4 Relating the modular sheaf with $\Omega_{\mathfrak{X}/R}$

Let $x\in\mathfrak{X}$ be a generic point of the special fiber. Recall the
value of $d_{x}$ is equal to the valuation of the different ideal
corresponding to the induced map on stalks

${\rm pr}_{x}:{\cal O}_{\mathfrak{X}(N),{\rm pr}(x)}\rightarrow{\cal
O}_{\mathfrak{X},x}.$

We will use the discussion in Section A.1 to provide an explicit description
of ${\rm pr}_{x}$. Recall in Theorem A.21 we have a commutative diagram

$\textstyle{{\rm
Ig}(p^{r},N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{\rho}$$\textstyle{{\rm
ExIg}(p^{r},i,N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{\prime}}$$\textstyle{\mathfrak{X}(N)_{\mathbb{F}_{q}}^{(\sigma^{-i})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F^{i}}$$\textstyle{\mathfrak{X}(N)_{\mathbb{F}_{q}}}$

where $\rho$ and $\rho^{\prime}$ are the usual projection maps. By Theorem
A.25, the restriction of ${\rm pr}$ to any irreducible component of
$\bar{\mathfrak{X}}$ is the map $\rho^{\prime}$. We get a commutative diagram:

$\displaystyle\begin{gathered}\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
24.71025pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{{\cal O}_{\mathfrak{X}(N)_{/\mathbb{F}_{q}},{\rm
pr}(x)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
29.3562pt\raise 5.82222pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.82222pt\hbox{$\scriptstyle{F^{r}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 48.71025pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
0.0pt\raise-20.52887pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.8264pt\hbox{$\scriptstyle{{\rm
pr}_{x}}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern
0.0pt\raise-31.22444pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
48.71025pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{{\cal
O}_{\mathfrak{X}(N)_{/\mathbb{F}_{q}}^{(p^{-r})},{\rm
pr}(x)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
72.10176pt\raise-20.52887pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{{\rm\rho}_{x}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
72.10176pt\raise-31.22444pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\cal
O}_{{\rm
ExIg}(p^{r},r,N),x}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
30.32864pt\raise-36.43463pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.62312pt\hbox{$\scriptstyle{\simeq}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
52.7381pt\raise-41.05775pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
52.7381pt\raise-41.05775pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{{\cal O}_{{\rm
Ig}(p^{r},N),x}}$}}}}}}}\ignorespaces}}}}\ignorespaces\end{gathered}$ (23)

In particular, the map ${\rm pr}_{x}$ is the same for each generic point $x$
of the special fiber. Hence, the value of $d_{x}$ is independent of $x$ in
this situation.

######  Lemma 4.17.

We have

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal
O}_{\mathfrak{X}}((p^{r}-1)\mathfrak{C}(Np^{r})).$

###### Proof.

Recall from equation (17) and the paragraphs proceeding it, we have

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal
O}_{\mathfrak{X}}\left(\sum_{x\in\mathfrak{X}{}_{/\mathbb{Q}_{p}(\zeta_{Np^{r}})}}d_{x}\cdot\overline{\left\\{x\right\\}}+\sum_{x\in\mathfrak{X}_{/\mathbb{F}_{q}}}d_{x}\cdot\overline{\left\\{x\right\\}}\right)$

where the first sum is over closed points of the generic fiber and the second
sum is over the generic points of the irreducible components of the special
fiber. By Proposition 4.16, we have

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal
O}_{\mathfrak{X}}\left(\sum_{x\in\mathfrak{C}(Np^{r})}(p^{r}-1)\cdot\overline{\left\\{x\right\\}}+\sum_{x\in\mathfrak{X}_{/\mathbb{F}_{q}}}d_{x}\cdot\overline{\left\\{x\right\\}}\right)$

where the first sum is over all the cusps in the generic fiber. As discussed
above, the $d_{x}$ values appearing in the second sum are independent of $x$.
Using the fact that the special fiber is reduced, we have

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal
O}_{\mathfrak{X}}((p^{r}-1)\mathfrak{C}(Np^{r})+d_{x}\mathfrak{X}_{/\mathbb{F}_{q}}).$

Note that $\mathfrak{X}_{/\mathbb{F}_{q}}$, when viewed as a divisor, is
principal (see Proposition 2.13). Thus

$\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\simeq{\cal
O}_{\mathfrak{X}}((p^{r}-1)\mathfrak{C}(Np^{r}))$

as desired. ∎

Recall the map from (15):

$\mathfrak{X}\overset{{\rm
pr}}{\longrightarrow}\mathfrak{X}(N)_{/R}\overset{g}{\longrightarrow}{\rm
Spec}(R)$

Again, for convenience, we let $\mathfrak{X}(N)$ denote the base change
$\mathfrak{X}(N)_{/R}$. To compute
$\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes 2}|_{\Lambda})$, our strategy
is to first prove that $\underline{\omega}_{\mathfrak{X}}^{\otimes
2}\simeq\Omega_{\mathfrak{X}/R}(\mathfrak{C}(Np^{r}))$ using the isomorphism
in (16):

$\Omega_{\mathfrak{X}/R}\simeq\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\otimes_{{\cal
O}_{\mathfrak{X}}}{\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}(-\mathfrak{C}(N)).$

Our next step is to provide a better description of ${\rm
pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes 2}(-\mathfrak{C}(N))$.

######  Lemma 4.18.

We have an isomorphism of sheaves on $\mathfrak{X}(Np^{r})$

${\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}(-\mathfrak{C}(N))\simeq\underline{\omega}_{\mathfrak{X}}^{\otimes
2}\otimes_{{\cal O}_{\mathfrak{X}}}{\cal
O}_{\mathfrak{X}}(-p^{r}\mathfrak{C}(Np^{r})).$

###### Proof.

We have

${\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}(-\mathfrak{C}(N))\simeq{\rm
pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes 2}\otimes_{{\cal
O}_{\mathfrak{X}(Np^{r})}}{\rm pr}^{*}{\cal
O}_{\mathfrak{X}(N)}(-\mathfrak{C}(N)).$

By Proposition A.11 and Proposition A.13, we have

${\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}=\underline{\omega}_{\mathfrak{X}(Np^{r})}^{\otimes 2}.$

Since ${\rm pr}$ maps $\mathfrak{C}(Np^{r})$ onto $\mathfrak{C}(N)$ and is
ramified at the cusps of degree $p^{r}$ by __ Proposition 4.16, we have

${\rm pr}^{*}{\cal O}_{\mathfrak{X}(N)}(-\mathfrak{C}(N))\simeq{\cal
O}_{\mathfrak{X}(Np^{r})}(-p^{r}\mathfrak{C}(Np^{r})).$

Thus

${\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}(-\mathfrak{C}(N))\simeq\underline{\omega}_{\mathfrak{X}}^{\otimes
2}\otimes_{{\cal O}_{\mathfrak{X}}}{\cal
O}_{\mathfrak{X}}(-p^{r}\mathfrak{C}(Np^{r}))$

as desired. ∎

######  Theorem 4.19.

We have $\underline{\omega}_{\mathfrak{X}}^{\otimes
2}\simeq\Omega_{\mathfrak{X}/R}(\mathfrak{C}(Np^{r})).$

###### Proof.

Recall the isomorphism in (16) states

$\Omega_{\mathfrak{X}/R}\simeq\Omega_{\mathfrak{X}/\mathfrak{X}(N)}\otimes_{{\cal
O}_{\mathfrak{X}}}{\rm pr}^{*}\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}(-\mathfrak{C}(N)).$

By Lemma 4.17 and Lemma 4.18, we have

$\Omega_{\mathfrak{X}/R}\simeq{\cal
O}_{\mathfrak{X}}((p^{r}-1)\mathfrak{C}(Np^{r}))\otimes\underline{\omega}_{\mathfrak{X}}^{\otimes
2}\otimes{\cal
O}_{\mathfrak{X}}(-p^{r}\mathfrak{C}(Np^{r}))\simeq\underline{\omega}_{\mathfrak{X}}^{\otimes
2}(-\mathfrak{C}(Np^{r})).\qed$

######  Corollary 4.20.

Let $\Lambda$ be an irreducible component of the special fiber of
$\mathfrak{X}$. We have

$\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2}|_{\Lambda})=\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda})+\deg(\mathfrak{C}(Np^{r})|_{\Lambda}).$

The following will allow us to calculate
$\deg(\mathfrak{C}(Np^{r})|_{\Lambda})$.

######  Lemma 4.21.

$\deg(\mathfrak{C}(Np^{r})|_{\Lambda})$ is equal to the number of cusps
$\overline{\left\\{x\right\\}}$ which intersect $\Lambda$.

###### Proof.

Viewed as a divisor, $\mathfrak{C}(Np^{r})$ is equal to the closure of all the
cusps of the generic fiber of $\mathfrak{X}$ by Proposition A.15. By Theorem
2.11(d),

$\deg(\mathfrak{C}(Np^{r})|_{\Lambda})=\mathfrak{C}(Np^{r}).\Lambda=\sum_{x}\overline{\left\\{x\right\\}}.\Lambda$

where the sum is indexed over
$x\in\mathfrak{C}(\mathfrak{X}{}_{/\mathbb{Q}_{p}(\zeta_{Np^{r}})})$ _i.e._
over the cusps of the generic fiber.

All the cusps are rational, so by Corollary 2.15,
$\overline{\left\\{x\right\\}}$ intersects at a single irreducible component
of the special fiber. Therefore

$\overline{\left\\{x\right\\}}.\Lambda=\begin{cases}1&\mbox{if
}\overline{\left\\{x\right\\}}\cap\Lambda\neq\emptyset\\\
0&\mbox{otherwise}\end{cases}$

so $\deg(\mathfrak{C}(Np^{r})|_{\Lambda})$ is equal to the number of cusps
which intersect $\Lambda$ as desired. ∎

For convenience, we let
$C(Np^{r})=\mathfrak{C}(\mathfrak{X}{}_{/\mathbb{Q}_{p}(\zeta_{Np^{r}}})$, the
set of all cusps of the generic fiber. Using Lemma 4.21 and Proposition A.31,
we obtain the following:

######  Proposition 4.22.

We have

$\deg(\mathfrak{C}(Np^{r})|_{\Lambda})=\varphi(p^{r})\\#C(N).$

To compute $\deg(\Omega_{\mathfrak{X}(Np^{r})/R}|_{\Lambda})$, we will use the
following result in [Liu02, Theorem 9.1.37].

######  Theorem 4.23.

Let $X\rightarrow S$ be a regular fibered surface, $s\in S$ a closed point,
and $E\in{\rm Div}_{s}(X)$ such that $0<E\leq X_{s}$. Then we have

$\Omega_{E/k(s)}\simeq({\cal O}_{X}(E)\otimes\Omega_{X/S})|_{E}.$

######  Corollary 4.24.

We have

$\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda})=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}-\varphi(p^{r})\\#C(N)+\deg{\rm S}(N)\cdot
p^{2r-1}$

###### Proof.

Applying Theorem 4.23 to $\mathfrak{X}\rightarrow{\rm Spec}(R)$ and
$E=\Lambda$, we get

$\Omega_{\Lambda/\mathbb{F}_{q}}\simeq({\cal
O}_{\mathfrak{X}}(\Lambda)\otimes\Omega_{\mathfrak{X}/R})|_{\Lambda}.$

Therefore

$\displaystyle\deg(\Omega_{\Lambda/\mathbb{F}_{q}})$
$\displaystyle=\deg(({\cal
O}_{\mathfrak{X}}(\Lambda)\otimes\Omega_{\mathfrak{X}/R})|_{\Lambda})$
$\displaystyle=\deg({\cal
O}_{\mathfrak{X}}(\Lambda)|_{\Lambda})+\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda})$
$\displaystyle=\Lambda.\Lambda+\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda}).$

Thus

$\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda})=\deg(\Omega_{\Lambda/\mathbb{F}_{q}})-\Lambda.\Lambda.$

By [Liu02, Corollary 7.3.31(a)],

$\deg(\Omega_{\Lambda/\mathbb{F}_{q}})=2\rho_{a}(\Lambda)-2$

where $\rho_{a}(\Lambda)$ is the arithmetic genus of $\Lambda$. By [KM85,
12.9.4] along with [KM85, Corollary 10.13.12], we have

$\displaystyle 2\rho_{a}(\Lambda)$ $\displaystyle=2\rho_{a}({\rm
Ig}(p^{r},N))$ $\displaystyle=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}+2-\varphi(p^{r})\\#C(N)$

We have already shown $\Lambda.\Lambda=-\deg{\rm S}(N)\cdot p^{2r-1}$ in
Proposition 3.5. Thus

$\displaystyle\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda})$
$\displaystyle=(2\rho_{a}(\Lambda)-2)-\Lambda.\Lambda$
$\displaystyle=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}-\varphi(p^{r})\\#C(N)+\deg{\rm S}(N)\cdot
p^{2r-1}.$

as desired. ∎

Recall that the space of cusp forms, by definition, are the global sections of
$\underline{\omega}_{\mathfrak{X}}^{\otimes 2}(-\mathfrak{C}(Np^{r}))$.

######  Theorem 4.25.

Let $k\geq 1$ be an integer. We have

$\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes 2k}|_{\Lambda})=k\cdot\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})\left[\frac{1}{12}(p-1)p^{2r-1}\right]$

and

$\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2k}(-\mathfrak{C}(Np^{r}))|_{\Lambda})=k\cdot\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})(p-1)\left[\frac{p^{2r-1}}{12}-\frac{p^{r-1}}{2N}\right].$

###### Proof.

We first focus on the case $k=1$. By Corollary 4.20, Proposition 4.22, and
Corollary 4.24, we have

$\displaystyle\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes 2}|_{\Lambda})$
$\displaystyle=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}-\varphi(p^{r})\\#C(N)+\deg{\rm S}(N)\cdot
p^{2r-1}+\varphi(p^{r})\\#C(N)$
$\displaystyle=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}+\deg{\rm S}(N)\cdot p^{2r-1}$

By [KM85, 12.4.5], $\deg{\rm S}(N)=\frac{(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}$. Continuing, we have

$\displaystyle\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes 2}|_{\Lambda})$
$\displaystyle=p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}+\frac{(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}\cdot p^{2r-1}$ $\displaystyle=\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})\left[\frac{p^{r}\varphi(p^{r})}{24}+\frac{p^{2r-1}(p-1)}{24}\right]$
$\displaystyle=\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})\left[\frac{1}{12}(p-1)p^{2r-1}\right].$

By Proposition 4.22 along with Lemma A.29, we have

$\displaystyle\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2}(-\mathfrak{C}(Np^{r}))|_{\Lambda})$
$\displaystyle=\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2})-\deg(\mathfrak{C}(Np^{r})|_{\Lambda})$
$\displaystyle=\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2})-\varphi(p^{r})\\#C(N)$ $\displaystyle=\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})\left[\frac{1}{12}(p-1)p^{2r-1}\right]-\left[\varphi(p^{r})\frac{{\rm\\#SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{2N}\right]$
$\displaystyle=\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})(p-1)\left[\frac{p^{2r-1}}{12}-\frac{p^{r-1}}{2N}\right].$

In general, taking tensor powers commutes with pullback of sheaves (see
[Sta21, Tag01CD]). Hence $\underline{\omega}_{\mathfrak{X}}^{\otimes
2k}|_{\Lambda}\simeq\left(\underline{\omega}_{\mathfrak{X}}^{\otimes
2}|_{\Lambda}\right)^{\otimes k}$ and consequently

$\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2k}|_{\Lambda})=\deg\left(\left(\underline{\omega}_{\mathfrak{X}}^{\otimes
2}|_{\Lambda}\right)^{\otimes
k}\right)=k\cdot\deg(\underline{\omega}_{\mathfrak{X}}^{\otimes
2}|_{\Lambda})$

which gives our desired result. ∎

###  4.5 An Upper Bound

Recall at the end of Remark 2.20 we arrived at the following expression for
the exponent

$e=\max_{\begin{subarray}{c}1\leq i\leq n\\\ f\in
H^{0}(\mathfrak{X},\underline{\omega}^{\otimes
2})\end{subarray}}\left\\{\nu_{\Lambda_{0}}(f)-\nu_{\Lambda_{i}}(f)\right\\}.$

At the end of Section 2.4 we established

$\nu_{\Lambda}(f)-\nu_{\Lambda_{(1,0)}}(f)=\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}\left(\deg(\underline{\omega}^{\otimes
k}|_{\Lambda^{\prime}})-H_{f}.\Lambda^{\prime}\right)c^{\Lambda,\Lambda^{\prime}}$

where the sum is over all irreducible components of the special fiber
excluding $\Lambda_{(1,0)}$. We will now provide an upper bound for $e$. First
we need to compute the sums $\sum c^{\Lambda,\Lambda^{\prime}}$, where
$c^{\Lambda,\Lambda^{\prime}}$ is the entry of $T^{-1}$ corresponding to row
label $\Lambda$ and column label $\Lambda^{\prime}$ (see Section 3.2).

