\mychapter{Integrality Properties of Singular Moduli}
In this chapter, we prove Theorem \ref{mainthm1}, which is joint work with Michael Griffin in \cite{Griffin-Rolen}. The chapter is organized as follows. In $\S\ref{proofspecdecomp}$ we use the Maass lowering and raising operators to prove our ``spectral decomposition''. This has the effect of relating our problem to the study of \emph{traces} of singular moduli for negative weight modular forms. In $\S\ref{IntegralityDukeJenkins}$, we recall the important work of Duke-Jenkins on integrality for traces of modular forms of negative weight. In $\S\ref{RCBrackets}$, we use Rankin-Cohen brackets and a basic binomial sum identity to study which forms in the decomposition actually appear. In $\S\ref{Padic_Poincare}$ we use the theory of Poincar\'e series and develop a $p$-adic theory using special families of Hecke operators to deal with cusp forms arising in the ``dual weight'' and take care of complicated denominators introduced in the spectral decomposition of $\S\ref{specdecomp}$.

Thanks to \emph{Newton's identities}, one can express elementary symmetric functions in terms of sums of powers and vice versa.  We recall that if \\ $e_k(x_1,\ldots,x_n)$ is the usual elementary symmetric polynomial in $x_1,\ldots,x_n$ and \\ $p_k(x_1,\ldots,x_n):=\displaystyle\sum_{i=1}^nx_i^k$ is the $k^{th}$ \emph{power sum}, then the Newton-Girard formulae state that \begin{equation}\label{Newton}ke_k(x_1,\ldots,x_n)=\displaystyle\sum_{i=1}^k(-1)^{i-1}e_{k-i}(x_1,\ldots,x_n)p_i(x_1,\ldots,x_n).\end{equation}
Thus, our problem is reduced to the of study of traces of singular moduli for powers of Maass forms.  Unfortunately, products and powers of Maass forms typically are not Maass forms, or even finite sums of Maass forms. However we are only studying a special class of Maass forms which are simply derivatives of modular forms. In this case, the problem has a straightforward solution. 
\\

We note that products and sums of raisings of modular forms must be in the kernal of a finite power of $L$. This allows us to decompose such forms as per the following theorem. This theorem is originally due to Shimura  (see \cite{ShimuraNearlyHol} Proposition 3.4, or \cite{Hida} Section 10.1), however we give a short proof which gives explicit components which we will need later.
\begin{thm}[Shimura \cite{ShimuraNearlyHol}]\label{specdecomp} Suppose $F$ smooth function n $\mathbb H$ which is in the kernal of $L^{E+1}$ (and not in the kernal of $L^E$). Then there exist uniquely determined modular forms $g_j\in M_{k-2j}^!$ such that \begin{equation}\label{specinduct}F=\displaystyle\sum_{j=0}^{E}R^jg_j.\end{equation} 
\end{thm}

\section{The Spectral Decomposition}\label{proofspecdecomp}
Here we prove Theorem \ref{specdecomp}.
\begin{proof}
Since $L^{E+1}F=0$, we have that $L^{E}F$ is weakly holomorphic. We define the $g_i$ recursively beginning with $g_E$.
Let \[g_E=\frac{L^EF}{c_{E,E}},\] and for each $i$ with $0\leq i< E$, let 
\begin{equation}
g_i:=\frac{1}{c_{i,i}}\left( L^iF-\sum_{j=i+1}^E c_{i,j} R^{j-i}g_j \right),
\end{equation}

\noindent where \begin{equation}c_{i,j}:=\frac{j!(-k+j+i)!}{(j-i)!(-k+j)!}.\end{equation} By assumption, $k\leq0$, so $c_{i,j}$ is defined for all $j\geq i$. Note that each $g_i$ is modular of weight $k-2i$. By rearranging the definition of $g_0$, we see that $F=\sum_{i=0}^ER_ig_i$. Therefore we need only prove that each $g_i$ is weakly holomorphic. We do so by inductively showing that $Lg_i=0$ for each $g_i$.

We have already seen that $Lg_E=0$. Suppose that $i<E$ is fixed, and that it is known that $g_j$ is weakly holomorphic for each $i<j\leq E$. By construction, we have that for any fixed $i$,
\[L^iF=\sum_{j=i}^E c_{i,j} R^{j-i}g_j.\]
Applying the lowering operator gives
\[L^{i+1}F=\sum_{j=i}^E c_{i,j}LR^{j-i}g_j=\sum_{j=i+1}^E c_{i+1,j} R^{j-i-1}g_j.\]
This rearranges to 
\[c_{i,i}Lg_i=\sum_{j=i+1}^E \left(c_{i+1,j} R^{j-i-1}g_j-c_{i,j}(LR)R^{j-i-1}g_j\right).\]
Since for $j>i$ we have that $g_i$ is weakly holomorphic, a short calculation shows that \[(LR)R^{j-i-1}g_j=(j-i)(-k+j+i+1)R^{j-i-1}g_j.\] Hence we have that
\[c_{i,i}Lg_i=\sum_{j=i+1}^E \left(c_{i+1,j}-c_{i,j}(j-i)(-k+j+i+1) \right)R^{j-i-1}g_j.\]
By the definition of the $c_{i,j}$, we see that this sum is zero.
\end{proof}

\section{Integrality Results of Duke and Jenkins}\label{IntegralityDukeJenkins}
In this section we describe the important work of Duke and Jenkins on integrality of traces of singular moduli in \cite{DukeJenkins}. In particular, their results allow us to predict the correct denominators of the traces of singular moduli of each summand arising in the relevant case of Theorem \ref{specdecomp} in our spectral decomposition. Following their paper, consider any $f=\displaystyle\sum_{n\gg-\infty}a(n)q^n\in M_{2-2s}^!$ for $1\leq s\in\Z$. For convenience, set 
%1 deleted upper oo bound.
\begin{equation}
\widehat{s}:=\begin{cases}s &\mbox{ if } (-1)^sD>0\\1-s &\mbox{ otherwise}.\end{cases}
\end{equation} They also let
\begin{equation}
\label{tr*}\operatorname{Tr}^*_{d,D}(f):=(-1)^{\lfloor\frac{\widehat{s}-1}{2}\rfloor}|d|^{\frac{-\widehat{s}}{2}}|D|^{\frac{\widehat{s}-1}{2}}\operatorname{Tr}_{d,D}((-1)^{s-1}\partial f).
\end{equation}%2 put in (-1)^{s-1}
 Then for any fundamental discriminant $D$, they define the \emph{$D^{th}$ Zagier lift} of $f$ to be:

 \begin{equation}\label{ZagierLift}\begin{split}
\mathfrak{Z}_D(f):=\displaystyle\sum_{m>0}a(-m)m^{s-\widehat{s}}\displaystyle\sum_{n|m}\chi_D(n)n^{\widehat{s}-1}q^{-\frac{m^2|D|}{n^2}}+\frac{1}{2}L(1-s,\chi_D)a(0)\\ 
+\displaystyle\sum_{dD<0}\operatorname{Tr}^*_{d,D}(f)q^{|d|}.
\end{split}
\end{equation}

 Their main theorem states that the linear map $\mathfrak{Z}(\cdot)$ preserves integrality of Fourier coefficients. More specifically, they show the following.

\begin{thm*}[Duke-Jenkins \cite{DukeJenkins}] Suppose that $f\in M_{2-2s}^!$ for an integer $s\geq 2$. If $D$ is a fundamental discriminant with $(-1)^sD>0$, then we have that $\mathfrak{Z}_D(f)\in M_{3/2-s}^!$, whereas if $(-1)^sD<0$, then $\mathfrak{Z}_D(f)\in M_{s+1/2}^!$. Furthermore, if $f$ has integral Fourier coefficients, so does $\mathfrak{Z}_D(f)$. 
\end{thm*}
This builds on Zagier's original work for $s=1$. In that case the theorem and (\ref{ZagierLift}) hold, as long as the constant term of $f$ is $0$. Note that $\operatorname{Tr}_{d,1}(1)$ is the Hurwitz-Kronecker class number for $d$. It is well known that this is integral for $d$ fundamental, unless $d=-3$ or $-4$. In these cases we have that $\operatorname{Tr}_{-3,1}(1)=\frac{1}{3}$ and $\operatorname{Tr}_{-4,1}(1)=\frac{1}{2}$. Therefore for the remainder of this paper we will assume that the constant term of the weight $0$ component in future decompositions is zero.

\section{A Useful Vanishing Condition}\label{RCBrackets}
Using the Newton identities, Theorem \ref{specdecomp}, and the theorem of Duke and Jenkins, one can prove a bound on the denominators of symmetric functions in our singular moduli by bounding each summand in Theorem \ref{specdecomp} individually. This falls short of Theorem \ref{mainthm1}. There are two obstructions. Firstly, the bounds for the denominators given by Duke-Jenkins applied to forms of certain weights in the allowed range exceed the bound given in Theorem \ref{mainthm1}. Secondly, the coefficients $c_{i,j}$ which appear in Theorem \ref{specdecomp} could potentially contribute to the denominators. Here we address the first obstruction by showing that forms of certain ``bad weight'' in the decomposition must be identically zero. When $k=0$, there are no bad weights. Otherwise, we make the following definition:
\begin{dfn} Let $k$ be a negative, even integer and $n\in\N$. Then we say that $m$ is a \emph{bad weight} for the pair $(k,n)$ if $m$ is an integer of the form $kn+4i+2$ for $0\leq i\leq-\frac{k}{2}-1$.
\end{dfn}
We have the following theorem:
\begin{thm}\label{vanishing} Let $f\in M_k^!$ and consider the product $F=(\partial f)^n$. Write $F=\sum \partial(g_i)$, where each $g_i$ is a weakly holomorphic modular form as in Theorem \ref{specdecomp}. Then if $m$ is a bad weight for $k,n$, we have that $g_m\equiv0$.%9 change f to g_m
\end{thm}
This theorem arises from a combinatorial fact which can be applied to arbitrary power series. In the next subsection, we define a family of bilinear operators on modular forms known as the  \emph{Rankin-Cohen brackets}. These will provide a convenient basis for expressing the combinatorics of the spectral decomposition of the product of two forms.
\subsection{ Rankin-Cohen Brackets}
In \cite{rankin} and [13], Rankin and Cohen studied the general question of which polynomials in derivatives of two modular forms are again modular. In doing so, they described the so-called \emph{Rankin-Cohen Brackets}, which are defined as follows. Let $f$ be a modular form of weight $k$, $g$ a modular form of weight $\ell$, and $n$ a non-negative integer. Then the $n^{th}$ Rankin-Cohen bracket is defined as:
\begin{equation}
[f,g]_n^{(k,\ell)}:=\displaystyle\sum_{r+s=n}(-1)^r\binom{n+k-1}{s}\binom{n+\ell-1}{r}f^{(r)}\cdot g^{(s)}.\end{equation}%10 added a bang on the last one
%4 put in $f$ in equation below
Here $f^{(n)}:=\left(\frac{1}{2\pi i}\frac{d}{dz}\right)^nf$. We will also suppress the dependence on the weights $k$ and $\ell$ and write simply $[f,g]_n$ when the dependence is clear from context. The key fact is that $[f,g]_n$ gives a map: \begin{equation}[\cdot,\cdot]_n^{(k),(\ell)}:\ M_k^!\otimes M_{\ell}^!\rightarrow M^!_{k+\ell+2n}.\end{equation} This can be seen, for example, using the Cohen-Kuznetsov lifting to Jacobi-like forms. Moreover, it can be shown that these operators are essentially the \emph{unique} universal bilinear differential operators between these spaces of modular forms. That is, given a polynomial in the derivatives of $f$ and $g$ of degree at most $2$ which is modular of weight $k+\ell+2n$, it must be a multiple of $[f,g]_n$.