######  Lemma 4.26.

For $1\leq a^{\prime}\leq p^{r}-1$, we have

$\sum_{\begin{subarray}{c}a=1\\\ a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(\frac{1}{a}-\frac{1}{a^{\prime}}\right)=-(p^{r}-2)\nu_{p}(a^{\prime}).$

###### Proof.

We have

$\sum_{\begin{subarray}{c}a=1\\\ a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(\frac{1}{a}-\frac{1}{a^{\prime}}\right)=\sum_{\begin{subarray}{c}a=1\\\
a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(\frac{a^{\prime}-a}{aa^{\prime}}\right)=\sum_{\begin{subarray}{c}a=1\\\
a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}(a^{\prime}-a)-\sum_{\begin{subarray}{c}a=1\\\
a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(a\right)-\sum_{\begin{subarray}{c}a=1\\\
a\neq a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(a^{\prime}\right)$
$=\sum_{\begin{subarray}{c}a=1\\\ a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}(a-a^{\prime})-\left(\sum_{a=1}^{p^{r}-1}\nu_{p}(a)-\nu_{p}(a^{\prime})\right)-(p^{r}-2)\nu_{p}(a^{\prime})$
(24)

Note that

$\displaystyle\sum_{\begin{subarray}{c}a=1\\\ a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}(a-a^{\prime})$
$\displaystyle=\sum_{a=1}^{a^{\prime}-1}\nu_{p}(a-a^{\prime})+\sum_{a=a^{\prime}+1}^{p^{r}-1}\nu_{p}(a-a^{\prime})$
$\displaystyle=\sum_{a=1}^{a^{\prime}-1}\nu_{p}(a)+\sum_{a=1}^{p^{r}-1-a^{\prime}}\nu_{p}(a)$
$\displaystyle=\sum_{a=-a^{\prime}+1}^{-1}\nu_{p}(a)+\sum_{a=1}^{p^{r}-1-a^{\prime}}\nu_{p}(a)$
$\displaystyle=\sum_{a=p^{r}-a^{\prime}+1}^{p^{r}-1}\nu_{p}(a)+\sum_{a=1}^{p^{r}-a^{\prime}-1}\nu_{p}(a)$
$\displaystyle=-\nu_{p}(a^{\prime})+\sum_{a=1}^{p^{r}-1}\nu_{p}(a).$

Continuing equation (24), we have

$\displaystyle=-\nu_{p}(a^{\prime})+\sum_{a=1}^{p^{r}-1}\nu_{p}(a)-\left(\sum_{a=1}^{p^{r}-1}\nu_{p}(a)-\nu_{p}(a^{\prime})\right)-(p^{r}-2)\nu_{p}(a^{\prime})$
$\displaystyle=-(p^{r}-2)\nu_{p}(a^{\prime}).$ ∎

######  Proposition 4.27.

We have

$\deg{\rm
S}(N)\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}c^{\Lambda,\Lambda^{\prime}}=\begin{cases}{\displaystyle-p^{-r}(pr-r+1)+p^{-r}(p-1)\nu_{p}(a)}&\mbox{
if }\Lambda=\Lambda_{(1,-a)}\\\ \\\ {\displaystyle-p^{-r}(pr-r+1)}&\mbox{ if
}\Lambda=\Lambda_{(-pb,1)}\end{cases}$

###### Proof.

We split into two cases, depending on $\Lambda$.

Case 1: Suppose $\Lambda=\Lambda_{(1,-a^{\prime})}$. Using Theorem 3.20, we
compute

$\displaystyle\sum_{\Lambda\neq\Lambda_{(1,0)}}c^{\Lambda_{(1,a^{\prime})},\Lambda}$
$\displaystyle=\sum_{a=1}^{p^{r}-1}c^{\Lambda_{(1,-a^{\prime})},\Lambda_{(1,-a)},}+\sum_{b=0}^{p^{r-1}-1}c^{\Lambda_{(1,-a^{\prime})},\Lambda_{(-pb,1)}}$
$\displaystyle=c^{\Lambda_{(1,a^{\prime})},\Lambda_{(1,a^{\prime})}}+\sum_{\begin{subarray}{c}a=1\\\
a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}\left(\frac{1}{a}-\frac{1}{a^{\prime}}\right)\right)$
$\displaystyle+\sum_{b=0}^{p^{r-1}-1}\left(-\frac{p^{1-2r}(pr-r+1)}{p+1}+\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})\right)$
$\displaystyle=c^{\Lambda_{(1,a^{\prime})},\Lambda_{(1,a^{\prime})}}+(p^{r}-2)\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}\right)-\frac{p^{1-2r}(p-1)}{p+1}\sum_{\begin{subarray}{c}a=1\\\
a\neq
a^{\prime}\end{subarray}}^{p^{r}-1}\nu_{p}\left(\frac{1}{a}-\frac{1}{a^{\prime}}\right)$
$\displaystyle-(p^{r-1})\frac{p^{1-2r}(pr-r+1)}{p+1}+p^{r-1}\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})$

Using Lemma 4.26 on the sum $\sum
v_{p}(\nicefrac{{1}}{{a}}-\nicefrac{{1}}{{a^{\prime}}})$, we have

$\displaystyle=c^{\Lambda_{(1,a^{\prime})},\Lambda_{(1,a^{\prime})}}+(p^{r}-2)\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}\right)-\frac{p^{1-2r}(p-1)}{p+1}\left(-(p^{r}-2)\nu_{p}(a^{\prime})\right)$
$\displaystyle-(p^{r-1})\frac{p^{1-2r}(pr-r+1)}{p+1}+p^{r-1}\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})$
$\displaystyle=c^{\Lambda_{(1,a^{\prime})},\Lambda_{(1,a^{\prime})}}+(p^{r}-2+p^{r-1})\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}\right)-\frac{p^{1-2r}(p-1)}{p+1}\left(-(p^{r}-2)\nu_{p}(a^{\prime})\right)+p^{r-1}\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})$
$\displaystyle=c^{\Lambda_{(1,a^{\prime})},\Lambda_{(1,a^{\prime})}}+(p^{r}-2+p^{r-1})\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}\right)+(p^{r}+p^{r-1}-2)\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})$
$\displaystyle={\displaystyle-\frac{2p^{1-2r}(pr-r+1)}{p+1}+\frac{2p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})}+(p^{r}-2+p^{r-1})\left(-p^{1-2r}\frac{(pr-r+1)}{p+1}\right)$
$\displaystyle+(p^{r}+p^{r-1}-2)\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a^{\prime})$
$\displaystyle=-p^{-r}(pr-r+1)+p^{-r}(p-1)\nu_{p}(a^{\prime}).$

Case 2: Suppose $\Lambda^{\prime}=\Lambda_{(-pb^{\prime},1)}$. Then

$\displaystyle\sum_{\Lambda\neq\Lambda_{(1,0)}}c^{\Lambda_{(-pb^{\prime},1)},\Lambda}$
$\displaystyle=\sum_{a=1}^{p^{r}-1}c^{\Lambda_{(-pb^{\prime},1)},\Lambda_{(1,-a)},}+\sum_{b=0}^{p^{r-1}-1}c^{\Lambda_{(-pb^{\prime},1)},\Lambda_{(-pb,1)}}$
$\displaystyle=\sum_{a=1}^{p^{r}-1}\left(-\frac{p^{1-2r}(pr-r+1)}{p+1}+\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(a)\right)$
$\displaystyle-\frac{2p^{1-2r}(pr-r+1)}{p+1}+\sum_{\begin{subarray}{c}b=0\\\
b\neq
b^{\prime}\end{subarray}}^{p^{r-1}-1}\left(-\frac{p^{1-2r}(pr+p-r)}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\nu_{p}(b^{\prime}-b)\right)$
$\displaystyle=-(p^{r}-1)\frac{p^{1-2r}(pr-r+1)}{p+1}+\frac{p^{1-2r}(p-1)}{p+1}\sum_{a=1}^{p^{r}-1}\nu_{p}(a)$
$\displaystyle-\frac{2p^{1-2r}(pr-r+1)}{p+1}-(p^{r-1}-1)\frac{p^{1-2r}(pr+p-r)}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\sum_{\begin{subarray}{c}b=0\\\
b\neq b^{\prime}\end{subarray}}^{p^{r-1}-1}\nu_{p}(b^{\prime}-b)$

Using Lemma 3.16, we get

$\displaystyle=-(p^{r}-1)\frac{p^{1-2r}(pr-r+1)}{p+1}+\frac{p^{1-2r}(p-1)}{p+1}\left(\frac{p^{r}-pr+r-1}{p-1}\right)$
$\displaystyle-\frac{2p^{1-2r}(pr-r+1)}{p+1}-(p^{r-1}-1)\frac{p^{1-2r}(pr+p-r)}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\left(-\nu_{p}(b^{\prime})+\frac{p^{r-1}-p(r-1)+r-2}{p-1}+\nu_{p}(b^{\prime})\right)$
$\displaystyle=-\frac{(p-1)rp^{1-r}}{p+1}-\frac{2p^{1-2r}(pr-r+1)}{p+1}-(p^{r-1}-1)\frac{p^{1-2r}(pr+p-r)}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\left(\frac{p^{r-1}-p(r-1)+r-2}{p-1}\right)$
$\displaystyle=\frac{p^{-2r}(-p^{r+1}-rp^{r+2}+rp^{r}-p^{2}(r-1)+p(r-2))}{p+1}-\frac{p^{1-2r}(p-1)}{p+1}\left(\frac{p^{r-1}-p(r-1)+r-2}{p-1}\right)$
$\displaystyle=-p^{-r}(pr-r+1).$ ∎

######  Theorem 4.28.

Let $k\geq 1$, $N\geq 3$, and $r\geq 1$ be integers and $p\geq 2$ be a prime
such that $p\nmid N$. Let $e$ be the exponent of $\pi$ in the annihilator of
$M_{2k}(\Gamma(Np^{r}),\mathbb{Z}_{p}[\zeta_{Np^{r}}])/H^{0}(\mathfrak{X},\underline{\omega}^{\otimes
2k})$. Then

$e\leq 2kp^{r-1}(pr-r+1).$

###### Proof.

For any $f\in M_{2k}(Np^{r},\mathbb{Z}_{p}[\zeta_{Np^{r}}])$, we always have
$\nu_{\Lambda_{(1,0)}}(f)\geq 0$. Thus

$\displaystyle-\nu_{\Lambda}(f)$
$\displaystyle\leq\nu_{\Lambda_{(1,0)}}(f)-\nu_{\Lambda}(f)$
$\displaystyle=\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}\left(H_{f}.\Lambda^{\prime}-\deg(\underline{\omega}^{\otimes
2k}|_{\Lambda^{\prime}})\right)c^{\Lambda,\Lambda^{\prime}}$
$\displaystyle=\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}(H_{f}.\Lambda^{\prime})c^{\Lambda,\Lambda^{\prime}}-\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}\deg(\underline{\omega}^{\otimes
2k}|_{\Lambda^{\prime}})c^{\Lambda,\Lambda^{\prime}}.$

By Corollary 3.21, $c^{\Lambda,\Lambda^{\prime}}<0$. Also note that $H_{f}$ is
an effective horizontal divisor since $f$ has no poles while
$\Lambda^{\prime}$ is an effective vertical divisor. Since $H_{f}$ and
$\Lambda^{\prime}$ do not have any common components, the intersection number
$H_{f}.\Lambda^{\prime}$ is positive. Thus
$\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}(H_{f}.\Lambda^{\prime})c^{\Lambda,\Lambda^{\prime}}$
will always be negative. Continuing, we have

$\leq-\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}\deg(\underline{\omega}^{\otimes
2k}|_{\Lambda^{\prime}})c^{\Lambda,\Lambda^{\prime}}$

Using Theorem 4.25, the above expression becomes

$-\frac{k(p-1)p^{2r-1}\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{12}\frac{24}{(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}\deg{\rm
S}(N)\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}c^{\Lambda,\Lambda^{\prime}}$
$=-kp^{2r-1}\deg{\rm
S}(N)\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}c^{\Lambda,\Lambda^{\prime}}.$

Using Proposition 4.27, we have

$=\begin{cases}{\displaystyle
2kp^{r-1}(pr-r+1)-2kp^{r-1}(p-1)\nu_{p}(a)}&\mbox{ if
}\Lambda=\Lambda_{(1,-a)}\\\ \\\ {\displaystyle 2kp^{r-1}(pr-r+1)}&\mbox{ if
}\Lambda=\Lambda_{(-pb,1)}\end{cases}$

which is maximized whenever $a$ is coprime to $p$ and attains a value of
$2kp^{r-1}(pr-r+1)$. ∎

By replacing $\underline{\omega}^{\otimes 2}$ with
$\underline{\omega}^{\otimes 2}(-\mathfrak{C}(Np^{r})$ in the proof of Theorem
4.28, we obtain an upper bound for the exponent in the situation of cusp
forms.

######  Corollary 4.29.

Let $k\geq 1,N\geq 3$, and $r\geq 1$ be integers and $p\geq 2$ a prime such
that $p\nmid N$. The exponent $e$ of $\pi$ in the annihilator of
$S_{2k}(\Gamma(Np^{r}),\mathbb{Z}_{p}[\zeta_{Np^{r}}])/H^{0}(\mathfrak{X},\underline{\omega}^{\otimes
2k}(-\mathfrak{C}(Np^{r}))$ is bounded above by

$2kp^{r-1}(pr-r+1)-\frac{12k}{Np}(pr-r+1).$

###### Proof.

Using Theorem 4.25 to compute $\deg(\underline{\omega}^{\otimes
2k}(-\mathfrak{C}(Np^{r}))|_{\Lambda})$, we get an upper bound of

$-k\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})(p-1)\left[\frac{p^{2r-1}}{12}-\frac{p^{r-1}}{2N}\right]\frac{24}{(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}\deg{\rm
S}(N)\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}c^{\Lambda,\Lambda^{\prime}}$

$\displaystyle=k\left[\frac{12p^{r-1}}{N}-2p^{2r-1}\right]\deg{\rm
S}(N)\sum_{\Lambda^{\prime}\neq\Lambda_{(1,0)}}c^{\Lambda,\Lambda^{\prime}}$
$\displaystyle\leq-k\left[\frac{12p^{r-1}}{N}-2p^{2r-1}\right]p^{-r}(pr-r+1)$
$\displaystyle=2kp^{r-1}(pr-r+1)-\frac{12k}{Np}(pr-r+1).$ ∎

######  Remark 4.30.

We compare our result in Corollary 4.29 with Edixhoven’s method in [Edi06]. He
considers the situation of weight $2$ and level $\Gamma_{0}(N)$ cusp forms __
where ${\rm ord}_{p}(N)=1$, and bounds $e$ via the inequality

$e<\frac{\deg(\Omega_{\mathfrak{X}_{0}(N)}|_{\Lambda})}{\deg{\rm S}(N)}.$

We will show a similar inequality holds our situation. Let
$\Lambda_{0},\Lambda_{1},\dots,\Lambda_{n}$ denote the irreducible components
of $\bar{\mathfrak{X}}$ where $\Lambda_{0}$ contains the cusp $\infty$. Let
$f\in M_{2}(Np^{r},\mathbb{Z}_{p}[\zeta_{Np^{r}}])$ be a non-zero cusp form
such that $\nu_{\Lambda_{0}}(f)=0$ and, without loss of generality, let
$-m:=\nu_{\Lambda_{1}}(f)<0$ be the minimum among the values in
$\left\\{\nu_{\Lambda_{i}}(f)\right\\}_{i=1}^{n}$. By scaling, we can assume
$\nu_{\Lambda_{0}}(f)=m$, $\nu_{\Lambda_{1}}(f)=0$, and
$\nu_{\Lambda_{i}}(f)\geq 0$ for $i=2,\dots,n$ so that $f$ has non-negative
valuation along every irreducible component _i.e._ $f\in
H^{0}(\mathfrak{X},\underline{\omega}^{\otimes 2})$. We can write the divisor
associated to $f$ as

${\rm
div}(f)=H_{f}+m\Lambda_{0}+\sum_{i=2}^{n}\nu_{\Lambda_{i}}(f)\Lambda_{i}$

where $H_{f}$ is the horizontal part of the divisor ${\rm div}(f)$. By Theorem
4.19, the sheaf of cusp forms $\underline{\omega}^{\otimes
2}(-\mathfrak{C}(Np^{r}))$ is isomorphic to the relative dualizing sheaf
$\Omega_{\mathfrak{X}/R}$. Since $\mathfrak{X}$ is an arithmetic surface, we
can use intersection theory (see Theorem 2.11), along with Proposition 2.17,
to compute:

$\displaystyle\deg\Omega_{\mathfrak{X}/R}|_{\Lambda_{1}}(-m{\rm S}(N))$
$\displaystyle=\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda_{1}})-m\deg{\rm S}(N)$
$\displaystyle=\deg({\rm div}(f)|_{\Lambda_{1}})-m\deg{\rm S}(N)$
$\displaystyle={\rm div}(f).\Lambda_{1}-m\deg{\rm S}(N)$
$\displaystyle=H_{f}.\Lambda_{1}+m\Lambda_{0}.\Lambda_{1}+\sum_{i=2}^{n}\nu_{\Lambda_{i}}(f)\Lambda_{i}.\Lambda_{1}-m\deg{\rm
S}(N)$ $\displaystyle=H_{f}.\Lambda_{1}+m(\Lambda_{0}.\Lambda_{1}-\deg{\rm
S}(N))+\sum_{i=2}^{n}\nu_{\Lambda_{i}}(f)\Lambda_{i}.\Lambda_{1}.$

Since $f$ has no poles, $H_{f}$ is effective so $H_{f}.\Lambda_{1}\geq 0$. By
assumption, $\nu_{\Lambda_{i}}(f)\geq 0$ for $i=2,\dots,n$; we also have
$\Lambda_{i}.\Lambda_{1}\geq 0$ since $\Lambda_{i}$ and $\Lambda_{1}$ have no
common components. Lastly, $\Lambda_{0}$ and $\Lambda_{1}$ intersect precisely
at the supersingular points, so $\Lambda_{0}.\Lambda_{1}\geq\deg{\rm S}(N)$.
Therefore we conclude

$0\leq\deg\Omega_{\mathfrak{X}/R}|_{\Lambda_{1}}(-m{\rm
S}(N))=\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda_{1}})-m\deg{\rm S}(N).$

Using Corollary 4.24, we have:

$\displaystyle m$
$\displaystyle<\frac{\deg(\Omega_{\mathfrak{X}/R}|_{\Lambda_{1}})}{\deg{\rm
S}(N)}$ $\displaystyle=\frac{p^{r}\varphi(p^{r})\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24}-\varphi(p^{r})\\#C(N)+\deg{\rm S}(N)\cdot
p^{2r-1}}{\deg{\rm S}(N)}$ $\displaystyle=\frac{24p^{r}\varphi(p^{r})\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{24(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}-\frac{24\varphi(p^{r})\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{2N(p-1)\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}+p^{2r-1}$
$\displaystyle=\frac{p^{r}p^{r-1}(p-1)}{p-1}-\frac{12p^{r-1}(p-1)}{N(p-1)}+p^{2r-1}$
$\displaystyle=2p^{2r-1}-\frac{12p^{r-1}}{N}$

which provides an upper bound for $e$. When $r=1$, this bound agrees with the
bound in Corollary 4.29, but in general is larger by a factor of
$p^{r}/(pr-r+1)$.