\section{The Vanishing Lemma}
Using the Rankin-Cohen brackets as defined above, we are now in position to prove Theorem \ref{vanishing}. Using an inductive argument and the spectral decomposition of Theorem \ref{specdecomp}, it suffices to prove the following lemma for the product of raisings of just two forms of possibly different weights.
\begin{lemm}\label{vanishing2}
Let $f\in M_k^!$ and $g\in M_{\ell}^!$ have negative even weight. Set $F:=\partial f\cdot\partial g$ and write $F=\sum \partial(g_i)$ for the modular forms $g_i$ defined in Theorem $\ref{specdecomp}$. Suppose $m=k+\ell+4i+2$ where $0\leq i\leq-\frac{\min{\{k,\ell\}}}{2}-1$. Then if $g_i$ has weight $m$, we have that $g_i\equiv0$. \end{lemm}
\begin{proof} In Proposition 2.3 of \cite{B-J-T-X}, the authors consider a similar combinatorial expansion which is given in terms of Rankin-Cohen brackets. Using their proposition, it suffices to prove the following (setting $k=-2r$, $\ell=-2s$) whenever $j<r$ and $j$ is odd:
\begin{equation}S(j):=\displaystyle\sum_{m=0}^s(-1)^{(j+m)}\cdot\frac{\binom{m+r}{j}\binom{s}{m}\binom{m-r-1}{r+m-j}}{\binom{-r-2s+m+j-1}{m+r-j}}=0.\end{equation}
Using the Wilf-Zeilberger method \cite{A=B}, one finds that the function $S(j)$ satisfies the following recursion in the range $j<r$:
\[(2+j)(1+j-2r)(1+j-2s)(j-2r-2s)\cdot S(j+2)\]\[-4(1+2j-2r-2s)(3+2j-2r-2s)(j-r-s)(1+j-r-s)\cdot S(j)=0.\] For the base case, $j=1$, we must show that $g_{E-1}$ vanishes in the notation of Theorem \ref{specdecomp}. A calculation shows that $L^{E-1}[(\partial f)\cdot(\partial g)]$ is some nonzero multiple of $R(f\cdot g)$, so that by (\ref{specinduct}), we have that $g_{E-1}$ is a multiple of $Rg_E$. But $g_{E-1}$ is holomorphic, whereas $Rg_E$ is not, so in fact we must have \noindent$g_{E-1}\equiv0$. Alternatively, the identity can be proven automatically for $j=1$ with another application of the Wilf-Zeilberger method.
\end{proof}


\section{Hecke Structure of the Zagier Lift, and a Special Family of Operators.}
In this section we will complete the proof of Theorem \ref{mainthm1}. We will need to define a notion of integrality for coefficients of the nonholomorphic modular forms we have been studying. We say that a nonholomorphic modular form $F$ of integral weight (resp. half-integral weight) of level $1$ (resp. level $4$) has integral coefficients if $F\in \Z((q))\left[\frac1{4\pi y}\right]$ $\left(\textrm{resp. }F\in \Z((q))\left[\frac1{16\pi y}\right]\right)$. We may also refer to the coefficient of $\left(\frac1{4\pi y}\right)^0$ as the holomorphic part of $F$. We define rational coefficients for such forms similarly. Given this notation, we have the following:

\begin{theorem}\label{Integrality_of_Trace}
Suppose \[F=\displaystyle \sum_{n=0}^NR^nf_{-2n}\] has integral coefficients, where $f_{-2n}\in M^!_{-2n}$  If $d<0$ is a fundamental discriminant, which is not divisible by any bad primes for the pair $(k,n$), then $d^e\operatorname{Tr}_{d,1}(F)\in \Z,$
 where $e=\min\{\ell \mid f_{-4\ell}\not \equiv 0\}$ or $0$ if no such $\ell$ exist.%5 fixed def of e
\end{theorem}

The proof of this theorem, requires an analog of the Zagier lift for the function $F$. This may be done by taking the Zagier lift of each of the $f_{-2n}$ and combining them in an appropriate way.  Let $e_1=\max\{0\}\cup\{\ell \mid f_{-4\ell}\not \equiv 0\},$ and $e_2=\max\{0\}\cup\{\ell \mid f_{2-4\ell}\not \equiv 0\}.$ Then we define 
\begin{equation}\label{Z_F}
\mathcal Z_D(F):=\sum R^{i+e_1} \mathfrak Z_D(f_{2-4i})+\sum R^{e_1-i} \mathfrak Z_D(f_{-4i}),
\end{equation}
which has weight $\frac32 +2e_1$.  If $d$ is a negative fundamental discriminant, then the coefficient of $q^{|d|}$ in $\mathcal Z_1(F)$ is \begin{equation}(-1)^{\floor{\frac{\widehat s-1}{2}}}|d|^{-\frac{\widehat s}{2}}\operatorname{Tr}_{d,1}(F),\end{equation} where $\widehat s=-2e_1$. Thus, Theorem \ref{Integrality_of_Trace} is equivalent to the statement that the holomorphic part of $\mathcal Z_1(F)$ has integral coefficients. Note that, up to a sign, the coefficient of $q^d$ in $\mathcal Z_1(F)$ is the coefficient of $q^1$ in $\mathcal Z_d(F)$. We prove this in two parts using the following lemmas:

\begin{Lemma}\label{Integrality_from_principal_part}
Suppose $F$ is a nonholomorphic modular function with rational coefficients and whose holomorphic part has $p$-integral principal part for some good prime $p$. Then $F$ has $p$-integral coefficients.
\end{Lemma}%.. modular function arising from the Hecke algebra of the Poincar\'e series and raising operators

As noted in the remark following Theorem \ref {mainthm1}, we need only worry about the case where $e_2>e_1$. Therefore, we can conclude the proof with the following:

\begin{prop}\label{Integrality_of_princ_part}
Suppose $F$ has integral coefficients for a good prime $p$ and a decomposition with $e_2>e_1$. Then for any negative discriminant $d$, the principal part of $\mathcal Z_d(F)$ is $p$-integral.
\end{prop}

In the following subsection, we set-up the structure of the Hecke algebra acting on Zagier lifts of certain families of Poincar\'e series in preparation for the proof of Lemma \ref{Integrality_from_principal_part}. This is used in Section \ref{Padic_Poincare} to construct a certain family of special family of Hecke operators. The final sections complete the proofs of Lemma \ref{Integrality_from_principal_part} and Proposition \ref{Integrality_of_princ_part} respectively.



\section{Poincar\'e Series and the Hecke Structure of Zagier Lifts}\label{Padic_Poincare}


The work in this paper relies heavily on the construction of the Zagier lift given by Duke and Jenkins in their proof of integrality of Zagier lifts mentioned above \cite{DukeJenkins}. Their original construction of the lift is given as explicit linear combinations of Maass-Poincar\'e series, which are then expressed via the Hecke algebra. 

The Maass-Poincar\'e series we need were studied by Fay~\cite{Fay} in the  integral weight case, and by Bringmann and Ono~\cite{ono96} in the half-integral weight case. In the half-integral weight case, we restrict our attention to series in the Kohnen plus-space. That is, we assume they are supported on coefficients $q^n$ where $n\equiv 0,$ or $(-1)^{k+\frac 12}\pmod 4$. For each $k\in \frac12+\Z$, Let $f_{k,m}$ be the unique Maass-Poincar\'e series in one of these families, of weight $k$ and of the form $q^{-m}+O(1)$ if $k< 0$ or $\frac 32$ or of the form $q^{-m}+O(q)$ otherwise. The main results we need from these works may be summarized as follows:

\begin{prop}
If $M_k$ is empty, then any form $f\in M_k^!$ may be expressed as a linear combination of the (harmonic) Poincar\'e series $\{f_{k,m}\}$. If $S_{2-k}$ is empty, $\{f_{k,m}\}\subset M_k^!$
\end{prop} 

Here, if $k\in\frac12+\Z$, then $M^!_k$ denotes the subspace of $M^!_k\left(\Gamma_0(4)\right)$ in the Kohnen plus-space.
The Hecke operators preserve theses spaces of Maass-Poincar\'e series. Since $\{f_k\}$ spans $M_k^!$ for negative $k$, we have the following proposition.

\begin{prop}\label{Hecke-exp}
Let $k$ be a negative even integer, and let$F=\displaystyle \sum_{n\in \Z} a(n)q^n\in M^!_k$. \vspace{-0.15in} Then we have that \begin{equation}F=\sum_{n<0} a(n)n^{1-k}f_{k,1}|_kT(n).
\end{equation}
\end{prop}

The same is true if $k=0$ and $F$ has no constant term. The $n^{k-1}$ in the formula above is necessary to clear denominators introduced by the standard Hecke operator on non-positive weights.

Proposition \ref{Hecke-exp} does not extend directly to half-integral weights. In particular, the Hecke operators with non-square index map half-integral weight forms identically to $0$. The action of $T(p^2)$ on half-integral forms is given by
\begin{equation}F|_kT(p^2)=F|U(p^2)+ \left(\frac{(-1)^{k-\frac12}}{p}\right)p^{k-\frac32}F+ p^{2k-2}F|V(p^2).\end{equation}
(In the case $p=2$, we have taken an involution of the standard Hecke operator which would just be $U_4$ as we are working in level $4$). 

Duke and Jenkins showed that if $d$ is a fundamental discriminant and $k$ a negative even integer, then 
\begin{equation}
\mathfrak Z_d( f_{k,1})=f_{\widehat k,d},
\end{equation} 
where $\widehat k:=\frac{k+1}{2}$ or $\frac{-k+3}{2}$. They also show that the Zagier lift commutes with the Hecke algebra. More precisely, we have 

\begin{eqnarray}\label{Zagier-Hecke1}
\mathfrak Z_D\left(n^{1-k}f|_k T(n)\right)=n^{1-k}(\mathfrak Z_D f)|_{\frac{k+1}{2}} T\left(n^2\right)
\end{eqnarray}
or 
\begin{eqnarray} \label{Zagier-Hecke2}
\mathfrak Z_D\left(n^{1-k}f|_k T(n)\right)=(\mathfrak Z_D f)|_{\frac{-k+3}{2}} T\left(n^2\right),
\end{eqnarray} depending on the sign of $D$.

To ease notation, we introduce a few conventions. Suppose $H$ is a Hecke operator 
\begin{equation}
H=\displaystyle\sum_n a_nT(n).
\end{equation} 
Then we define 
\begin{equation}
\widehat H:=\displaystyle\sum_n a_nT(n^2).
\end{equation} 
\begin{equation}H^{(m)}:=\displaystyle\sum_n a_nn^{-m}T(n).\end{equation}
 This notation is chosen so that if $m$ is a positive integer, then 
\begin{equation}\label{Deriv_of_Hecke}
D^m(F|_kH)= D^m(F)|_{k+2m}H^{(m)}.
\end{equation}
In this notation, we do not require $m$ to be a positive integer. For convenience, if the weight $k$ is clear from context, then we also define $H^*:=H^{(2-k)}.$ Then (\ref{Zagier-Hecke1}) and (\ref{Zagier-Hecke2}) may be extended to compound Hecke operators as follows:

\begin{eqnarray}
\mathfrak Z_D\left(f|_k H\right)=(\mathfrak Z_D f)|_{\frac{k+1}{2}}  \widehat H
\end{eqnarray}
or 
\begin{eqnarray}
\mathfrak Z_D\left(f|_k H\right)=(\mathfrak Z_D f)|_{\frac{-k+3}{2}}  \widehat H^*,
\end{eqnarray} again depending on the sign of $D$.