##  5 A Lower Bound

In this section, we will use Klein forms to build an explicit modular form in
$M_{2}(\Gamma_{1}(p^{r}),\mathbb{Z}_{p}[\zeta_{p}])$. By viewing these modular
forms at level $\Gamma(Np^{r})$ with coefficients in
$\mathbb{Z}_{p}[\zeta_{Np^{r}}]$, we will obtain a lower bound for the
exponent.

######  Definition 5.1.

A nearly holomorphic modular form is a modular form which is allowed to be
meromorphic at the cusps.

Fix $(r_{1},r_{2})\in\mathbb{Q}^{2}-\mathbb{Z}^{2}$. Let $\tau\in{\cal H}$,
$q=e^{2\pi i\tau}$, and $q_{z}=e^{2\pi iz}$ where $z=r_{1}\tau+r_{2}$. Define
the Klein form

$\kappa_{(r_{1},r_{2})}(\tau)=e^{\pi
ir_{2}(r_{1}-1)}q^{\frac{1}{2}r_{1}(r_{1}-1)}(1-q_{z})\prod_{n=1}^{\infty}(1-q^{n}q_{z})(1-q^{n}q_{z}^{-1})(1-q^{n})^{-2}.$
(25)

The following result, which is [EKS11, Theorem. 2.6], builds upon results in
[KL81, §2.1, §3.4] which establish a criterion for when a product of Klein
forms is a nearly holomorphic modular form. For $r\in\mathbb{R}$, we let
$\left\langle r\right\rangle$ denote the fractional part of $r$. Note
$\left\langle r\right\rangle=0$ precisely when $r\in\mathbb{Z}$.

######  Theorem 5.2.

For an integer $N\geq 2$, let $\left\\{m(t)\right\\}_{t=1}^{N-1}$ be a family
of integers. Then the product

$\kappa(\tau)=\prod_{t=1}^{N-1}\kappa_{(t/N,0)}(N\tau)^{m(t)}$

is a nearly holomorphic modular form for $\Gamma_{1}(N)$ of weight
$k=-\sum_{t=1}^{N-1}m(t)$ if

$\sum_{t=1}^{N-1}m(t)t^{2}\equiv 0\;({\rm mod\,}\gcd(2,N)\cdot N).$

Furthermore, for $\alpha=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in{\rm SL}_{2}(\mathbb{Z})$ we have

${\rm
ord}_{q}\left(\kappa(\tau)|[\alpha]_{k}\right)=\frac{\gcd(c,N)^{2}}{2N}\sum_{t=1}^{N-1}m(t)\left\langle\frac{at}{\gcd(c,N)}\right\rangle\left(\left\langle\frac{at}{\gcd(c,N)}\right\rangle-1\right).$

Using this result when $N=p^{r}>3$, we will choose a family of integers
$\left\\{m(t)\right\\}_{t=1}^{N-1}$ such that $k=2$ and ${\rm
ord}_{q}\left(\kappa(\tau)|[\alpha]_{2}\right)\geq 0$ for all $\alpha\in{\rm
SL}_{2}(\mathbb{Z})$. This guarantees that $\kappa(\tau)$ is a weight 2
(holomorphic) modular form of level $\Gamma_{1}(p^{r})$. As
$\Gamma(Np^{r})\leq\Gamma(p^{r})\leq\Gamma_{1}(p^{r})$, this also gives us a
modular form of level $\Gamma(Np^{r})$ as originally desired.

It may be natural to choose $m(t)$ to be zero for most values of $t$ in order
to simplify the quadratic condition in Theorem 5.2 and the expression for the
order. Consider the situation when $m(t_{1}),m(t_{2}),$ and $m(t_{3})$ are the
only non-zero values for some distinct
$t_{1},t_{2},t_{3}\in\left\\{1,2,\dots,p^{r}-1\right\\}$. By Theorem 5.2, we
seek $m(t_{1}),m(t_{2}),$ and $m(t_{3})$ such that

$m(t_{1})+m(t_{2})+m(t_{3})=-2$ (26)
$m(t_{1})t_{1}^{2}+m(t_{2})t_{2}^{2}+m(t_{3})t_{3}^{2}\equiv 0\,({\rm
mod\,}2p^{r})$ (27)

and

$\frac{\gcd(c,p^{r})^{2}}{2p^{r}}\left(\sum_{i=1}^{3}m(t_{i})\left\langle\frac{at_{i}}{\gcd(c,p^{r})}\right\rangle\left(\left\langle\frac{at_{i}}{\gcd(c,p^{r})}\right\rangle-1\right)\right)\geq
0$ (28)

for all $a,c\in\mathbb{Z}$.

We will choose the $m(t_{i})$ satisfying these conditions in three separate
cases depending on the level: $p>5$, $p=5$, and $p^{r}>3$ with $r\geq 2$. It
is not clear how one constructs similar modular forms of the remaining levels
$p=2$ and $p=3$. We first consider the case of level $p>5$. To satisfy
equation (26), we can take $m(t_{1})=m(t_{2})=-2$ and $m(t_{3})=2$. The next
equation (27) becomes

$-t_{1}^{2}-t_{2}^{2}+t_{3}^{2}\equiv 0\,({\rm mod\,}p)\mbox{ or equivalently
}t_{1}^{2}+t_{2}^{2}\equiv t_{3}^{2}\,({\rm mod\,}p)$

which can be satisfied if we take $(t_{1},t_{2},t_{3})$ to be a Pythagorean
triple. As we now show, taking $t_{1}=3,t_{2}=4,$ and $t_{3}=5$ suffices if
$p>5$. For convenience, we let $(x,y)$ denote $\gcd(x,y)$ for
$x,y\in\mathbb{Z}$.

######  Proposition 5.3.

The product of Klein forms

$\kappa(\tau)=\kappa_{(3/p,0)}(p\tau)^{-2}\kappa_{(4/p,0)}(p\tau)^{-2}\kappa_{(5/p,0)}(p\tau)^{2}$

is a weight 2 modular form of level $\Gamma_{1}(p)$ for $p>5$.

###### Proof.

Define $m:\\{1,\dots,p-1\\}\rightarrow\mathbb{Z}$ by

$m(t)=\begin{cases}-2&\mbox{if }t=3,4\\\ 2&\mbox{if }t=5\\\
0&\mbox{otherwise}\end{cases}$

By Theorem 5.2, the product $\kappa(\tau)$ defined by $m(t)$ is a weight 2
nearly holomorphic modular form of level $\Gamma_{1}(p)$. It remains to show
$\kappa(\tau)$ is holomorphic at the cusps. The order of $\kappa(\tau)$ is
given by

${\rm ord}_{q}(\kappa(\tau)|[\alpha]_{2})$
$=\frac{(c,p)^{2}}{2p}\left(-2\left\langle\frac{3a}{(c,p)}\right\rangle\left(\left\langle\frac{3a}{(c,p)}\right\rangle-1\right)-2\left\langle\frac{4a}{(c,p)}\right\rangle\left(\left\langle\frac{4a}{(c,p)}\right\rangle-1\right)+2\left\langle\frac{5a}{(c,p)}\right\rangle\left(\left\langle\frac{5a}{(c,p)}\right\rangle-1\right)\right)$

where $\alpha=\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\in{\rm
SL}_{2}(\mathbb{Z})$. We will show this expression is always non-negative.

Since $(c,p)^{2}/2p>0$, we can ignore this factor. Dividing through by 2, our
goal is to show

$-\left\langle\frac{3a}{(c,p)}\right\rangle\left(\left\langle\frac{3a}{(c,p)}\right\rangle-1\right)-\left\langle\frac{4a}{(c,p)}\right\rangle\left(\left\langle\frac{4a}{(c,p)}\right\rangle-1\right)+\left\langle\frac{5a}{(c,p)}\right\rangle\left(\left\langle\frac{5a}{(c,p)}\right\rangle-1\right)$

or equivalently

$\left\langle\frac{3a}{(c,p)}\right\rangle\left(1-\left\langle\frac{3a}{(c,p)}\right\rangle\right)+\left\langle\frac{4a}{(c,p)}\right\rangle\left(1-\left\langle\frac{4a}{(c,p)}\right\rangle\right)-\left\langle\frac{5a}{(c,p)}\right\rangle\left(1-\left\langle\frac{5a}{(c,p)}\right\rangle\right)$

is non-negative for all $a,c\in\mathbb{Z}$.

Let

$f(x)=\left\langle 3x\right\rangle\left(1-\left\langle
3x\right\rangle\right)+\left\langle 4x\right\rangle\left(1-\left\langle
4x\right\rangle\right)-\left\langle 5x\right\rangle\left(1-\left\langle
5x\right\rangle\right),$

noting that $f(a/(c,p))$ is the expression we are showing is non-negative.
Since $\left\langle x\right\rangle$ is periodic with period $0\leq x<1$, it
suffices to show $f(x)\geq 0$ for all $0\leq x<1$. Since

$\left\langle-x\right\rangle(1-\left\langle-x\right\rangle)=(1-\left\langle
x\right\rangle)(1-(1-\left\langle x\right\rangle))=\left\langle
x\right\rangle(1-\left\langle x\right\rangle)$

we can conclude $f(-x)=f(x)$. Therefore it suffices to show $f(x)\geq 0$ for
all $0\leq x\leq\nicefrac{{1}}{{2}}$.

We accomplish this by finding an explicit piecewise defined expression for
$f(x)$ and then computing its derivative. Note that

$\displaystyle\left\langle 3x\right\rangle\left(1-\left\langle
3x\right\rangle\right)$ $\displaystyle=\begin{cases}3x(1-3x)&\mbox{if }0\leq
x<\nicefrac{{1}}{{3}}\\\
3(x-\nicefrac{{1}}{{3}})(1-3(x-\nicefrac{{1}}{{3}}))&\mbox{if
}\nicefrac{{1}}{{3}}\leq x<\nicefrac{{2}}{{3}}\\\
3(x-\nicefrac{{2}}{{3}})(1-3(x-\nicefrac{{2}}{{3}}))&\mbox{if
}\nicefrac{{2}}{{3}}\leq x<1\end{cases}$
$\displaystyle=\begin{cases}3x(1-3x)&\mbox{if }0\leq x<\nicefrac{{1}}{{3}}\\\
(3x-1)(2-3x)&\mbox{if }\nicefrac{{1}}{{3}}\leq x<\nicefrac{{2}}{{3}}\\\
3(3x-2)(1-x)&\mbox{if }\nicefrac{{2}}{{3}}\leq x<1\end{cases}$

Similar expressions can be obtained for $\left\langle
4x\right\rangle\left(1-\left\langle 4x\right\rangle\right)$ and $\left\langle
5x\right\rangle\left(1-\left\langle 5x\right\rangle\right)$. Putting these
together, we have

$f(x)=\begin{cases}3x(1-3x)+4x(1-4x)-5x(1-5x)&\mbox{if }0\leq
x<\nicefrac{{1}}{{5}}\\\ 3x(1-3x)+4x(1-4x)-(5x-1)(2-5x)&\mbox{if
}\nicefrac{{1}}{{5}}\leq x<\nicefrac{{1}}{{4}}\\\
3x(1-3x)+2(4x-1)(1-2x)-(5x-1)(2-5x)&\mbox{if }\nicefrac{{1}}{{4}}\leq
x<\nicefrac{{1}}{{3}}\\\ (3x-1)(2-3x)+2(4x-1)(1-2x)-(5x-1)(2-5x)&\mbox{if
}\nicefrac{{1}}{{3}}\leq x<\nicefrac{{2}}{{5}}\\\
(3x-1)(2-3x)+2(4x-1)(1-2x)-(5x-2)(3-5x)&\mbox{if }\nicefrac{{2}}{{5}}\leq
x<\nicefrac{{2}}{{4}}\end{cases}.$

Thus we can compute the derivative directly:

$f^{\prime}(x)=\begin{cases}2x&\mbox{if }0\leq x<\nicefrac{{1}}{{5}}\\\
2-8x&\mbox{if }\nicefrac{{1}}{{5}}\leq x<\nicefrac{{1}}{{4}}\\\ 0&\mbox{if
}\nicefrac{{1}}{{4}}\leq x<\nicefrac{{1}}{{3}}\\\ 6x-2&\mbox{if
}\nicefrac{{1}}{{3}}\leq x<\nicefrac{{2}}{{5}}\\\ 2-4x&\mbox{if
}\nicefrac{{2}}{{5}}\leq x<\nicefrac{{2}}{{4}}\end{cases}$

This shows that the pieces of $f(x)$ are either strictly increasing, strictly
decreasing, or constant in their appropriate interval. For each interval, the
following table shows if $f(x)$ is increasing, decreasing, or constant based
on $f^{\prime}(x)$. The value of $f$ at the rightmost endpoint is also
calculated for each interval.

$a\leq x<b$ | $0\leq x<\nicefrac{{1}}{{5}}$ | $\nicefrac{{1}}{{5}}\leq x<\nicefrac{{1}}{{4}}$ | $\nicefrac{{1}}{{4}}\leq x<\nicefrac{{1}}{{3}}$ | $\nicefrac{{1}}{{3}}\leq x<\nicefrac{{2}}{{5}}$ | $\nicefrac{{2}}{{5}}\leq x<\nicefrac{{1}}{{2}}$
---|---|---|---|---|---
Inc, Dec, Con | increasing | decreasing | constant | increasing | decreasing
$f(b)$ | $\nicefrac{{2}}{{5}}$ | $0$ | $0$ | $\nicefrac{{2}}{{5}}$ | $0$

Thus we conclude $f(x)\geq 0$ for all $x$. Consequently, $\kappa(\tau)$ is
holomorphic at each cusp so is indeed a modular form. ∎

We compute the valuation of its $q$-expansion at each cusp. The $q$-expansion
at $\infty$ for $\kappa_{(r_{1},r_{2})}(\tau)$ was given earlier in equation
(5.2). In particular, the $q$-expansion of
$\kappa_{(\nicefrac{{a}}{{p}},0)}(p\tau)$ is

$\kappa_{(\nicefrac{{a}}{{p}},0)}(p\tau)=q^{p\frac{1}{2}\frac{a}{p}(\frac{a}{p}-1)}(1-q^{a})\prod_{n=1}^{\infty}(1-q^{pn}q^{a})(1-q^{pn}q^{-a})(1-q^{pn})^{-2}.$

For convenience, let
$H_{a}(q):=\prod_{n=1}^{\infty}(1-q^{pn}q^{a})(1-q^{pn}q^{-a})(1-q^{pn})^{-2}$
and note $1/H_{a}(q)\in 1+\mathbb{Z}[[q]]$. The $q$-expansion of
$\kappa(\tau)$ is

$\displaystyle\kappa(\tau)$
$\displaystyle=\kappa_{(\nicefrac{{3}}{{p}},0)}(p\tau)^{-2}\kappa_{(\nicefrac{{4}}{{p}},0)}(p\tau)^{-2}\kappa_{(\nicefrac{{5}}{{p}},0)}(p\tau)^{2}$
$\displaystyle=\frac{q^{p\frac{1}{2}\frac{5}{p}(\frac{5}{p}-1)2}(1-q^{5})^{2}}{q^{p\frac{1}{2}\frac{3}{p}(\frac{3}{p}-1)2}(1-q^{3})^{2}q^{p\frac{1}{2}\frac{4}{p}(\frac{4}{p}-1)2}(1-q^{4})^{2}}\left(\frac{H_{5}(q)}{H_{3}(q)H_{4}(q)}\right)$
$\displaystyle=q^{2}\frac{(1-q^{5})^{2}}{(1-q^{3})^{2}(1-q^{4})^{2}}\left(\frac{H_{5}(q)}{H_{3}(q)H_{4}(q)}\right)$

which has all integral coefficients.

To compute the $q$-expansion of $\kappa(\tau)$ at the other cusps, we use the
following identity for Klein forms, which can be found in [EKS11, Prop. 2.1].
For any $\alpha=\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\in{\rm
SL}_{2}(\mathbb{Z})$, we have

$\kappa_{(r_{1},r_{2})}(\alpha\tau)=\kappa\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^{-1}\kappa_{\alpha(r_{1},r_{2})}(\tau).$

We let $\pi=1-\zeta_{p^{r}}$ which is a uniformizer of
$\mathbb{Z}_{p}[\zeta_{Np^{r}}]$.

######  Proposition 5.4.

The $\pi$-adic valuation of $\kappa(\tau)$ at its $q$-expansions around the
cusp $0$ is $-2p$ i.e. $\nu_{0}(\kappa(\tau))=-2p$.