We will require the following theorem and corollary, which is proved using standard techniques in the theory of Maass forms.
\begin{theorem}\label{Kernel}
Let $k$ be a negative even integer. If $H$ is a compound Hecke operator, then $f_{k,1}|_{k}H\in M^!_{k}$ if and only if $S|_{2-k}H^*=0$ for every $S\in S_{2-k}$.
\end{theorem}

\begin{corollary}
If $k$ and $H$ are as in Theorem \ref{PadicPoincare}, then $f_{\frac{k+1}{2},m}|_{\frac{k+1}{2}}H\in M^!_{\frac{k+1}{2}}$ for all $m\equiv 0$, or $1 \pmod {4}$ and $S_{\frac{-k+3}{2}}$ is in the kernel of $\widehat {H}^*$.
\end{corollary}

\begin{proof} For the proof, we require the $\xi$-operator which is given by
\begin{equation}
\xi_k:=2iy^k\overline {\frac{\partial}{\partial\overline{z}}}
\end{equation}
(see Section 7.3 of \cite{OnoCDM}). As before, we drop the dependence on $k$ when it is clear from context. We recall that a harmonic Maass form is actually weakly holomorphic if and only if it is in the kernel of the $\xi$-operator. It is not difficult to see that the Hecke operator $H$ commutes with the period integral, becoming $H^*$. Therefore $f_{k,1}|_{k}H\in M^!_{k}$ if and only if $\xi f_{k,1}|_{2-k}H^*=0$. Moreover, $\{\xi f_{k,n}\}$ spans $S_{2-k},$ and $\xi (f_{k,n})|H^*=\xi(f_{k,1})|T(n)H^*.$ Since Hecke operators commute, this is $0$ if $\xi(f_{k,1})|H^*=0$.

The proof of the corollary is similar, however we first note that from Duke and Jenkins we have that any Zagier lift takes weakly holomorphic forms to weakly holomorphic forms. Thus $f_{\frac{k+1}{2},D}|_{\frac{k+1}{2}}\widehat H\in M^!_{\frac{k+1}{2}}$. This gives us $\xi (f_{\frac{k+1}{2},D})|_{\frac{-k+3}{2}}\widehat H^*=0$ for all $D$ of appropriate sign. As these span $S_{\frac{-k+3}{2}}$, we are finished.
\end{proof}

We will use this theorem and corollary in the following section to describe a $p$-adic structure for Poincar\'e series under the action of certain Hecke operators. 

\section{A Special Family of Hecke Operators}
\newcommand{\HH}{\mathcal H}
In this section, we construct a family of Hecke operators and demonstrate some basic properties. These will allow us to prove Theorem \ref{Integrality_from_principal_part} by studying the $p$-adic properties of forms coming from Poincar\'e series via the Hecke algebra.

\begin{theorem}\label{PadicPoincare}
Let $k\in \frac12 \Z$ and let $p$ be ordinary for all eigenforms in a basis of $S_k$. Then there is a compound Hecke operator $\HH_{k, N}$ such that $f_{2-k,1}|\HH_{k, N}$ is weakly holomorphic with integral coefficients, and $f_{k,1}|\HH_{k, N}^*$ has integral coefficients with $f_{k,1}|\HH_{k, N}\equiv q^{-1}+O(q)\pmod {p^N}$.
 Moreover, any such $\HH_{k,N}$ satisfies the following properties:

\begin{enumerate}
\item If $H$ is a compound Hecke operator such that $f_{2-k,1}|H^*$ is weakly holomorphic and $f_{k,1}|H$ has integer coefficients, then \[\left(f_{k,1}|\HH_{k,N}\right)|H \equiv f_{k,1}|H \pmod{p^N}.\]
\item If $\HH_{k,N}$ and $\HH_{k,N'}$ are two such operators, then \[f_{k,1}|\HH_{k,N}\equiv f_{k,1}|\HH_{k,N'}\pmod {p^N}.\]
\item If $H$ is as above and $\left(f_{k,1}|\HH_{k,N}\right)| H\equiv 0+O(q) \pmod{p^M}$ for some $M\leq N$, then\\
 $\left(f_{k,1}|\HH_{k,N}\right)| H\equiv 0 \pmod{p^M}$.
%\item If $\widehat k=\frac{k+1}{2}$, and $d$ is a square free integer such that $f_{\widehat k,d}\not \equiv 0$, then $f_{\widehat k,d}|\widehat \HH_n \equiv q^{-d}+O(q)\pmod{p^{n-k/2}}$ and also satisfies properties $(1)-(3),$ substituting $n-k/2$ for $n$. 
\end{enumerate}
\end{theorem}

For our purposes, we are especially interested in the following:
\begin{corollary}
If $H$ is a compound Hecke operator such that $f_{k,1}|H$ has integer coefficients, $p$ is ordinary for all eigenforms in a basis of $S_k$, and $f_{k,1}|H \equiv 0+O(q)\pmod {p^N}$, then $f_{k,1}|H\equiv 0\pmod {p^N}$ 
\end{corollary}
As before, when $k$ is clear from context, we may omit it from the notation, and write $\HH_N$.

\begin{proof}[Construction of the Hecke operator]
Let $k$ be a fixed, positive, even integer. If $S_k$ is empty, we can let $\HH_N=T(1)$. Otherwise, let $\ell$ be the dimension of $S_k$. Let $F_{k,m}$ be the elements of the Miller basis for $S_k$, so that  $F_{k,m}$ is the unique cusp form in $S_k$ with Fourier expansion
\[F_{k,m}=q^m+\sum_{n>\ell} c_{k}(m,n)q^n\]
where $c_k(m,n)$ are integers. We represent this form by the $m$-th elementary column vector $b_m$ which has a $1$ in the $m^{\text{th}}$ row, and all other entries are $0$. The Hecke operators act on $S_k$ linearly with trivial kernel, so we may represent the action of $T_k(N)$ by left multiplication by a matrix $\mathbb T_N$.