###### Proof.

Let $\sigma=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right)\in{\rm
SL}_{2}(\mathbb{Z})$; note $\sigma\cdot\infty=0$. The $q$-expansion of
$\kappa(\tau)$ at the cusp $0$ is given by

$\kappa(\tau)|[\sigma]_{2}=(-\tau)^{-2}\kappa(\sigma\cdot\tau)=\frac{\kappa_{(\nicefrac{{5}}{{p}},0)}(p\cdot\sigma\tau)^{2}}{\tau^{2}\kappa_{(3/p,0)}(p\cdot\sigma\tau)^{2}\kappa_{(4/p,0)}(p\cdot\sigma\tau)^{2}}.$

Observe that

$\kappa_{(r_{1},r_{2})}(p\cdot\sigma\tau)=\kappa_{(r_{1},r_{2})}(-p/\tau)=\kappa_{(r_{1},r_{2})}(\sigma\cdot(\tau/p))$
$=(-\tau/p)^{-1}\kappa_{\sigma(r_{1},r_{2})}(\tau/p)=-(\tau/p)^{-1}\kappa_{(r_{2},-r_{1})}(\tau/p).$

Thus

$\displaystyle\kappa(\tau)|[\sigma]_{2}$
$\displaystyle=\frac{(\tau/p)^{-2}\kappa_{(0,-\nicefrac{{5}}{{p}})}(\tau/p)^{2}}{\tau^{2}(\tau/p)^{-2}\kappa_{(0,-\nicefrac{{3}}{{p}})}(\tau/p)^{2}(\tau/p)^{-2}\kappa_{(0,-\nicefrac{{4}}{{p}})}(\tau/p)^{2}}$
$\displaystyle=\frac{\kappa_{(0,-\nicefrac{{5}}{{p}})}(\tau/p)^{2}}{p^{2}\kappa_{(0,-\nicefrac{{3}}{{p}})}(\tau/p)^{2}\kappa_{(0,-\nicefrac{{4}}{{p}})}(\tau/p)^{2}}.$

Using equation (25), the $q$-expansion of
$\kappa_{(0,-\nicefrac{{a}}{{p}})}(\tau/p)$ is

$\kappa_{(0,-\nicefrac{{a}}{{p}})}(\tau/p)=e^{\pi
i\frac{a}{p}}(1-\zeta_{p}^{-a})\prod_{n=1}^{\infty}(1-q^{n/p}\zeta_{p}^{-a})(1-q^{n/p}\zeta_{p}^{a})(1-q^{n/p})^{-2}$

where $\zeta_{p}=e^{2\pi i/p}$. Therefore

$\displaystyle\kappa(\tau)|[\sigma]_{2}$ $\displaystyle=\frac{e^{\pi
i\frac{5}{p}}(1-\zeta_{p}^{-5})^{2}}{p^{2}e^{\pi
i\frac{3}{p}}(1-\zeta_{p}^{-3})^{2}e^{\pi
i\frac{4}{p}}(1-\zeta_{p}^{-4})}H(q^{1/n})$
$\displaystyle=\frac{\zeta_{p}^{-1}(1-\zeta_{p}^{-5})^{2}}{p^{2}(1-\zeta_{p}^{-3})^{2}(1-\zeta_{p}^{-4})^{2}}H(q^{1/n})$

for some $H(q^{1/n})\in 1+\mathbb{Z}[[q^{1/n}]]$. Note that $1-\zeta_{p}^{a}$
is a uniformizer for $\mathbb{Z}_{p}[\zeta_{Np}]$ for $a$ coprime to $p$.
Furthermore,

$(1-\zeta_{p}^{a})^{(p-1)}\mathbb{Z}_{p}[\zeta_{Np}]=p\mathbb{Z}_{p}[\zeta_{Np}].$

Thus the minimal valuation among the coefficients of the $q$-expansion of
$\kappa(\tau)|[\sigma]_{2}$ is

$\nu_{\pi}\left(\frac{\zeta_{p}^{-1}(1-\zeta_{p}^{-5})^{2}}{p^{2}(1-\zeta_{p}^{-3})^{2}(1-\zeta_{p}^{-4})^{2}}\right)=\nu_{\pi}\left(\frac{1}{p^{2}(1-\zeta_{p}^{-4})^{2}}\right)=-2\left(\nu_{\pi}(p)+\nu_{\pi}(1-\zeta_{p}^{-4})\right)$
$=-2\left((p-1)+1\right)=-2p.\qed$

Next we will handle the case of $p=5$.

######  Proposition 5.5.

The product of Klein forms

$\kappa(\tau)=\kappa_{(\nicefrac{{1}}{{5}},0)}(5\tau)^{4}\kappa_{(\nicefrac{{2}}{{5}},0)}(5\tau)^{-2}\kappa_{(\nicefrac{{3}}{{5}},0)}(5\tau)^{-4}$

is a weight 2 modular form of level $\Gamma_{1}(5)$. Furthermore,
$\nu_{0}(\kappa(\tau))=-2\cdot 5$.

###### Proof.

Define $m:\\{1,2,3,4\\}\rightarrow\mathbb{Z}$ by

$m(t)=\begin{cases}4&\mbox{if }t=1\\\ -2&\mbox{if }t=2\\\ -4&\mbox{if }t=3\\\
0&\mbox{if }t=4\end{cases}$

Note that

$m(1)+m(2)+m(3)=-2$

and

$m(1)\cdot 1^{2}+m(2)\cdot 2^{2}+m(3)\cdot 3^{2}=4\cdot 1^{2}-2\cdot
2^{2}-4\cdot 3^{2}$ $=-40\equiv 0\;({\rm mod}\,10).$

By Theorem 5.2, the product $\kappa(\tau)$ defined by $m(t)$ is a weight 2
nearly holomorphic modular form of level $\Gamma_{1}(5)$. It remains to show
$\kappa(\tau)$ is holomorphic at the cusps. Similar to the proof in
Proposition 5.3, it suffices to show

$4\left\langle\frac{a}{(c,5)}\right\rangle\left(\left\langle\frac{a}{(c,5)}\right\rangle-1\right)-2\left\langle\frac{2a}{(c,5)}\right\rangle\left(\left\langle\frac{2a}{(c,5)}\right\rangle-1\right)-4\left\langle\frac{3a}{(c,5)}\right\rangle\left(\left\langle\frac{3a}{(c,5)}\right\rangle-1\right)\geq
0$

for all $a,c\in\mathbb{Z}$. Let

$f(x)=4\left\langle x\right\rangle\left(\left\langle
x\right\rangle-1\right)-2\left\langle 2x\right\rangle\left(\left\langle
2x\right\rangle-1\right)-4\left\langle 3x\right\rangle\left(\left\langle
3x\right\rangle-1\right).$

As before, it suffices to show $f(x)\geq 0$ for $0\leq
x\leq\nicefrac{{1}}{{2}}$. Since we consider $x$ of the form $a/(c,5)$, it
suffices to show $f(x)\geq 0$ when
$x=0,\nicefrac{{1}}{{5}},\nicefrac{{2}}{{5}}$ or $\nicefrac{{1}}{{2}}$. We
compute directly:

$f(0)=f(\nicefrac{{1}}{{2}})=0$
$f(\nicefrac{{1}}{{5}})=-4\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)=\frac{4}{5}$
$f(\nicefrac{{2}}{{5}})=-4\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+6\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)=0.$

Consequently, $\kappa(\tau)$ is holomorphic at each cusp so is indeed a
modular form.

The same proof holds from Proposition 5.4 to show the $q$-expansion around the
cusp 0 of our modular form $\kappa(\tau)$ from Proposition 5.5 has $\pi$-adic
valuation equal to $-2\cdot 5$. ∎

Lastly we handle the case $p^{r}>3$ with $r\geq 2$.

######  Proposition 5.6.

Suppose $p^{r}>3$ and $r\geq 2$. The product of Klein forms

$\kappa(\tau)=\kappa_{(\nicefrac{{p^{r}-1}}{{p^{r}}},0)}(p^{r}\tau)^{-2}\kappa_{(\nicefrac{{p^{r-1}}}{{p^{r}}},0)}(p^{r}\tau)^{-2}\kappa_{(\nicefrac{{1}}{{p^{r}}},0)}(p^{r}\tau)^{2}$

is a weight 2 modular form of level $\Gamma_{1}(p^{r})$. Furthermore,
$\nu_{0}(\kappa(\tau))=-2p^{r-1}(pr-r+1)$.

###### Proof.

Define $m(t):\\{1,\dots,p^{r}-1\\}\rightarrow\mathbb{Z}$ by

$m(t)=\begin{cases}-2&\mbox{if }t=p^{r-1},p^{r}-1\\\ 2&\mbox{if }t=1\\\ 0&{\rm
otherwise}\end{cases}$

Since $r\geq 2$, the values $p^{r-1},p^{r}-1,$ and $1$ are all distinct. We
have

$\displaystyle\sum_{t=1}^{p^{r}-1}m(t)t^{2}$
$\displaystyle=-2\cdot(p^{r-1})^{2}-2\cdot(p^{r}-1)^{2}+2\cdot 1^{2}$
$\displaystyle=-2p^{2r-2}-2(p^{r}-1)^{2}+2$
$\displaystyle\equiv-2(p^{r}-1)^{2}+2\mod p^{r}$ $\displaystyle\equiv-2+2\mod
p^{r}$ $\displaystyle\equiv 0\mod 2p^{r}$

noting that $r\geq 2$ so $2r-2\geq r$. Furthermore,
$-\sum_{t=1}^{p^{r-1}}m(t)=2$ so $\kappa(\tau)$ is a weight 2 level
$\Gamma_{1}(p^{r})$ nearly holomorphic modular form. It remains to show
$\kappa(\tau)$ has non-negative order at each cusp.

The order of $\kappa(\tau)$ is given by

${\rm ord}_{q}(\kappa(\tau)|[\alpha]_{2})$
$=\frac{(c,p^{r})^{2}}{2p^{r}}\left(-2\left\langle\frac{(p^{r}-1)a}{(c,p^{r})}\right\rangle\left(\left\langle\frac{(p^{r}-1)a}{(c,p^{r})}\right\rangle-1\right)-2\left\langle\frac{p^{r-1}a}{(c,p^{r})}\right\rangle\left(\left\langle\frac{p^{r-1}a}{(c,p)}\right\rangle-1\right)+2\left\langle\frac{a}{(c,p^{r})}\right\rangle\left(\left\langle\frac{a}{(c,p^{r})}\right\rangle-1\right)\right)$

where $\alpha=\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\in{\rm
SL}_{2}(\mathbb{Z})$. We will show this expression is always non-negative.
Similar to the proof of Proposition 5.3, we let

$f(x)=\left\langle(p^{r}-1)x\right\rangle\left(1-\left\langle(p^{r}-1)x\right\rangle\right)+\left\langle
p^{r-1}x\right\rangle\left(1-\left\langle
p^{r-1}x\right\rangle\right)-\left\langle x\right\rangle\left(1-\left\langle
x\right\rangle\right).$

Writing $(c,p^{r})=p^{d}$ for some $0\leq d\leq r$, it suffices to show
$f(a/p^{d})\geq 0$ for all $0\leq a/p^{d}\leq\nicefrac{{1}}{{2}}$. Note that

$\frac{(p^{r}-1)a}{p^{d}}=\frac{p^{r}}{p^{d}}-\frac{a}{p^{d}}.$

Since $p^{r}/p^{d}\in\mathbb{Z}$ and $a/p^{d}\leq 1$, we have

$\left\langle\frac{(p^{r}-1)a}{(c,p^{r})}\right\rangle=1-\frac{a}{p^{d}}.$

Therefore

$\displaystyle f(a/p^{d})$
$\displaystyle=\left(1-\frac{a}{p^{d}}\right)\left(\frac{a}{p^{d}}\right)+\left\langle\frac{p^{r-1}a}{p^{d}}\right\rangle\left(1-\left\langle\frac{p^{r-1}a}{p^{d}}\right\rangle\right)-\left(\frac{a}{p^{d}}\right)\left(1-\frac{a}{p^{d}}\right)$
$\displaystyle=\left\langle\frac{p^{r-1}a}{p^{d}}\right\rangle\left(1-\left\langle\frac{p^{r-1}a}{p^{d}}\right\rangle\right)$

which is always non-negative. Thus $\kappa(\tau)$ is holomorphic at each cusp.

We compute $\nu_{0}(\kappa(\tau))$ in the same way as in Proposition 5.4. Let
$\sigma=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right)$; the
$q$-expansion of $\kappa(\tau)$ at the cusp $0$ is given by

$\displaystyle\kappa(\tau)|_{2}[\sigma]$
$\displaystyle=(-\tau)^{-2}\kappa(\sigma\cdot\tau)$
$\displaystyle=\frac{\kappa_{(\nicefrac{{1}}{{p^{r}}},0)}(p^{r}\cdot\sigma\tau)^{2}}{\kappa_{(\nicefrac{{p^{r}-1}}{{p^{r}}},0)}(p^{r}\cdot\sigma\tau)^{2}\kappa_{(\nicefrac{{p^{r-1}}}{{p^{r}}},0)}(p^{r}\cdot\sigma\tau)^{2}}.$

Observe that

$\displaystyle\kappa_{(a/p^{r},0)}(p^{r}\cdot\sigma\tau)$
$\displaystyle=\kappa_{(a/p^{r},0)}(-p^{r}/\tau)$
$\displaystyle=\kappa_{(a/p^{r},0)}(\sigma\cdot(\tau/p^{r}))$
$\displaystyle=(\tau/p^{r})^{-1}\kappa_{\sigma(a/p^{r},0)}(\tau/p^{r})$
$\displaystyle=(\tau/p^{r})^{-1}\kappa_{(0,-a/p^{r})}(\tau/p^{r}).$

Similar to our computation in the proof of Proposition 5.4, we have

$\displaystyle\kappa(\tau)|_{2}[\sigma]$
$\displaystyle=\frac{\kappa_{(0,-\nicefrac{{1}}{{p^{r}}})}(\tau/p^{r})^{2}}{\tau^{2}(\tau/p^{r})^{-2}\kappa_{(0,-\nicefrac{{p^{r}-1}}{{p^{r}}})}(\tau/p^{r})^{2}\kappa_{(0,-\nicefrac{{p^{r-1}}}{{p^{r}}})}(\tau/p^{r})^{2}}$
$\displaystyle=\frac{\kappa_{(0,-\nicefrac{{1}}{{p^{r}}})}(\tau/p^{r})^{2}}{p^{2r}\kappa_{(0,-\nicefrac{{p^{r}-1}}{{p^{r}}})}(\tau/p^{r})^{2}\kappa_{(0,-\nicefrac{{p^{r-1}}}{{p^{r}}})}(\tau/p^{r})^{2}}$
$\displaystyle=\frac{e^{-2\pi
i/p^{r}}(1-\zeta_{p^{r}}^{-1})^{2}}{p^{2r}e^{-2\pi i(p^{r}-1)/p^{r}}e^{-2\pi
i(p^{r-1})/p^{r}}(1-\zeta_{p^{r}}^{-(p^{r}-1)})^{2}(1-\zeta_{p^{r}}^{-p^{r-1}})^{2}}F(q^{1/p^{r}})$
$\displaystyle=\frac{\zeta_{p^{r}}^{-1}(1-\zeta_{p^{r}}^{-1})^{2}}{p^{2r}\zeta_{p^{r}}^{1-p^{r}}\zeta_{p^{r}}^{p^{r-1}}(1-\zeta_{p^{r}}^{-(p^{r}-1)})^{2}(1-\zeta_{p^{r}}^{-p^{r-1}})^{2}}F(q^{1/p^{r}}).$

for some explicit $F(q^{1/p^{r}})\in
1+\mathbb{Z}[\zeta_{p^{r}}][[q^{1/p^{r}}]]$. Let $\nu_{\pi_{r}}$ denote the
normalized valuation in $\mathbb{Z}_{p}[\zeta_{p^{r}}]$ so that
$\nu_{\pi_{r}}(1-\zeta_{p^{r}})=1$. Then we have

$\displaystyle\nu_{0}(\kappa(\tau))$
$\displaystyle=\nu_{\pi_{r}}\left(\frac{\zeta_{p^{r}}^{-1}(1-\zeta_{p^{r}}^{-1})^{2}}{p^{2r}\zeta_{p^{r}}^{1-p^{r}}\zeta_{p^{r}}^{p^{r-1}}(1-\zeta_{p^{r}}^{-(p^{r}-1)})^{2}(1-\zeta_{p^{r}}^{-p^{r-1}})^{2}}\right)$
$\displaystyle=\nu_{\pi_{r}}\left(\frac{1}{p^{2r}(1-\zeta_{p^{r}}^{-p^{r-1}})^{2}}\right)$
$\displaystyle=-2\nu_{\pi_{r}}(p^{r}(1-\zeta_{p^{r}}^{-p^{r-1}})).$

Note that
$(1-\zeta_{p^{r}}^{-p^{r-1}})\mathbb{Z}_{p}[\zeta_{p^{r}}]=(1-\zeta_{p}^{-1})\mathbb{Z}_{p}[\zeta_{p^{r}}]$
and
$(1-\zeta_{p}^{-1})^{\varphi(p)}\mathbb{Z}_{p}[\zeta_{p^{r}}]=p\mathbb{Z}_{p}[\zeta_{p^{r}}].$
Hence

$\nu_{\pi_{r}}(1-\zeta_{p^{r}}^{-p^{r-1}})=\varphi(p^{r})/\varphi(p)=p^{r-1}.$

Continuing,

$\displaystyle\nu_{0}(\kappa(\tau))$ $\displaystyle=-2(rp^{r-1}(p-1)+p^{r-1})$
$\displaystyle=-2p^{r-1}(pr-r+1)$

as desired. ∎

######  Theorem 5.7.

Let $k\geq 1,N\geq 3$ and $r\geq 1$ be integers and $p\geq 2$ be a prime such
that $p\nmid N$ and $p^{r}>3$. The exponent $e$ of $\pi$ in the annihilator of
$M_{2k}(\Gamma(Np^{r}),\mathbb{Z}_{p}[\zeta_{Np^{r}}])/H^{0}(\mathfrak{X}(Np^{r}),\underline{\omega}_{\mathfrak{X}(Np^{r})}^{\otimes
2k})$ is equal to $2kp^{r-1}(pr-r+1)$.

###### Proof.

In Theorem 4.28 we showed $2kp^{r-1}(pr-r+1)$ was an upper bound for $e$. Let
$\kappa(\tau)\in M_{2}(\Gamma(Np^{r}),\mathbb{Z}_{p}[\zeta_{Np^{r}}])$ be the
modular form as in Propositions 5.3, 5.5, and 5.6. We have shown that
$\nu_{0}(\kappa)=-2p^{r-1}(pr-r+1)$. In general, we can consider the $k$th
power $\kappa(\tau)^{k}\in
M_{2k}(\Gamma(Np^{r}),\mathbb{Z}_{p}[\zeta_{Np^{r}}])$ so that
$\nu_{0}(\kappa^{k})=-2kp^{r-1}(pr-r+1)$. Then
$\pi^{2kp^{r-1}(pr-r+1)}\kappa(\tau)^{k}$ has integral $q$-expansion at the
cusp $0$ so $2kp^{r-1}(pr-r+1)$ is a lower bound for $e$. Hence
$e=2kp^{r-1}(pr-r+1)$. ∎

##  Appendix A Appendix – The modular curve and modular forms

We will summarize the formulation of the regular integral model of the modular
curve of full level, as presented in [KM85]. Let $R$ be a ring and let ${\rm
Ell}_{R}$ denote the category whose objects are elliptic curves $E\rightarrow
S$ where $S$ is an $R$-scheme, and whose morphisms $(E\rightarrow
S)\rightarrow(E^{\prime}\rightarrow S^{\prime})$ are Cartesian squares.

We will concern ourselves with the representability of moduli problems ${\cal
F}:{\rm Ell}_{R}\rightarrow{\rm Sets}$. The functor ${\cal F}$ induces a
functor $\tilde{{\cal F}}:{\rm Sch}_{R}\rightarrow{\rm Sets}$ defined by

$S\mapsto\left\\{\left[(E\rightarrow S,\gamma)\right]:\gamma\in{\cal
F}(E\rightarrow S)\right\\}$

where $\left[(E\rightarrow S,\gamma)\right]$ denotes the isomorphism class of
the pair $(E\rightarrow S,\gamma)$. If ${\cal F}$ is representable by ${\cal
E}\rightarrow\mathfrak{M}({\cal F})$, then $\tilde{{\cal F}}$ is representable
by $\mathfrak{M}({\cal F})$. Indeed, from the bijection

$\Psi:{\cal F}(E\rightarrow S)\rightarrow{\rm Hom}_{{\rm
Ell}_{R}}(E\rightarrow S,{\cal E}\rightarrow\mathfrak{M}({\cal F}))$

we can associate to $\gamma\in{\cal F}(E\rightarrow S)$ a morphism

$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\pi}$$\textstyle{{\cal
E}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\prime}}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{\mathfrak{M}({\cal
F})}$

in ${\rm Ell}_{R}$. We define the map $\tilde{\Psi}:\tilde{{\cal
F}}(S)\rightarrow{\rm Hom}_{{\rm Sch}_{R}}(S,\mathfrak{M}({\cal F}))$ by
sending $[(E\overset{\pi}{\rightarrow}S,\gamma)]$ to the map
$f:S\rightarrow\mathfrak{M}({\cal F})$. This correspondence is bijective and
functorial in $S$ by the properties of $\Psi$.

######  Definition A.1.

Let ${\cal F},{\cal F}^{\prime}:{\rm Ell}_{R}\rightarrow{\rm Sets}$ be two
moduli problems for elliptic curves. A morphism between moduli problems (over
${\rm Ell}_{R}$) is a natural transformation $\eta:{\cal F}\Rightarrow{\cal
F}^{\prime}$.

Let $\eta:{\cal F}\Rightarrow{\cal F}^{\prime}$ be a morphism between two
representable moduli problems over ${\rm Ell}_{R}$. We have an induced natural
transformation $\tilde{\eta}:\tilde{{\cal F}}\Rightarrow\tilde{{\cal F}}$. The
map $\eta$ induces a map $\eta:\mathfrak{M}({\cal
F})\rightarrow\mathfrak{M}({\cal F}^{\prime})$ which on $T$-points coincides
with the map $\tilde{\eta}_{T}$ in the following diagram:

$\textstyle{\mathfrak{M}({\cal
F})(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta}$$\textstyle{\mathfrak{M}({\cal
F}^{\prime})(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\rm
Hom}_{{\rm Sch}_{R}}(T,\mathfrak{M}({\cal
F}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{{\rm
Hom}_{{\rm Sch}_{R}}(T,\mathfrak{M}({\cal
F}^{\prime}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\textstyle{\tilde{{\cal
F}}(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta_{T}}$$\textstyle{\tilde{{\cal
F}}^{\prime}(T)}$

######  Definition A.2.

Let ${\cal F}$ and ${\cal F}^{\prime}$ be two moduli problems for elliptic
curves over $R$. We define the product, or simultaneous, moduli problem ${\cal
F}\times{\cal F}^{\prime}$ by

$(E\rightarrow S)\mapsto{\cal F}(E\rightarrow S)\times{\cal
F}^{\prime}(E\rightarrow S).$

Suppose ${\cal F}$ is representable by an elliptic curve ${\cal
E}\rightarrow\mathfrak{M}({\cal F})$ and ${\cal F}^{\prime}$ is relatively
representable. According to [KM85, 4.3.4], ${\cal F}\times{\cal F}^{\prime}$
is representable and $\mathfrak{M}({\cal F}\times{\cal F}^{\prime})={\cal
F}^{\prime}_{{\cal E}/\mathfrak{M}({\cal F})}$ so we naturally have a map
$\mathfrak{M}({\cal F}\times{\cal F}^{\prime})\rightarrow\mathfrak{M}({\cal
F})$. Following the notation in [KM85], we will usually denote
$\mathfrak{M}({\cal F}\times{\cal F}^{\prime})$ by $\mathfrak{M}({\cal
F},{\cal F}^{\prime})$.

######  Definition A.3.

Let $E$ be an elliptic curve over a scheme $S$. A section $s\in E(S)$
corresponds to a morphism333In general, a section of a morphism $X\rightarrow
Y$ will be a locally closed embedding $s:Y\rightarrow X$. When $X\rightarrow
Y$ is separated, as in the case of an elliptic curve, the map $s:Y\rightarrow
X$ becomes a closed immersion. The associated divisor $\left[s\right]$ is in
general defined to be the closure of the scheme-theoretic image of $s$ in $E$.
$s:S\rightarrow E$ whose composition with the structural morphism
$E\rightarrow S$ is the identity on $S$. Since $E\rightarrow S$ is separated,
$s$ is a closed immersion. We denote by $\left[s\right]$ the scheme-theoretic
image of $s$ in $E$ which is an effective Cartier divisor on $E$. Consider the
multiplication-by-$N$ isogeny $\left[N\right]:E\rightarrow E$ whose kernel we
denote by $E[N]$, an $S$-group scheme.

A $\Gamma(N)$-structure on $E\rightarrow S$ is a group homomorphism
$\phi:(\mathbb{Z}/N\mathbb{Z})^{2}\rightarrow E[N](S)$ such that we have an
equality of effective Cartier divisors

$\sum_{v\in(\mathbb{Z}/N\mathbb{Z})^{2}}[\phi(v)]=E[N].$

In this case, we call $P=\phi(1,0)$ and $Q=\phi(0,1)$ the corresponding
Drinfeld basis of $E[N]$.

We define $\left[\Gamma(N)\right]:{\rm Ell}_{\mathbb{Z}}\rightarrow{\rm Sets}$
to be the moduli problem which assigns to each elliptic curve $E\rightarrow S$
the set of all $\Gamma(N)$-structures on $E\rightarrow S$. Let
$(\alpha,f):(E\rightarrow S)\rightarrow(E^{\prime}\rightarrow S^{\prime})$ be
a morphism in the category ${\rm Ell}_{R}$. Then

$\left[\Gamma(N)\right](\alpha,f):\left[\Gamma(N)\right](E^{\prime}/S^{\prime})\rightarrow\left[\Gamma(N)\right](E/S)$

is defined by sending a Drinfeld basis $(P,Q)$ of $E^{\prime}[N]$ to
$(\alpha^{*}P,\alpha^{*}Q)$. Indeed, $(\alpha^{*}P,\alpha^{*}Q)$ is a Drinfeld
basis of $E[N]$ (see the paragraph proceeding [KM85, 1.4.1.2]).

We moreover consider the full level $N$ moduli problem over the cyclotomic
integers $\mathbb{Z}[\zeta_{N}]:=\mathbb{Z}[X]/(\Phi_{N}(X))$, where
$\Phi_{N}(X)\in\mathbb{Z}[X]$ is the $N$th cyclotomic polynomial, by
constructing a new moduli problem $\left[\Gamma(N)\right]^{{\rm can}}$ from
$\left[\Gamma(N)\right]$ as follows (see also [KM85, 9.4.3.1]). For a
$\mathbb{Z}[\zeta_{N}]$-algebra $A$, let $\zeta_{N}$ denote the image of $X$
mod $\Phi_{N}(X)$ under the map $\mathbb{Z}[\zeta_{N}]\rightarrow A$,
following the convention in [KM85, 9.1.5]. Section 2.8 of [KM85] associates to
the isogeny $\left[N\right]$ a bilinear pairing $\mbox{$e_{N}:E[N]\times
E[N]\rightarrow\mu_{N}$}.$ Define the moduli problem
$\left[\Gamma(N)\right]^{{\rm can}}:{\rm
Ell}_{\mathbb{Z}[\zeta_{N}]}\rightarrow{\rm Sets}$ which assigns to each
elliptic curve $E\rightarrow S$ the set of all Drinfeld bases $(P,Q)$ such
that $e_{N}(P,Q)=\zeta_{N}$.

######  Theorem A.4.

Suppose $N\geq 3$. The moduli problem $\left[\Gamma(N)\right]^{{\rm can}}$ is
represented by a regular scheme
$\mathfrak{Y}(N):=\mathfrak{M}([\Gamma(N)]^{{\rm can}})$ which is flat over
${\rm Spec}(\mathbb{Z}[\zeta_{N}])$ of dimension 2. Moreover,
$\mathfrak{Y}(N)$ is smooth over $\mathbb{Z}[\zeta_{N},1/N]$.

###### Proof.

The moduli problem $\left[\Gamma(N)\right]$ is relatively representable and
finite, flat over ${\rm Ell}_{\mathbb{Z}}$ by [KM85, 5.1.1]. Furthermore,
$\left[\Gamma(N)\right]$ is rigid whenever $N\geq 3$ by [KM85, 2.7.2].
Therefore, by Proposition [KM85, 4.7.0] and [KM85, 5.1.1],
$\left[\Gamma(N)\right]$ is representable by a regular scheme flat over
$\mathbb{Z}$ of dimension 2. By [KM85, 9.1.8, 9.1.9], the same things hold
true for the associated canonical moduli problem $\left[\Gamma(N)\right]^{{\rm
can}}$. Using [KM85, §5.1.1] again, $\left[\Gamma(N)\right]$ is étale over
${\rm Ell}_{\mathbb{Z}[1/N]}$, which, along with [KM85, 4.7.1], implies
smoothness. ∎

######  Remark A.5.

Theorem A.4 also holds for the moduli problem $\left[\Gamma(N)\right]^{{\rm
can}}$ over ${\rm Ell}_{R}$, for any ring extension
$\mathbb{Z}[\zeta_{N}]\rightarrow R$. For any moduli problem ${\cal F}:{\rm
Ell}_{\mathbb{Z}}\rightarrow{\rm Sets}$ and ring $R$, let ${\cal F}_{R}:{\rm
Ell}_{R}\rightarrow{\rm Sets}$ denote the moduli problem obtained by composing
${\cal F}$ with the forgetful functor ${\rm Ell}_{R}\rightarrow{\rm
Ell}_{\mathbb{Z}}$. According to [KM85, 4.13], if ${\cal F}$ is relatively
representable, then ${\cal F}_{R}$ is also relatively representable by the
same morphism ${\cal F}_{E/S}\rightarrow S$ for any $R$-scheme $S$.
Furthermore, if ${\cal F}$ is representable by ${\cal
E}\rightarrow\mathfrak{M}({\cal F})$, then ${\cal F}_{R}$ is representable by
the base change ${\cal E}_{R}\rightarrow\mathfrak{M}({\cal F})_{R}$. In other
words, we have $\mathfrak{M}({\cal F}_{R})=\mathfrak{M}({\cal F})_{R}$.

By a process called "normalizing near infinity", as described in [KM85, §8.6],
$\mathfrak{Y}(N)$ extends to a scheme $\mathfrak{X}(N)$ which is proper over
$\mathbb{Z}[\zeta_{N}]$.

######  Definition A.6.

Let $\mathfrak{C}(\mathfrak{X}(N))$ denote the closed subscheme
$\mathfrak{X}(N)-\mathfrak{Y}(N)$ of $\mathfrak{X}(N)$ endowed with the
reduced scheme structure, called the cuspidal locus of $\mathfrak{X}(N)$. If
the modular curve is clear from context, we denote
$\mathfrak{C}(N)=\mathfrak{C}(\mathfrak{X}(N))$.

The following definition captures many of the desirable properties of
$\mathfrak{X}(N)$.

######  Definition A.7.

Let $R$ be a Dedekind domain. We call a regular, integral, projective, flat
$R$-scheme $X$ of dimension 2 an arithmetic surface.

######  Theorem A.8.

Let $N\geq 3$. The scheme $\mathfrak{X}(N)$ is an arithmetic surface over
$\mathbb{Z}[\zeta_{N}]$. Moreover, $\mathfrak{X}(N)$ is geometrically
connected with reduced closed fibers and smooth over
$\mathbb{Z}[\zeta_{N},1/N]$.

###### Proof.

We have already established that $\mathfrak{Y}(N)$ is regular and flat over
$\mathbb{Z}[\zeta_{N}]$ of dimension 2. By [KM85, 10.9.1(2)], there exists an
open neighborhood of $\mathfrak{C}(N)$ which is smooth over
$\mathbb{Z}[\zeta_{N}]$. In particular, $\mathfrak{C}(N)$ is regular and flat
over $\mathbb{Z}[\zeta_{N}]$ so $\mathfrak{X}(N)$ is regular and flat over
$\mathbb{Z}[\zeta_{N}]$. Since $\mathfrak{X}(N)$ is connected and regular,
$\mathfrak{X}(N)$ is integral.

By construction, $\mathfrak{X}(N)$ is proper over $\mathbb{Z}[\zeta_{N}]$, so
by [Liu02, Theorem 8.3.16], we can conclude $\mathfrak{X}(N)$ is projective
over $\mathbb{Z}[\zeta_{N}]$ hence is an arithmetic surface. By [KM85,
10.9.2(2)], $\mathfrak{X}(N)$ is geometrically connected and by [KM85,
13.8.4], $\mathfrak{X}(N)$ has reduced closed fibers. Smoothness follows from
the smoothness of $\mathfrak{Y}(N)$ over $\mathbb{Z}[\zeta_{N},1/N]$. ∎

Let $N\geq 3$ and let $R$ be a noetherian, regular, excellent
$\mathbb{Z}[\zeta_{N}]$-algebra. We will construct an invertible sheaf on
$\mathfrak{X}(N)_{/R}$ whose global sections will be defined as the space of
modular forms. Let $f:{\cal E}\rightarrow\mathfrak{Y}(N)_{/R}$ denote the
universal elliptic curve. We define $\underline{\omega}:=f_{*}\Omega_{{\cal
E}/\mathfrak{Y}(N)_{/R}}^{1}$ to be the pushforward of the sheaf of Kahler
differentials on ${\cal E}$. If $e:\mathfrak{Y}(N)_{/R}\rightarrow{\cal E}$ is
the identity section of our elliptic curve, then we have
$\underline{\omega}\simeq e^{*}\Omega_{{\cal E}/\mathfrak{Y}(N)_{/R}}^{1}$.
Hence $\underline{\omega}$, being the pullback of an invertible sheaf, is an
invertible sheaf on $\mathfrak{Y}(N)_{/R}$. According to [KM85, 10.13.2],
there is a canonical way to extend $\underline{\omega}^{\otimes 2}$ to an
invertible sheaf on $\mathfrak{X}(N)_{/R}$, which we denote by
$\underline{\omega}_{\mathfrak{X}(N)_{/R}}^{\otimes 2}$ or simply
$\underline{\omega}^{\otimes 2}$ if the modular curve is clear from context.

######  Proposition A.9.

The invertible sheaf $\underline{\omega}^{\otimes 2}$ canonically extends to
an invertible sheaf $\underline{\omega}_{\mathfrak{X}(N)_{/R}}^{\otimes 2}$ on
$\mathfrak{X}(N)_{/R}$.

######  Definition A.10.

We call the invertible sheaf $\underline{\omega}^{\otimes 2k}$ on
$\mathfrak{X}(N)_{/R}$ the modular sheaf (of weight 2k). The global sections
$H^{0}(\mathfrak{X}(N)_{/R},\underline{\omega}^{\otimes 2k})$ are known as
modular forms of weight $2k$ and level $\Gamma(N)$. The global sections
$H^{0}(\mathfrak{X}(N)_{/R},\underline{\omega}^{\otimes 2k}(-\mathfrak{C}(N))$
are known as cusp forms of weight 2k and level $\Gamma(N)$.