Without loss of generality, assume that $p^n>\ell$. We then define the matrix
\begin{equation}C_n:=
\begin{pmatrix}c_k(1,p^n)&c_k(2,p^n)&\dots&c_k(\ell,p^n)\\
c_k(1,2p^n)&c_k(2,2p^n)&\dots&c_k(\ell,2p^n)\\
\vdots&\vdots&&\vdots\\
c_k(1,\ell p^n)&c_k(2,\ell p^n)&\dots&c_k(\ell,\ell p^n) \end{pmatrix}.
\end{equation}
Considering the action of $\mathbb T_{p^n}$ on the identity, which represents the elements of the basis, we see that $\mathbb T_{p^n}\equiv C_n \pmod {p^{k-1}}$. The determinant of $\mathbb T_{p}$ is the product of the eigenvalues of $T_k(p)$, which by assumption, is not divisible $p$. Moreover,  $\mathbb T_{p^n}\equiv \mathbb T_{p}^n\pmod{p^{k-1}}$, so $\mathbb T_{p^n},$ and hence $C_n$ has an inverse with $p$-integral coefficients. Take $A\equiv C_n^{-1}$ such that $A$ has integral coefficients.

By the Zagier duality studied by Duke and Jenkins \cite{DukeJenkins}, the rows of the matrix $C_n$ give coefficients of the forms in an integral basis of $M^!_{2-k}$. The elements of this basis have the form 
\begin{equation}
F_{2-k,m}=q^{-m}+\sum_{i=-\ell}^{\infty} c_{2-k}(m,i)q^{-i},
\end{equation}

where $\ell$ is as above. Using this notation, Zagier duality gives us that 
\begin{equation}\label{Duality}
c_{2-k}(a,b)= -c_k(b,-a).
\end{equation}

If we now let $C_n$ represent these modular forms of weight $2-k$, and multiplication on the left by $A$ to take linear combinations of these forms, we find see there is a form 

\begin{equation}
\mathcal F_{k,n}=\sum_{m=0}^\ell\left(- a(m)q^{-p^nm}\right)-q^{-1}+O(1),
\end{equation}
where $a(m)$ is the coefficient $a_{(1,m)}$ from $A$. As this is a integral linear combination of elements from an integral basis of $M^!_{2-k}$, we see that $\mathcal F_{k,n}$ has integral coefficients. We then take 
\begin{equation}
\HH_n=\sum_{m=0}^\ell \left(a(m)T(mp^n)\right) +T(1).
\end{equation}
A short calculation shows that $f_{2-k,1}|\HH_n^*=\mathcal F_{k,n}$, and that \\ $\allowbreak f_{k,1}|\HH_n= D^{k-1}(f_{2-k,1}|\HH_n^*)\equiv q^{-1}+O(q)\pmod{p^n}$. Another short calculation shows that $\widehat{\HH_N}$ satisfies the properties for dual weights $\frac{k+1}{2}$ and $\frac{3-k}{2}$.
\end{proof}

\begin{proof}[Proof of Property (1)]
We can write 
\begin{equation}
f_{k,1}|\HH_n=p^nF+f_{k,1}+s,
\end{equation} 
where $F$ is a linear combination of elements from the integral basis. Thus by Theorem \ref{Kernel}, \begin{equation}f_{k,1}|\HH_n|H=p^nF|H+f_{1,k}|H\equiv f_{1,k}|H\pmod{p^n}.\end{equation}
\noqed
\end{proof}
\begin{proof}[Proof of Property (2)]
From Part (1), we have
\begin{equation}f_{k,1}|\HH_n\equiv f_{k,1}|\HH_n|\HH'_n=f_{k,1}|\HH'_n|\HH_n\equiv f_{k,1}|\HH'_n.\end{equation}
\noqed
\end{proof}
\begin{proof}[Proof of Property (3) (and corollary)]
Suppose $f_{k,1}|H\equiv s$ where $s\in S_k^!$. Then, \begin{equation}s\equiv f_{k,1}|H\equiv f_{k,1}|\HH_n|H=f|H|\HH_n\equiv s|\HH_n=0.\end{equation}
\noqed
\end{proof}


\section{Integrality of the Coefficients }\label{S6}
We now prove Proposition \ref{Integrality_of_princ_part}, which is an immediate corollary of the following:
\begin{prop}
Suppose $F=\sum a(m,n)q^n\left(\frac{1}{4\pi y}\right)^m$ has a decomposition with $e_2>e_1$. If $D>0$, then the principal part of $\mathcal Z_D(F)$ (up to constant multiples of $\frac{1}{4\pi y}$) is given by
\begin{equation}\label{Z_F_Principal_Part}
\sum R^{e_2-\lceil m/2\rceil}_{1/2+2\lceil m/2 \rceil}\left(\left.q^{-1}\right|_{1/2 -2\floor{ m/2}}H_m  \left(\frac{1}{16\pi y}\right)^m\right)
\end{equation}
where $H_m=\sum_n a(m,n)n^{1-4\floor{m/2}}T(n^2)$.
\end{prop}
%\{(Clarify, we're ignoring constants, but they don't matter since they just follow everything else. should specify that though).\}

\begin{proof}
The coefficients of the Hecke operators $H_m$ above come from the coefficients of the form $F$, and thus are the sum of contributions from each piece in the decomposition of $F$. We consider the contribution from each piece independently, beginning with those of weight divisible by $4$. Standard formulas  for the iterated raising operator (for example see \cite[Ch. 1]{123MF}) give us that the contribution of $f_{-4s}$ to $F$ is 
\begin{equation}\label{Raise_of_f_-4s}
R^{2s}f_{-4s}=\sum_{j=0}^{2s}\begin{pmatrix}2s\\j\end{pmatrix}\frac{(2s+j)!}{(2s)!}\left(\frac{1}{4\pi y}\right)^j\mathcal D^{2s-j}f_{-4s}.
\end{equation}