The formation of the modular sheaf $\underline{\omega}^{\otimes 2}$ behaves
well under base change. The following proposition is [KM85, 10.13.6].

######  Proposition A.11.

Let $R\rightarrow R^{\prime}$ be an extension of noetherian, regular,
excellent rings and let

$j:\mathfrak{X}(N)_{/R^{\prime}}\rightarrow\mathfrak{X}(N)_{/R}$

denote the induced base change map. Then we have an isomorphism of invertible
sheaves

$j^{*}\underline{\omega}_{\mathfrak{X}(N)_{/R}}^{\otimes
2}\simeq\underline{\omega}_{\mathfrak{X}(N)_{/R^{\prime}}}^{\otimes 2}.$

Let $f:E\rightarrow S$ be an elliptic curve over a smooth $R$-scheme $S$.
According to [KM85, 10.13.10] (see also [Kat73, A1.4]), we have map
$(f_{*}\Omega^{1}{}_{E/S})^{\otimes 2}\rightarrow\Omega_{S/R}^{1}$ of ${\cal
O}_{S}$-modules, known as the Kodaira-Spencer map, which becomes an
isomorphism precisely when $E\rightarrow S$ represents a moduli problem which
is étale over ${\rm Ell}_{R}$. In particular, if $\mathfrak{Y}(N)_{/R}$ is
smooth over $R$ (_e.g._ if $N$ is a unit in $R$), we get an isomorphism
$\underline{\omega}_{\mathfrak{Y}(N)}^{\otimes
2}\simeq\Omega_{\mathfrak{Y}(N)/R}^{1}$.

######  Theorem A.12.

Let $R$ be a noetherian, regular, excellent $\mathbb{Z}[\zeta_{N}]$-algebra
containing $1/N$. The Kodaira-Spencer isomorphism
$\underline{\omega}_{\mathfrak{Y}(N)}^{\otimes
2}\simeq\Omega_{\mathfrak{Y}(N)/R}^{1}$ on $\mathfrak{Y}(N)$ extends to an
isomorphism on $\mathfrak{X}(N)$

$\underline{\omega}_{\mathfrak{X}(N)}^{\otimes
2}\simeq\Omega_{\mathfrak{X}(N)/R}^{1}(\mathfrak{C}(N)).$

###### Proof.

By [KM85, 5.1.1], the moduli problem $\left[\Gamma(N)\right]^{{\rm can}}$ is
finite étale over ${\rm Ell}_{R}$. The result then follows from [KM85,
10.13.11]. ∎

Suppose ${\cal F},{\cal F}^{\prime}$ are two representable moduli problems
that are finite over ${\rm Ell}_{R}$ and normal near infinity. Any morphism
$\eta:{\cal F}\Rightarrow{\cal F}^{\prime}$ of moduli problems is compatible
with the usual morphism ${\cal F},{\cal
F}^{\prime}\Rightarrow\left[\Gamma(1)\right]$. Thus the induced map
$\eta:\mathfrak{M}({\cal F})\rightarrow\mathfrak{M}({\cal F}^{\prime})$ fits
in the commutative diagram

---
$\textstyle{\mathfrak{M}({\cal
F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta}$$\scriptstyle{j}$$\textstyle{\mathfrak{M}({\cal
F}^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\textstyle{\mathbb{A}_{R}^{1}}$

Hence $\eta$ extends to a map $\bar{\eta}:\overline{\mathfrak{M}}({\cal
F})\rightarrow\overline{\mathfrak{M}}({\cal F}^{\prime})$ on the
normalizations near infinity.

######  Proposition A.13.

Let $R$ be an excellent, noetherian, regular ring and let ${\cal F}$ and
${\cal F}^{\prime}$ be two representable moduli problems, both finite over
${\rm Ell}_{R}$ and both normal near infinity. Let $\eta:{\cal
F}\Rightarrow{\cal F}^{\prime}$ be a morphism of moduli problems over ${\rm
Ell}_{R}$. Under the induced map $\bar{\eta}:\overline{\mathfrak{M}}({\cal
F})\rightarrow\overline{\mathfrak{M}}({\cal F}^{\prime})$ we have

$\bar{\eta}^{*}(\underline{\omega}_{\overline{\mathfrak{M}}({\cal
F}^{\prime})}^{\otimes 2})=\underline{\omega}_{\overline{\mathfrak{M}}({\cal
F})}^{\otimes 2}.$

###### Proof.

This is [KM85, 10.13.5(2)]. ∎

###  A.1 Cusps and the Special Fiber

In this section we will investigate the cusps and special fiber of
$\mathfrak{X}(N)$ and recall the theory of $q$-expansions for modular forms.
First we establish that the formation of the cuspidal locus $\mathfrak{C}(N)$
and the formation of $\mathfrak{X}(N)$ behaves well under base change. The
following is [KM85, 8.6.6, 8.6.7].

######  Proposition A.14.

Let $R$ and $R^{\prime}$ be excellent, noetherian, regular
$\mathbb{Z}[\zeta_{N}]$-algebras. For any extension of scalars $R\rightarrow
R^{\prime}$, we have

$\mathfrak{Y}(N)_{/R}\times_{{\rm Spec}(R)}{\rm
Spec}(R^{\prime})\simeq\mathfrak{Y}(N)_{/R^{\prime}}$
$\mathfrak{C}(\mathfrak{X}(N)_{/R})\times_{{\rm Spec}(R)}{\rm
Spec}(R^{\prime})\simeq\mathfrak{C}(\mathfrak{X}(N)_{/R^{\prime}})$
$\mathfrak{X}(N)_{/R}\times_{{\rm Spec}(R)}{\rm
Spec}(R^{\prime})\simeq\mathfrak{X}(N)_{/R^{\prime}}$

For two groups $G_{1}$ and $G_{2}$, let ${\rm HomSurj}(G_{1},G_{2})$ denote
the set of surjective group homomorphisms from $G_{1}$ to $G_{2}$. Any
subgroup $\Gamma\leq{\rm GL}_{2}(\mathbb{Z}/N\mathbb{Z})$ acts on the set

${\rm HomSurj}((\mathbb{Z}/N\mathbb{Z})^{2},\mathbb{Z}/N\mathbb{Z})$

via $\gamma\cdot\Lambda=\Lambda\circ\gamma$ for any $\gamma\in\Gamma$. Let

${\rm HS}(N)={\rm
HomSurj}((\mathbb{Z}/N\mathbb{Z})^{2},\mathbb{Z}/N\mathbb{Z})/\left\\{\pm
I\right\\}$

where $I$ is the identity matrix in ${\rm GL}_{2}(\mathbb{Z}/N\mathbb{Z})$.

According to [KM85, 10.9.1], the cusps $\mathfrak{C}(N)$ of $\mathfrak{X}(N)$
is a disjoint union of $\\#{\rm HS}(N)$ many sections of $\mathfrak{X}(N)$.
Furthermore, the formal completion of $\mathfrak{X}(N)$ along
$\mathfrak{C}(N)$ is the disjoint union of $\\#{\rm HS}(N)$ copies of the
formal spectrum ${\rm Spf}(\mathbb{Z}[\zeta_{N}]\left\llbracket
q^{1/h_{\Lambda}}\right\rrbracket)$ for some integer $h_{\Lambda}\geq 1$
dividing $N$, dependent on the index $\Lambda\in{\rm HS}(N)$. Refer to [KM85,
10.2.5] which provides a canonical bijection between a cuspidal section and
the corresponding label in ${\rm HS}(N)$.

######  Proposition A.15.

Let $R$ be a Dedekind domain with fraction field $K$. As Weil divisors, we
have

$\mathfrak{C}(\mathfrak{X}(N)_{/R})=\sum_{x}\overline{\left\\{x\right\\}}$

where the sum is indexed over $x\in\mathfrak{C}(\mathfrak{X}(N)_{/K})$.

###### Proof.

As a divisor, $\mathfrak{C}(\mathfrak{X}(N)_{/R})$ consists only of horizontal
components since $\mathfrak{C}(\mathfrak{X}(N)_{/R})$ is a finite disjoint
union of sections of $\mathfrak{X}(N)_{/R}$. By [Liu02, 8.3.4], these
horizontal components are necessarily the closure, in $\mathfrak{X}(N)_{/R}$,
of points closed in the generic fiber $\mathfrak{X}(N)_{/K}$. By Proposition
A.14,
$\mathfrak{C}(\mathfrak{X}(N)_{/R})_{/K}\simeq\mathfrak{C}(\mathfrak{X}(N)_{/K})$,
hence these points are precisely the cusps of $\mathfrak{X}(N)_{/K}$. ∎

Now we recall how the $q$-expansion map

$H^{0}(\mathfrak{X}(N)_{/R},\underline{\omega}^{\otimes 2})\rightarrow
R[[q^{1/N}]]$

is defined for a modular form over a $\mathbb{Z}[\zeta_{N}]$-algebra $R$ at a
specified cusp. Let $c$ be a cusp of $\mathfrak{X}(N)$, which corresponds to
the image of a section ${\rm
Spec}(\mathbb{Z}[\zeta_{N}])\rightarrow\mathfrak{X}(N)$ and is a connected
component $\overline{\left\\{c^{\prime}\right\\}}$ of $\mathfrak{C}(N)$ where
$c^{\prime}$ is a cusp of the generic fiber. According to [DR73, §VII, 2.3.2],
the cusp $c$ corresponds to a Tate curve ${\rm Tate}(q^{1/N})$ over
$\mathbb{Z}[\zeta_{N}][[q^{1/N}]$ (see [DR73, §VII, Définition 1.16]) and a
closed immersion ${\rm
Spec}(\mathbb{Z}[\zeta_{N}][[q^{1/N}]])\rightarrow\mathfrak{X}(N)$ fitting in
the diagram

$\textstyle{{\rm
Tate}(q^{1/N})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\rm
Spec}(\mathbb{Z}[\zeta_{N}][[q^{1/N}]]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{X}(N)}$

By [DR73, §VII, Corollaire 2.4] (see also [KM85, 10.9.1]), this closed
immersion induces an isomorphism of formal schemes

${\rm
Spf}(\mathbb{Z}[\zeta_{N}][[q^{1/N}]])\overset{\sim}{\longrightarrow}\hat{\mathfrak{X}}(N)_{c}$

where $\hat{\mathfrak{X}}(N)_{c}$ denotes the formal completion of
$\mathfrak{X}(N)$ along the cusp $c$. According to [KM85, 8.8(T.2)] (see also
[DR73, §VII, 1.16.1]) we have an isomorphism

$\phi:\hat{\mathrm{Tate}}(q^{1/N})\overset{\sim}{\rightarrow}\hat{\mathbb{G}}_{m}$

between formal Lie groups. Let $\omega_{{\rm can},c}:=\phi^{*}(dX/X)$, where
$dX/X$ is the standard invariant differential on $\mathbb{G}_{m}$. Pulling
back by ${\rm Tate}(q^{1/N})\rightarrow\mathfrak{X}(N)$ allows us to identify
$\hat{\omega}_{c}^{\otimes 2}$ with
$\mathbb{Z}[\zeta_{N}][[q^{1/N}]]\cdot\omega_{{\rm can},c}$.

######  Definition A.16.

Let $f\in H^{0}(\mathfrak{X}(N),\underline{\omega}^{\otimes 2})$ and let $c$
be a cusp of $\mathfrak{X}(N)$. Viewing $f$ in the completed stalk
$\underline{\hat{\omega}}{}_{c}^{\otimes 2}$, we write
$f=f_{c}\cdot\omega_{{\rm can},c}$ for some $f_{c}\in\hat{{\cal
O}}_{\mathfrak{X}(N),c}$. We call $f_{c}$ the $q$-expansion of $f$ at $c$.
This gives a map

$H^{0}(\mathfrak{X}(N),\underline{\omega}^{\otimes
2})\rightarrow\mathbb{Z}[\zeta_{N}][[q^{1/N}].$

More generally, for a $\mathbb{Z}[\zeta_{N}]$-algebra $R$, we obtain the
$q$-expansion map

$H^{0}(\mathfrak{X}(N)_{/R},\underline{\omega}^{\otimes
2})\rightarrow\mathbb{Z}[\zeta_{N}][[q^{1/N}]\otimes_{\mathbb{Z}[\zeta_{N}]}R\subset
R[[q^{1/N}]]$

by tensoring with $R$. We will always use $\omega_{{\rm can},c}$ as a local
generator for $\omega^{\otimes 2}$ to obtain the $q$-expansion at $c$.

The following result, which is [DR73, §VII, Théoréme 3.9] and known as the
$q$-expansion principle, essentially says $q$-expansions can detect the “ring
of definition”.

######  Proposition A.17.

Let $A$ be a $\mathbb{Z}[\zeta_{N}]$-algebra, let $B$ be a subalgebra of $A$,
and let $f\in H^{0}(\mathfrak{X}(N)_{/A},\underline{\omega}^{\otimes k})$. If
the $q$-expansion of $f$ at every cusp of $\mathfrak{X}(N)_{/A}$ lies in
$B[[q^{1/N}]]$, then $f$ is a modular form over $B$ i.e. $f\in
H^{0}(\mathfrak{X}(N)_{/B},\underline{\omega}^{\otimes k})$.

Let $N\geq 3$ and $r\geq 1$ be integers and let $p\geq 2$ be a prime such that
$p\nmid N$. We will describe the special fiber of $\mathfrak{X}(Np^{r})$. For
simplicity, we consider the modular curve $\mathfrak{X}(Np^{r})$ over a
$\mathbb{Z}[\zeta_{Np^{r}}]$-algebra $R$ which is a DVR of mixed
characteristic $(0,p)$ in which $N$ is invertible and with fraction field $K$
and perfect residue field $k$. Let $\bar{\mathfrak{X}}(Np^{r})$ denote the
special fiber of $\mathfrak{X}(Np^{r})$.

We will recall the theory of Igusa curves, as developed in Section 12 of
[KM85], to describe the irreducible components of
$\bar{\mathfrak{X}}(Np^{r})$. For $i\in\mathbb{Z}$, let $\sigma^{i}:{\rm
Spec}(k)\rightarrow{\rm Spec}(k)$ denote the map induced by the $i$th power of
the Frobenius automorphism on $k$. For any $k$-scheme $S$, we let
$S^{(\sigma^{i})}$ denote the pullback under $\sigma^{i}$ from which we also
obtain a map $F_{{\rm abs}}^{i}:S\rightarrow S$ called the absolute Frobenius.
More generally, for any scheme $X\rightarrow S$, we let $X^{(p^{i})}$ denote
the pullback under $F_{{\rm abs}}^{i}$ from which we obtain a morphism
$F_{X/S}:X\rightarrow X^{(p^{i})}$ of $S$-schemes, called the $i$th-fold
relative Frobenius.

Let ${\cal F}$ be a moduli problem on ${\rm Ell}_{k}$. We define the moduli
problem ${\cal F}^{(\sigma^{i})}$ on ${\rm Ell}_{k}$ by extending scalars via
$\sigma^{i}$ so that

${\cal F}(E\rightarrow S)={\cal F}^{(\sigma^{i})}(E^{(\sigma^{i})}\rightarrow
S^{(\sigma^{i})}).$

If ${\cal F}$ is representable and finite over ${\rm Ell}_{k}$ and normal near
infinity, then the same holds for ${\cal F}^{(\sigma^{i})}$ and we have

$\overline{\mathfrak{M}}({\cal
F}^{(\sigma^{i})})=\overline{\mathfrak{M}}({\cal F})^{(\sigma^{i})}.$

######  Definition A.18.

Let $E$ be an elliptic curve over an $k$-scheme $S$. An Igusa structure of
level $p^{r}$ on $E\rightarrow S$ is a point $P\in E^{(p^{r})}(S)$ which
generates the kernel of Verschiebung $V^{r}:E^{(p^{r})}\rightarrow E$ in the
sense of [KM85, 1.4.1]. Let $[{\rm Ig}(p^{r})]:{\rm Ell}_{k}\rightarrow{\rm
Sets}$ denote the moduli problem which assigns to each elliptic curve
$E\rightarrow S$ the set of all Igusa structures of level $p^{r}$ on
$E\rightarrow S$.

According to [KM85, 12.7.1], if ${\cal F}$ is a representable moduli problem
finite over ${\rm Ell}_{k}$ which is normal near infinity, the simultaneous
moduli problem $[{\rm Ig}(p^{r})]\times{\cal F}$ is representable over ${\rm
Ell}_{k}$ and normal near infinity. The following result is [KM85, 12.7.2]
applied to the simultaneous moduli problem $\left[{\rm
Ig}(p^{r})\right]\times\left[\Gamma(N)\right]^{{\rm can},(\sigma^{-i})}$ over
${\rm Ell}_{k}$ for any $i\in\mathbb{Z}$. We denote

${\rm Ig}(p^{r},N):=\overline{\mathfrak{M}}([{\rm
Ig}(p^{r})],\left[\Gamma(N)\right]^{{\rm can},(\sigma^{-i})}).$

######  Proposition A.19.

  1. a.

${\rm Ig}(p^{r},N)$ is a proper smooth curve over $k$.

  2. b.

The usual projection ${\rm
Ig}(p^{r},N)\rightarrow\mathfrak{X}(N)_{/k}^{(\sigma^{-i})}$ is finite and
étale outside the supersingular points of
$\mathfrak{X}(N)_{/k}^{(\sigma^{-i})}$ for all $i\in\mathbb{Z}$.

Next we define another moduli problem closely related to Igusa structures,
which will directly appear in the description of $\bar{\mathfrak{X}}(Np^{r})$.