We have that $\widehat f =\mathfrak Z(f_{-4s})$ is weight $\frac{-4s+1}{2}$. When $k$ is half-integral, the iterated raising operator can be written as 
\begin{equation}
R^{r}\widehat f=\sum_{j=0}^{r}\begin{pmatrix}r\\j\end{pmatrix}\frac{\Gamma\left(\frac{-4s+1}{2} +r\right)}{\Gamma\left(\frac{-4s+1}{2} +r-j\right)}\left(\frac{-4}{16\pi y}\right)^j\mathcal D^{r-j}\widehat f.
\end{equation}

If we set $r=s+m$ and consider the coefficient $\left(\frac{1}{16\pi y}\right)^{2m}$ above, this reduces to 
\begin{equation}\frac{(s+m)!(2s+2m)!(s-m)!}{(2m)!(s-m)!(s+m)!(2s-2m)!}=\begin{pmatrix}2s\\2m\end{pmatrix}\frac{(2s+2m)!}{(2s)!},\end{equation}
which is the coefficient of $\left(\frac{1}{4\pi y}\right)^{2m}$ in (\ref {Raise_of_f_-4s}). Similarly, the coefficient of $\left(\frac{1}{16\pi y}\right)^{2m+1}$ reduces to 
\begin{equation}\frac{(s+m)!(2s+2m+2)!(s-m)!}{(2m+1)!(s-m-1)!(s+m+1)!(2s-2m)!}=\begin{pmatrix}2s\\2m+1\end{pmatrix}\frac{(2s+2m+1)!}{(2s)!},\end{equation}
which is the coefficient of $\left(\frac{1}{4\pi y}\right)^{2m+1}$ in (\ref{Raise_of_f_-4s}).

This, along with the action of the differential operator suffices to prove the contributions to (\ref{Z_F_Principal_Part}) from the coefficients of forms of weight $-4s.$ The contribution from the forms $f_{2-4s}$ require more careful consideration. Note that
\begin{equation}
R^{2s-1}f_{2-4s}=\sum_{j=0}^{2s-1}\begin{pmatrix}2s-1\\j\end{pmatrix}\frac{(2s-1+j)!}{(2s-1)!}\left(\frac{-1}{4\pi y}\right)^j\mathcal D^{2s-1-j}f_{2-4s}.
\end{equation}

Let $\widehat f=\mathfrak Z_D(f_{2-4s})$, which is a modular form of weight $\frac{1+4s}{2}$. In order to describe the contribution from $\widehat f$ to (\ref{Z_F_Principal_Part}), we define $I$ to be an inverse operator to the derivative operator $\mathcal D$, whose action is on formal $q$-series $f=\sum_{n\neq 0}{a_nq^n}$, is given by 
\begin{equation}I(f):=\sum_{n\neq 0}a_nn^{-1}q^n.\end{equation}
Using this notation, the contribution from the coefficients of $\widehat f$ to (\ref{Z_F_Principal_Part}) is given by the principal part of 
\begin{equation}
\sum_{m=0}^{2s-1} R^{e_2-\lceil m/2\rceil}_{1/2+2\lceil m/2 \rceil}\left(a_m \left(I^{s+\lfloor m/2\rfloor}\widehat f(z)\right) \left(\frac{1}{16\pi y}\right)^m\right),
\end{equation}
where $a_m:=\begin{pmatrix}2s-1\\m\end{pmatrix}\frac{(2s-1+m)!}{(2s-1)!}$. Using (\ref{Z_F}), the contribution to $\Z_D(F)$ is 
\[R^{e_2-s}\widehat f.\]
We claim these are identical. It suffices to prove that the cancellation occurs in the non-holomorphic parts, giving
\begin{equation}\label{f_hat}
\sum_{m=0}^{2s-1} R^{s-\lceil m/2\rceil}_{1/2+2\lceil m/2 \rceil}\left(a_m \left(I^{s+\lfloor m/2\rfloor}\widehat f(z)\right) \left(\frac{1}{16\pi y}\right)^m\right) = \widehat f(z).
\end{equation}

Using the formulas for the iterated raising operator, and that 
\begin{equation}
R_{2m}\left(\frac{1}{16\pi y}\right)^m= -4m\left(\frac{1}{16\pi y}\right)^{m+1},
\end{equation}
we may rewrite the left hand side of (\ref{f_hat}) as
\begin{equation}
\begin{split}
\sum_{m=0}^{2s-1}\sum_{\ell=0}^{s-\left \lceil \frac m2 \right \rceil}&\begin{pmatrix}s-\left \lceil \frac m2 \right \rceil\\\ell\end{pmatrix}
\left[  \sum_{j=0}^{\ell}\begin{pmatrix}\ell\\j\end{pmatrix}\frac{\Gamma\left(\frac 12-2\left\lfloor\frac m2\right\rfloor+\ell\right) }{\Gamma\left(\frac 12-2\left\lfloor\frac m2\right\rfloor+\ell-j\right)}\left(\frac{-4}{16\pi y}\right)^j
\right]\\
&\frac{(2s-1+m)!(s+\left\lfloor\frac m2\right\rfloor-\ell-1)!}{m!(2s-1-m)!(m-1)!}\frac{(-4)^{s-\left\lceil\frac  m2\right\rceil}}{(16\pi y)^{s+\left\lfloor\frac m2\right\rfloor-\ell}}I^{s+\left\lfloor\frac m2\right\rfloor-\ell+j}\widehat f(z).
\end{split}
\end{equation}