######  Definition A.20.

Let $E$ be an elliptic curve over an $k$-scheme $S$ and fix $1\leq i\leq r$.
An exotic Igusa structure of level $(p^{r},i)$ on $E\rightarrow S$ is a point
$P\in E(S)$ such that $(O,P)$ is a Drinfeld $p^{i}$-basis of $E\rightarrow S$
along with a point $Q\in E^{(p^{r-i})}(S)$ such that $V^{r-i}(Q)=P$. Let

$[{\rm ExIg}(p^{r},i)]:{\rm Ell}_{k}\rightarrow{\rm Sets}$

denote the moduli problem which assigns to each elliptic curve $E\rightarrow
S$ the set of all exotic Igusa structures of level $(p^{r},i)$ on
$E\rightarrow S$.

The following result, which is [KM85, 12.10.6] in the situation ${\cal
F}=[\Gamma(N)]^{{\rm can},(\sigma^{-i})}$, relates Igusa structures with
exotic Igusa structures and also establishes the representability of $[{\rm
ExIg}(p^{r},i)]$. We denote

${\rm ExIg}(p^{r},i,N)=\overline{\mathfrak{M}}([{\rm
ExIg}(p^{r},i)],\left[\Gamma(N)\right]).$

######  Theorem A.21.

For $1\leq i\leq r$, we have a canonical isomorphism ${\rm
Ig}(p^{r},N)\simeq{\rm ExIg}(p^{r},i,N)$ sitting in the commutative diagram

$\textstyle{{\rm
Ig}(p^{r},N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{\rho}$$\textstyle{{\rm
ExIg}(p^{r},i,N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{\prime}}$$\textstyle{\mathfrak{X}(N)_{/k}^{(\sigma^{-i})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F^{i}}$$\textstyle{\mathfrak{X}(N)_{/k}}$

where $F^{i}$ is the $i$th-fold relative Frobenius
$F_{\mathfrak{X}(N)/k}^{i}$.

We introduce the following definition from [KM85, §13.1] which will allow us
to describe the special fiber of $\mathfrak{X}(Np^{r})$. Consider the general
situation with $k$ a field, $Y$ a smooth scheme over $k$, and $X\rightarrow Y$
a finite flat morphism of schemes. Suppose there exists a nonempty finite set
$S$ of $k$-rational points of $Y$ such that for each $y_{0}\in S$ there exists
a unique closed $k$-rational point $x_{0}\in X$ over $y_{0}$ such that
$\hat{{\cal O}}_{X,x_{0}}\simeq k[[x,y]]/(f)$ for some $f\in k[[x,y]]$. The
points of $S$ are referred to as the _supersingular points_ ; indeed in the
situation $X$ and $Y$ are modular curves, $S$ will be taken to be the
supersingular points of $Y$ which correspond to supersingular elliptic curves.
Furthermore, we assume there is a finite collection of $k$-schemes
$\left\\{Z_{i}\right\\}_{i\in I}$ with a morphism $\coprod_{i\in
I}Z_{i}\rightarrow X$ such that

  * •

for each $i\in I$ and $y_{0}\in S$ there exists a unique closed, $k$-rational
point $z_{i,0}\in Z_{i}$ over $y_{0}$.

  * •

$Z_{i}$ is finite flat over $Y$ and $(Z_{i})^{{\rm red}}$ is smooth over $k$.

  * •

$Z_{i}\rightarrow X$ is a closed immersion and $\coprod_{i\in
I}Z_{i}\rightarrow X$ is an isomorphism over the complement of $S$ in $Y$.

By [KM85, 13.1.3], known as the “Crossings Theorem”, if $Y$ is connected, then
the $Z_{i}$ are the irreducible components of $X$. Furthermore, if each
$Z_{i}$ is reduced, then $X$ is also reduced.

######  Definition A.22.

In the situation just discussed, we say $X$ is the disjoint union of the
$Z_{i}$’s with crossings at the supersingular points.

######  Remark A.23.

Following [KM85, 13.1.7], we can relax the $k$-rationality of $x_{0}$ and
$y_{0}$ and instead require $k$-rationality after extending scalars to a
separable closure of $k$. In this situation, we still say $X$ is the disjoint
union of the $Z_{i}$’s with crossings at the supersingular points.

Next we describe $\bar{\mathfrak{X}}(Np^{r})$. We identify
$(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}$ as the subgroup

$\left\\{\left(\begin{array}[]{cc}u&0\\\
0&u\end{array}\right):u\in(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}\right\\}$

of ${\rm GL}_{2}(\mathbb{Z}/p^{r}\mathbb{Z})$. According to [KM85, 13.7.1],
the moduli problem $\left[\Gamma(p^{r})\right]^{{\rm can}}\otimes{k}$ on ${\rm
Ell}_{k}$ assigns to each elliptic curve $E\rightarrow S$ the set of all
Drinfeld $p^{r}$-bases $(P,Q)$ with $e_{p^{r}}(P,Q)=1$. Let
$\phi:(\mathbb{Z}/p^{r}\mathbb{Z})^{2}\rightarrow E[p^{r}]$ denote the
homomorphism of $S$-schemes corresponding to $(P,Q)$. Consider the diagram

$\textstyle{(\mathbb{Z}/p^{r}\mathbb{Z})^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\scriptstyle{\Lambda}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ker
F^{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E[p^{r}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F^{r}}$$\textstyle{\ker(V^{r})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

where $\Lambda=F^{r}\circ\phi$. By [KM85, 13.7.2(3)], a choice of
$\mathbb{Z}/p^{r}\mathbb{Z}$-basis of
$(\mathbb{Z}/p^{r}\mathbb{Z})^{2}/\ker\Lambda$ defines an isomorphism
$\mathbb{Z}/p^{r}\mathbb{Z}\overset{\sim}{\rightarrow}\ker(V^{r})$, allowing
us to view $\Lambda$ as a surjective homomorphism
$(\mathbb{Z}/p^{r}\mathbb{Z})^{2}\rightarrow\mathbb{Z}/p^{r}\mathbb{Z}$.

######  Definition A.24.

The component label of $\phi$ is the class of $\Lambda$ in

$(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}/{\rm
HomSurj}((\mathbb{Z}/p^{r}\mathbb{Z})^{2},\mathbb{Z}/p^{r}\mathbb{Z}).$

By [KM85, 13.7.4, 13.7.5], this establishes a canonical bijection between the
irreducible components of the special fiber and the set of component labels.
The following is [KM85, 13.7.6].

######  Theorem A.25.

The special fiber of $\mathfrak{X}(Np^{r})$ is the disjoint union, with
crossings at the supersingular points of $\mathfrak{X}(N)_{/k}$, of the exotic
Igusa curves ${\rm ExIg}(p^{r},r,N)$ over $\mathfrak{X}(N)_{/k}$ indexed by

$(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}/{\rm
HomSurj}((\mathbb{Z}/p^{r}\mathbb{Z})^{2},\mathbb{Z}/p^{r}\mathbb{Z}).$

Furthermore, $\bar{\mathfrak{X}}(Np^{r})$ is reduced.

Note that the claim $\bar{\mathfrak{X}}(Np^{r})$ is reduced comes from the
fact that ${\rm ExIg}(p^{r},r,N)$ is reduced, together with the “Crossings
Theorem” of [KM85, 13.1.3] and [KM85, 13.1.4].

Recall the closed subscheme of cusps $\mathfrak{C}(Np^{r})$ of
$\mathfrak{X}(Np^{r})$ is a disjoint union of $\\#{\rm HS}(Np^{r})$ many
sections of $\mathfrak{X}(Np^{r})$ which, by Proposition A.15, can be viewed
as the closure $\overline{\left\\{x\right\\}}$ of points
$x\in\mathfrak{C}(\mathfrak{X}(N)_{/K})$. By Corollary 2.15,
$\overline{\left\\{x\right\\}}$ intersects precisely one irreducible component
of $\bar{\mathfrak{X}}(Np^{r})$. The following, which is [KM85, 13.9.3], tells
us precisely the component label of the irreducible component which
$\overline{\left\\{x\right\\}}$ intersects with, given the index of
$\overline{\left\\{x\right\\}}$ in ${\rm HS}(Np^{r})$ (see the paragraph
preceding Proposition A.15).

######  Theorem A.26.

The natural projection

} \HomSurj((Z/NZ)^2,Z/NZ)×HomSurj((Z/p^rZ)^2,Z/p^rZ)$\textstyle{\left\\{\pm
I\right\\}\backslash{\rm
HomSurj}((\mathbb{Z}/N\mathbb{Z})^{2},\mathbb{Z}/N\mathbb{Z})\times{\rm
HomSurj}((\mathbb{Z}/p^{r}\mathbb{Z})^{2},\mathbb{Z}/p^{r}\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
---

assigns to a component $\overline{\left\\{x\right\\}}$ of
$\mathfrak{C}(Np^{r})$ indexed by ${\rm HS}(Np^{r})$ the irreducible component
of $\bar{\mathfrak{X}}(Np^{r})$ that $\overline{\left\\{x\right\\}}$
intersects with.

######  Corollary A.27.

Each irreducible component of $\bar{\mathfrak{X}}(Np^{r})$ intersects with the
same number of cuspidal components of $\mathfrak{C}(Np^{r})$.

Every group homomorphism
$\Lambda:(\mathbb{Z}/p^{r}\mathbb{Z})^{2}\rightarrow\mathbb{Z}/p^{r}\mathbb{Z}$
is uniquely determined by its image on the basis vectors $(1,0)$ and $(0,1)$.
Let
$\Lambda_{(a,b)}:(\mathbb{Z}/p^{r}\mathbb{Z})^{2}\rightarrow\mathbb{Z}/p^{r}\mathbb{Z}$
denote the map defined by

$\Lambda_{(a,b)}(1,0)=a\mbox{ and }\Lambda_{(a,b)}(0,1)=b.$

A complete list of representatives in
$(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}/{\rm
HomSurj}((\mathbb{Z}/p^{r}\mathbb{Z})^{2},\mathbb{Z}/p^{r}\mathbb{Z})$ is
given by

$\begin{cases}\Lambda_{(1,-a)}&a\in\mathbb{Z}/p^{r}\mathbb{Z}\\\
\Lambda_{(-pb,1)}&b\in\mathbb{Z}/p^{r-1}\mathbb{Z}\end{cases}.$

With this labeling, the following is clear:

######  Corollary A.28.

The special fiber $\bar{\mathfrak{X}}(Np^{r})$ has $p^{r}+p^{r-1}$ many
irreducible components.

We will record the number of cuspidal components of $\mathfrak{C}(Np^{r})$ an
irreducible component of $\bar{\mathfrak{X}}(Np^{r})$ intersects with. For
convenience, we let
$C(Np^{r})=\mathfrak{C}(\mathfrak{X}{}_{/\mathbb{Q}_{p}(\zeta_{Np^{r}}})$, the
set of all cusps of the generic fiber. The following is [Miy89, 4.2.10].

######  Lemma A.29.

For $M\geq 3$, we have

$\\#C(M)=\frac{1}{2}M^{2}\prod_{p\mid M}(1-1/p^{2})=\frac{1}{2M}\\#{\rm
SL}_{2}(\mathbb{Z}/MZ).$

We will also need to know how to compute $\\#{\rm
SL}_{2}(\mathbb{Z}/M\mathbb{Z})$. The following is from [Miy89, 4.2.3, 4.2.4].

######  Lemma A.30.

  1. a.

Let $A$ and $B$ be coprime integers. Then

${\rm SL}_{2}(\mathbb{Z}/AB\mathbb{Z})\simeq{\rm
SL}_{2}(\mathbb{Z}/A\mathbb{Z})\times{\rm SL}_{2}(\mathbb{Z}/B\mathbb{Z}).$

  2. b.

Let $p$ be prime and $r\geq 1$ an integer. Then

$\\#{\rm SL}_{2}(\mathbb{Z}/p^{r}\mathbb{Z})=p^{3r}-p^{3r-2}.$

######  Proposition A.31.

Each irreducible component of $\bar{\mathfrak{X}}(Np^{r})$ intersects
precisely

$\frac{\\#C(Np^{r})}{p^{r}+p^{r-1}}=\varphi(p^{r})\\#C(N)$

many cuspidal components of $\mathfrak{C}(Np^{r})$.

###### Proof.

By Corollary A.27, this quantity is independent of irreducible component
$\Lambda$. Hence the number of cuspidal components $\Lambda$ intersects is
equal to the total number of cusps of the generic fiber divided by the total
number of irreducible components. By Corollary A.28, the number of irreducible
components is $p^{r}+p^{r-1}$.

Using Lemma A.29 and Lemma A.30a, we compute:

$\displaystyle\frac{\\#C(Np^{r})}{p^{r}+p^{r-1}}$ $\displaystyle=\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/Np^{r}\mathbb{Z})}{2Np^{r}(p^{r}+p^{r-1})}$
$\displaystyle=\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/p^{r}\mathbb{Z})}{p^{r}(p^{r}+p^{r-1})}\cdot\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/N\mathbb{Z})}{2N}$ $\displaystyle=\frac{\\#{\rm
SL}_{2}(\mathbb{Z}/p^{r}\mathbb{Z})}{p^{r}(p^{r}+p^{r-1})}\\#C(N)$

Lastly, using Lemma A.30b, we have

$\displaystyle=\frac{p^{3r}-p^{3r-2}}{p^{r}(p^{r}+p^{r-1})}\\#C(N)$
$\displaystyle=\varphi(p^{r})\\#C(N).$ ∎

##  Appendix B Appendix – Circulant Matrices

######  Definition B.1.

An $n\times n$ circulant matrix $C$ over a field is any matrix of the form

$C=\left(\begin{array}[]{ccccc}c_{0}&c_{n-1}&\cdots&c_{2}&c_{1}\\\
c_{1}&c_{0}&\cdots&c_{3}&c_{2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
c_{n-2}&c_{n-3}&\cdots&c_{0}&c_{n-1}\\\
c_{n-1}&c_{n-2}&\cdots&c_{1}&c_{0}\end{array}\right)$

where each column is equal to the previous column shifted downward by 1,
looping around as appropriate.

The entries $c_{i,j}$ of an $n\times n$ circulant matrix can be characterized
by the equation

$c_{i,j}=c_{i-j+1,1}$

for all $1\leq i,j\leq n$ where the index $i-j+1$ is taken modulo $n$ among
the residue classes in $\left\\{0,1,\dots,n-1\right\\}$. Let
$\zeta_{n}=e^{2\pi i/n}$ denote a primitive $n$th root of unity. The following
provides an explicit description of the eigenvalues and corresponding
eigenvectors of a circulant matrix.

######  Lemma B.2.

The eigenvalues of a circulant matrix $C=(c_{ij})$ are precisely

$\lambda_{j}=\sum_{k=0}^{n-1}c_{k}\zeta_{n}^{(j-1)(n-k)}$

for $j=1,\dots,n$. A corresponding eigenvector of $\lambda_{j}$ is given by

$\vec{v}_{j}=\frac{1}{\sqrt{n}}\left(1,\zeta_{n}^{j-1},\zeta_{n}^{2(j-1)},\dots,\zeta_{n}^{(n-1)(j-1)}\right).$

###### Proof.

We will verify that $C\vec{v}_{j}=\lambda_{j}\vec{v}_{j}$ for all $j$. For
$1\leq i\leq n$, the $i^{{\rm th}}$ entry of $C\vec{v}_{j}$ is

$\frac{1}{\sqrt{n}}\left(c_{i-1}+c_{i-2}\zeta_{n}^{j-1}+c_{i-3}\zeta_{n}^{2(j-1)}+\cdots+c_{0}\zeta^{(i-1)(j-1)}+c_{n-1}\zeta_{n}^{i(j-1)}\cdots+c_{i}\zeta_{n}^{(n-1)(j-1)}\right)$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=0}^{i-1}c_{k}\zeta_{n}^{(i-1-k)(j-1)}+\frac{1}{\sqrt{n}}\sum_{\ell=i}^{n-1}c_{\ell}\zeta_{n}^{(n-1+i-\ell)(j-1)}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=0}^{i-1}c_{k}\zeta_{n}^{(n+i-1-k)(j-1)}+\frac{1}{\sqrt{n}}\sum_{\ell=i}^{n-1}c_{\ell}\zeta_{n}^{(n-1+i-\ell)(j-1)}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}c_{k}\zeta^{(n+i-1-k)(j-1)}$

while the $i^{{\rm th}}$ entry of $\lambda_{j}\vec{v}_{j}$ is

$\displaystyle\frac{1}{\sqrt{n}}\zeta_{n}^{(i-1)(j-1)}\sum_{k=0}^{n-1}c_{k}\zeta_{n}^{(j-1)(n-k)}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}c_{k}\zeta_{n}^{(j-1)(n-k)+(i-1)(j-1)}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}c_{k}\zeta_{n}^{(j-1)(n+i-1-k)}$

which agrees with the $i^{{\rm th}}$ entry of $C\vec{v}_{j}$ for all $i$. ∎

Circulant matrices are always diagonalizable (see [KW01, §2]) so we can write
$C=PDP^{-1}$ where the columns of $P$ are the eigenvectors
$\vec{v}_{1},\dots,\vec{v}_{n}$ and $D$ is a diagonal matrix whose diagonal
entries are the corresponding eigenvalues $\lambda_{1},\dots,\lambda_{n}$.
Explicitly,

$P=\frac{1}{\sqrt{n}}\left(\begin{array}[]{cccccc}1&1&1&\cdots&1&1\\\
1&\zeta_{n}&\zeta_{n}^{2}&\cdots&\zeta_{n}^{n-2}&\zeta_{n}^{n-1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
1&\zeta_{n}^{n-2}&\zeta_{n}^{(n-2)2}&\cdots&\zeta_{n}^{(n-2)(n-2)}&\zeta_{n}^{(n-2)(n-1)}\\\
1&\zeta_{n}^{n-1}&\zeta_{n}^{(n-1)2}&\cdots&\zeta_{n}^{(n-1)(n-2)}&\zeta_{n}^{(n-1)(n-1)}\end{array}\right).$

We see the $(i,j)$-entry of $P$ is equal to $\zeta_{n}^{(i-1)(j-1)}/\sqrt{n}$
for $1\leq i,j\leq n$.