Setting $n=s+\left\lfloor\frac m2\right\rfloor-\ell+j$, we may rearrange the summations to get

\begin{equation}
\begin{split}
\sum_{n=0}^{2s-1}\sum_{m=2\textrm{Max}(0,n-s)}^{n}& \frac{(2s-1+m)!}{m!(2s-1-m)!(m-1)!}(-4)^{n-m}\left(\frac{1}{16\pi y}\right)^nI^{n}\widehat f(z)\\
&\hspace{-1.2in}\times\sum_{\ell=s-n+ \left \lfloor \frac m2 \right \rfloor}^{s-\left \lceil \frac m2 \right \rceil} \begin{pmatrix}s-\left \lceil \frac m2 \right \rceil\\\ell\end{pmatrix}
  \begin{pmatrix}\ell\\j\end{pmatrix}\frac{\Gamma\left(\frac 12-2\left\lfloor\frac m2\right\rfloor+\ell\right) (s+\left\lfloor\frac m2\right\rfloor-\ell-1)!    
}{\Gamma\left(\frac 12+s-n-\left\lfloor\frac m2\right\rfloor\right)}.
\end{split}
\end{equation}
Using \verb|Mathematica| software \cite{Mathematica}, we find the inner sum collapses algebraically into a closed form. The full expression becomes%11 reworded a little
\begin{equation}\label{2-sums}
\begin{split}
\sum_{n=0}^{2s-1}\left(\frac{1}{16\pi y}\right)^nI^{n}\widehat f(z)\cdot\sum_{m=2\max(0,n-s)}^{n}& \frac{(2s-1+m)!(2s-m)!(-1)^m}{m!(2s-1-m)!(2s-2n+m)!(n-m)!}
\end{split}
\end{equation}
Let $X_{s,n}$ be the inner sum. It is easy to see that $X_{s,0}=1$. Using the W-Z method as before, we find for $0\leq n\leq 2s-1$, that the $X_{s,n}$ satisfy the recursion relation
\begin{equation}
(n+1)X_{s,n+1}=n(3+4n-4s)X_{s,n}.
\end{equation}
It follows that $X_{s,n}=0$ if $0<n\leq 2s-1,$ and hence (\ref{2-sums}) is equal to $\widehat f(z).$
\end{proof}

\subsection{Integrality of $\mathcal Z_D(F)$}
Here we complete the proof of Theorem \ref{mainthm1} by proving cancellation of the denominators in the spectral decomposition. As before, assume we have cleared denominators, and we wish to show divisibility mod $p^e$ for a good prime $p$. Let $\widehat F=\mathcal Z_D(F)$, and suppose $\widehat F\not \equiv 0\pmod{p^e}.$ Then let $g_m$ be the coefficient of $\left(\frac{1}{16\pi y}\right)^m$ in $\widehat F$. We will refer to this as the component of $\widehat F$ at \emph{depth} $m$. Note that each $g_m$ is a quasi-modular form. If $\widehat F$ has weight $k$, and $M$ is maximal such that $g_{M}$ is not $0$, then $g_M$ is modular of weight $k-2M$ since it is some multiple of  $L^{M}\widehat F.$ If $g_M\equiv0\pmod {p^e}$, and $p$ is not $2$ or $3$, we may replace $\widehat F$ with \begin{equation}\widehat F-\left(\frac{E_2^*(4z)}{12}\right)^M\cdot g_M\equiv\widehat F\pmod{p^e},\end{equation} where $E_2^*:=E_2(z)-\frac{3}{\pi y}$. Therefore, if $m$ is chosen such that $g_{m'}\equiv 0\pmod{p^e}$ for all $m'>m$, then $g_{m}$ is congruent to a modular form of weight $k-2m.$ A similar construction can be shown to work when $p$ is $2$ or $3$.

Now return to $\mathcal Z_D(F)$. By Proposition \ref{Integrality_of_princ_part}, the principal part of $\mathcal Z_D(F)$ is congruent to $0\pmod {p^e}$, so $g_{m}$ is equivalent to a cusp form of weight $k-2m$. Suppose now that we replace each component $f$ of weight $\ell<k-2m$ in the decomposition of $\mathcal Z(F)$ with $f|\HH_{k-2m, e}^{(\ell-k+2m)}$. These are precisely the forms which contribute to the component at depth $m$. In fact, its contribution to  $g_m$ is a multiple of $\mathcal D^{(k-2m-\ell)} \left( f|\HH_{k-2m,e}^{(\ell-k+2m)}\right).$ By considering the action of the Hecke operators, we observe that this does not change the contribution to $g_{m}$ modulo $p^e$ (although it may alter components at other depths). 

Now by the definition of $\HH_{k-2m,e}^{(\ell-k+2m)}$, we have that 
\begin{equation}
\mathcal D^{(k-2m-\ell)} \left( f|\HH_{k-2m,e}^{(\ell-k+2m)}\right)=\left( \mathcal D^{(k-2m-\ell)} f\right)|\HH_{k-2m,e}.
\end{equation}
Therefore, we have that $g_{m}\equiv g_{m}|\HH_{k-2M',e} \pmod{p^e}$. As $\HH_{k-2m,e}$ sends cusp forms of weight $k-m$ to zero without introducing denominators, we see that $g_{m}\equiv 0\pmod{p^e}.$ Therefore, we have that the components of $\widehat F$ at every depth (and in particular at depth $0$) are congruent to $0\pmod {p^e}.$ From the discussion above, this concudes the proof of Theorem \ref{mainthm1}.