######  Lemma B.3.

The matrix $P$ is unitary i.e. the inverse of $P$ is equal to the conjugate
transpose $P^{*}$.

###### Proof.

Note that $P$ is symmetric and the conjugate of $\zeta_{n}$ is
$\bar{\zeta}_{n}=\zeta_{n}^{-1}$. The $(i,j)$-entry of $PP^{*}$ is equal to

$\frac{1}{n}\sum_{k=0}^{n-1}\zeta_{n}^{k(i-1)}\bar{\zeta}_{n}^{k(j-1)}=\frac{1}{n}\sum_{k=0}^{n-1}\zeta_{n}^{k(i-1)}\zeta_{n}^{-k(j-1)}$
$=\frac{1}{n}\sum_{k=0}^{n-1}\zeta_{n}^{k(i-j)}=\begin{cases}1&\mbox{if
}i=j\\\ 0&\mbox{if }i\neq j\end{cases}.$

Hence $PP^{*}=I_{n}$ so $P^{*}=P^{-1}$. ∎

Now that we know $C=PDP^{*}$, we can describe the $(i,j)$ entry of $C^{-1}$,
if it exists.

######  Proposition B.4.

Let $C$ be an $n\times n$ invertible circulant matrix whose entries along the
first column are $c_{0},\dots,c_{n-1}$ in that order. Then the $(i,j)$-entry
of $C^{-1}$ is equal to

$\frac{1}{n}\sum_{k=1}^{n}\lambda_{k-1}^{-1}\zeta_{n}^{(k-1)(i-j)}$

where $\lambda_{j}=\sum_{m=0}^{n-1}c_{m}\zeta_{n}^{(j-1)(n-m)}$ are the
eigenvalues of $C$ for $1\leq j\leq n$.

###### Proof.

We compute out the product $C^{-1}=PD^{-1}P^{*}$. The matrix $PD^{-1}$ will be
of the form

$PD^{-1}=\frac{1}{\sqrt{n}}\left(\begin{array}[]{ccccc}\lambda_{1}^{-1}&\lambda_{2}^{-1}&\lambda_{3}^{-1}&\cdots&\lambda_{n}^{-1}\\\
\lambda_{1}^{-1}&\lambda_{2}^{-1}\zeta_{n}&\lambda_{3}^{-1}\zeta_{n}^{2}&\cdots&\lambda_{n}^{-1}\zeta_{n}^{n-1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\lambda_{1}^{-1}&\lambda_{2}^{-1}\zeta_{n}^{n-2}&\lambda_{3}^{-1}\zeta_{n}^{(n-2)2}&\cdots&\lambda_{n}^{-1}\zeta_{n}^{(n-2)(n-1)}\\\
\lambda_{1}^{-1}&\lambda_{2}^{-1}\zeta_{n}^{n-1}&\lambda_{3}^{-1}\zeta_{n}^{(n-1)2}&\cdots&\lambda_{n}^{-1}\zeta_{n}^{(n-1)(n-1)}\end{array}\right).$

Thus the $(i,j)$-entry of $PD^{-1}$ is equal to

$\lambda_{j}^{-1}\zeta_{n}^{(i-1)(j-1)}/\sqrt{n}$

and so the $(i,j)$-entry of $PD^{-1}P^{*}$ is equal to

$\frac{1}{n}\sum_{k=1}^{n}\lambda_{k}^{-1}\zeta_{n}^{(i-1)(k-1)}\zeta_{n}^{-(k-1)(j-1)}=\frac{1}{n}\sum_{k=1}^{n}\lambda_{k}^{-1}\zeta_{n}^{(k-1)(i-j)}.\qed$

Using our explicit description of the entries of $C^{-1}$, we can establish
the following.

######  Corollary B.5.

The inverse of an invertible circulant matrix is circulant.

###### Proof.

Let $C=(c_{i,j})$ be an $n\times n$ circulant matrix whose entries along the
first column are ordered $c_{0},c_{1},\dots,c_{n-1}$ so that $c_{m+1,1}=c_{m}$
for $0\leq m\leq n-1$. By Lemma B.4, the $(i,j)$-entry of $C^{-1}$ is equal to

$c^{i,j}=\frac{1}{n}\sum_{k=1}^{n}\left(\sum_{m=0}^{n-1}c_{m}\zeta_{n}^{(k-1)(n-m)}\right)\zeta_{n}^{(k-1)(i-j)}.$

To show $C^{-1}$ is circulant, we will establish the relationship
$c^{i,j}=c^{i-j+1,1}$ where the index $i-j+1$ is taken modulo $n$ in the
residue class $\\{0,\dots,n-1\\}$. In the above expression for $c^{i,j}$, the
index $(i,j)$ only appears in the term $\zeta_{n}^{(k-1)(i-j)}$. Observe that

$\zeta_{n}^{(k-1)((i-j+1)-1)}=\zeta_{n}^{(k-1)(i-j)}.$

Hence $c^{i,j}=c^{i-j+1,1}$ so $C^{-1}$ is circulant. ∎

##  Appendix C Appendix – Inverse via the Woodbury Matrix Identity

Let $A$ be an $n\times n$ invertible matrix, $C$ an invertible $k\times k$
matrix where $k\leq n$, $U$ an $n\times k$ matrix, and $V$ a $k\times n$
matrix. The Woodbury matrix identity states

$(A+UCV)^{-1}=A^{-1}-A^{-1}U(C^{-1}+VA^{-1}U)^{-1}VA^{-1}.$

A proof of this identity can be found in [HS81, §1.3]. This identity allows us
to compute the inverse of $A+UCV$ provided we can easily compute the inverses
of $A$ and $C^{-1}+VA^{-1}U$.

In general, we can use the Woodbury Matrix Identity to compute the inverse of
$A+N$ where $A$ is an $n\times n$ invertible matrix and $N$ is an $n\times n$
matrix of rank $k$, provided $A+N$ is invertible. Let $U$ be the $n\times k$
matrix whose columns $\vec{v}_{1},\dots,\vec{v}_{k}$ are the $k$ linearly
independent columns of $N$. Let $\vec{u}_{i}$ denote the vector in the $i$th
column of $N$ which can be expressed as

$\vec{u}_{i}=c_{1}\vec{v}_{1}+\cdots+c_{k}\vec{v}_{k}.$

We define the $i$th column of the $k\times n$ matrix $V$ to consist of entries
$c_{1},\dots,c_{k}$ in that order. Consequently, $N=UCV$ where $C=I_{k}$ is
the $k\times k$ identity matrix.

Using the Woodbury Matrix Identity, we will provide a formula for the inverse
in the following situation as encountered in Section 3.3. Let $A$ be a block
diagonal matrix with two invertible blocks of sizes $n\times n$ and $m\times
m$. The inverse $A^{-1}$ is also block diagonal with blocks of the same size.
We write

$A^{-1}=\left(\begin{array}[]{cc}\begin{array}[]{ccc}a_{11}&\cdots&a_{1n}\\\
\vdots&\ddots&\vdots\\\ a_{n1}&\cdots&a_{nn}\end{array}&\mathbf{0}\\\
\mathbf{0}&\begin{array}[]{ccc}b_{11}&\cdots&b_{1m}\\\ \vdots&\ddots&\vdots\\\
b_{m1}&\cdots&b_{mm}\end{array}\end{array}\right).$

We also consider the case

$U=\left(\begin{array}[]{cc}0_{1}&1_{1}\\\ 0_{2}&1_{2}\\\ \vdots&\vdots\\\
0_{n}&1_{n}\\\ 1_{1}&0_{1}\\\ \vdots&\vdots\\\
1_{m}&0_{m}\end{array}\right)\mbox{ and
}V=\left(\begin{array}[]{ccccccc}1_{1}&1_{2}&\cdots&1_{n}&0_{1}&\cdots&0_{m}\\\
0_{1}&0_{2}&\cdots&0_{n}&1_{1}&\cdots&1_{m}\end{array}\right)$

where the subscripts on the $0$ and $1$ entries are there to help keep track
of their position.

######  Proposition C.1.

Let $T=A+UV$ where $A,U,$ and $V$ are the given matrices above. Let $c^{i,j}$
denote the $(i,j)$-entry of $T^{-1}$. We have

$c^{ij}=\begin{cases}{\displaystyle
a_{i,j}+\frac{\beta}{1-\alpha\beta}\left(\sum_{k=1}^{n}a_{i,k}\right)\left(\sum_{k=1}^{n}a_{k,j}\right)}&\mbox{if
}1\leq i,j\leq n\\\
{\displaystyle\frac{-1}{1-\alpha\beta}\left(\sum_{k=1}^{n}a_{i,k}\right)\left(\sum_{k=1}^{m}b_{k,j}\right)}&\mbox{if
}1\leq i\leq n\mbox{ and }n<j\leq n+m\\\
{\displaystyle\frac{-1}{1-\alpha\beta}\left(\sum_{k=1}^{m}b_{i,k}\right)\left(\sum_{k=1}^{n}a_{k,j}\right)}&\mbox{if
}1\leq j\leq n\mbox{ and }n<i\leq n+m\\\ {\displaystyle
b_{i,j}+\frac{\alpha}{1-\alpha\beta}\left(\sum_{k=1}^{m}b_{i,k}\right)\left(\sum_{k=1}^{m}b_{k,j}\right)}&\mbox{if
}n<i,j\leq n+m\end{cases}$

where $\alpha=\sum a_{i,j}$ is the sum of all entries in the first block in
$A^{-1}$ and $\beta=\sum b_{i,j}$ is the sum of all entries in the second
block in $A^{-1}$.

###### Proof.

By the Woodbury Matrix Identity,

$T^{-1}=A^{-1}-A^{-1}U(I_{2}+VA^{-1}U)^{-1}VA^{-1}.$

We first compute

$\displaystyle VA^{-1}U$
$\displaystyle=V\left(\begin{array}[]{cc}\begin{array}[]{ccc}a_{11}&\cdots&a_{1n}\\\
\vdots&\ddots&\vdots\\\ a_{n1}&\cdots&a_{nn}\end{array}&\mathbf{0}\\\
\mathbf{0}&\begin{array}[]{ccc}b_{11}&\cdots&b_{1m}\\\ \vdots&\ddots&\vdots\\\
b_{m1}&\cdots&b_{mm}\end{array}\end{array}\right)\left(\begin{array}[]{cc}0_{1}&1_{1}\\\
\vdots&\vdots\\\ 0_{n}&1_{n}\\\ 1_{1}&0_{1}\\\ \vdots&\vdots\\\
1_{m}&0_{m}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cccccc}1_{1}&\cdots&1_{n}&0_{1}&\cdots&0_{m}\\\
0_{1}&\cdots&0_{n}&1_{1}&\cdots&1_{m}\end{array}\right)\left(\begin{array}[]{cc}0&{\displaystyle{\displaystyle\sum_{i=1}^{n}a_{1i}}}\\\
\vdots&\vdots\\\ 0&{\displaystyle{\displaystyle\sum_{i=1}^{n}a_{n,i}}}\\\
{\displaystyle\sum_{i=1}^{m}b_{1,i}}&{\displaystyle 0}\\\ \vdots&\vdots\\\
{\displaystyle{\displaystyle\sum_{i=1}^{m}b_{m,i}}}&0\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}0&{\displaystyle{\displaystyle\sum_{i,j=1}^{n}a_{ij}}}\\\
{\displaystyle{\displaystyle\sum_{i,j=1}^{m}b_{ij}}}&0\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}0&\alpha\\\
\beta&0\end{array}\right).$

Hence

$(I_{2}+VA^{-1}U)^{-1}=\left(\begin{array}[]{cc}1&\alpha\\\
\beta&1\end{array}\right)^{-1}=\frac{1}{1-\alpha\beta}\left(\begin{array}[]{cc}1&-\alpha\\\
-\beta&1\end{array}\right).$

Next we will compute

$\displaystyle U(I_{2}^{-1}+VA^{-1}U)^{-1}V$
$\displaystyle=\frac{1}{1-\alpha\beta}U\left(\begin{array}[]{cc}1&-\alpha\\\
-\beta&1\end{array}\right)\left(\begin{array}[]{cccccc}1_{1}&\cdots&1_{n}&0_{1}&\cdots&0_{m}\\\
0_{1}&\cdots&0_{n}&1_{1}&\cdots&1_{m}\end{array}\right)$
$\displaystyle=\frac{1}{1-\alpha\beta}\left(\begin{array}[]{cc}0_{1}&1_{1}\\\
\vdots&\vdots\\\ 0_{n}&1_{n}\\\ 1_{1}&0_{1}\\\ \vdots&\vdots\\\
1_{m}&0_{m}\end{array}\right)\left(\begin{array}[]{cccccc}1_{1}&\cdots&1_{n}&-\alpha&\cdots&-\alpha\\\
-\beta&\cdots&-\beta&1_{1}&\cdots&1_{m}\end{array}\right)$
$\displaystyle=\frac{1}{1-\alpha\beta}\left(\begin{array}[]{cccccc}-\beta&\cdots&-\beta&1&\cdots&1\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
-\beta&\cdots&-\beta&-1&\cdots&-1\\\ 1&\cdots&1&-\alpha&\cdots&-\alpha\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
1&\cdots&1&-\alpha&\cdots&-\alpha\end{array}\right).$

Lastly, we compute

$\displaystyle A^{-1}U(I_{2}^{-1}+VA^{-1}U)^{-1}VA^{-1}$
$\displaystyle=\frac{1}{1-\alpha\beta}\left(\begin{array}[]{cccccc}a_{11}&\cdots&a_{1,n}&0&\cdots&0\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
a_{n,1}&\cdots&a_{n,n}&0&\cdots&0\\\ 0&\cdots&0&b_{11}&\cdots&b_{1,m}\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
0&\cdots&0&b_{m,1}&\cdots&b_{m,m}\end{array}\right)\left(\begin{array}[]{cccccc}-\beta&\cdots&-\beta&1&\cdots&1\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
-\beta&\cdots&-\beta&1&\cdots&1\\\ 1&\cdots&1&-\alpha&\cdots&-\alpha\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
1&\cdots&1&-\alpha&\cdots&-\alpha\end{array}\right)A^{-1}$
$\displaystyle=\frac{1}{1-\alpha\beta}\left(\begin{array}[]{cccccc}-\beta\sum
a_{1,\ell}&\cdots&-\beta\sum a_{1,\ell}&\sum a_{1,\ell}&\cdots&\sum
a_{1,\ell}\\\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\ -\beta\sum
a_{n,\ell}&\cdots&-\beta\sum a_{n,\ell}&\sum a_{n,\ell}&\cdots&\sum
a_{n,\ell}\\\ \sum b_{1,\ell}&\cdots&\sum b_{1,\ell}&-\alpha\sum
b_{1,\ell}&\cdots&-\alpha\sum b_{1,\ell}\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\ \sum b_{m,\ell}&\cdots&\sum
b_{m,\ell}&-\alpha\sum b_{m,\ell}&\cdots&-\alpha\sum
b_{m,\ell}\end{array}\right)A^{-1}.$

We now describe the $(i,j)$-entry of the above product in four different
cases.

Case 1($1\leq i,j\leq n$): The $(i,j)$-entry is equal to

$\frac{1}{1-\alpha\beta}\sum_{k=1}^{n}\left(-\beta\sum_{m=1}^{n}a_{i,m}\right)a_{k,j}=-\frac{\beta}{1-\alpha\beta}\left(\sum_{k=1}^{n}a_{i,k}\right)\left(\sum_{k=1}^{n}a_{k,j}\right).$

Case 2 ($1\leq i\leq n$ and $n<j\leq n+m$): The $(i,j)$-entry is equal to

$\frac{1}{1-\alpha\beta}\sum_{k=1}^{m}\left(\sum_{m=1}^{n}a_{i,m}\right)b_{k,j}=\frac{1}{1-\alpha\beta}\left(\sum_{k=1}^{n}a_{i,k}\right)\left(\sum_{k=1}^{m}b_{k,j}\right).$

Case 3 ($1\leq j\leq n$ and $n<i\leq n+m$): The $(i,j)$-entry is equal to

$\frac{1}{1-\alpha\beta}\sum_{k=1}^{n}\left(\sum_{m=1}^{m}b_{i,m}\right)a_{k,j}=\frac{1}{1-\alpha\beta}\left(\sum_{k=1}^{m}b_{i,k}\right)\left(\sum_{k=1}^{n}a_{k,j}\right).$

Case 4 ($n<i,j\leq n+m$): The $(i,j)$-entry is equal to

$\frac{1}{1-\alpha\beta}\sum_{k=1}^{p^{r-1}}\left(-\alpha\sum_{m=1}^{m}b_{i,m}\right)b_{k,j}=-\frac{\alpha}{1-\alpha\beta}\left(\sum_{k=1}^{m}b_{i,k}\right)\left(\sum_{k=1}^{m}b_{k,j}\right).$

Subtracting these quantities from corresponding $(i,j)$-entry of $A^{-1}$
gives our desired result. ∎

## Acknowledgements

Most of the results in this paper originated from my PhD thesis. I would like
to thank Bryden Cais for his guidance, support, and incredibly helpful
feedback. I would also like to thank Brandon Levin, Doug Ulmer, and Hang Xue
for their helpful comments. This research did not receive any specific grant
from funding agencies in the public, commercial, or not-for-profit sectors.

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