\mychapter{Basic Facts from the Theory of Harmonic Maass Forms}
\section{Raising and Lowering Operators}
In this section, we recall and sketch the proofs of some basic results from the theory of harmonic Maass forms.  The proofs of these theorems follow those in \cite{BruinierFunke}. Our first basic fact is that weakly holomorphic modular forms are harmonic Maass forms. To show this, we recall the Maass raising operator
\begin{equation}R_k:=2i\frac{\partial}{\partial z}+ky^{-1},\end{equation}
and the Maass lowering operator
\begin{equation}L_k:=-2iy^2\frac{\partial}{\partial\overline z}.\end{equation}
Our definition varies slightly from that found in, for example \cite{Bump} and in particular $L_k$ is independent of the weight as our definition shifts all the $k$-dependence into $R_k$. The terminology for these operators is apparent once we recall their key intertwining properties with the Petersson slash operator:
\begin{equation}
R_k(f|_k\gamma)=(R_kf)|_{k+2}\gamma,
\end{equation}
\begin{equation}
L_k(f|_k\gamma)=(L_kf)|_{k-2}\gamma,
\end{equation}
for any $\gamma\in\operatorname{SL}_2(\R)$. Thus, the Maass raising operator takes a function which transforms like a modular form of weight $k$ and gives a function which transforms like a modular form of weight $k+2$, and the lowering operator lowers the weight by $2$. To understand how these operators act on weak Maass forms, we need the following formulas.
\begin{proposition}
Let $f$ be an eigenfunction of $\Delta_k$ with eigenvalue $\lambda$. 
Then we have:
\begin{enumerate}
\item $\Delta_{k+2}R_kf=(\lambda+k)R_kf$
\item $\Delta_{k-2}L_kf=(\lambda-k-2)L_kf$.
\end{enumerate}
\end{proposition}
Thus, the lowering and raising operators take weak Maass forms to weak Maass forms with shifted weight and eigenvalue. Now we can show the following.
\begin{proposition}
For any $k\in\frac12\Z$, we have that $M_k^!\subseteq H_k$. 
\end{proposition}
\begin{proof}
It suffices to show that an analytic function on $\H$ is harmonic, as the modularity transformation condition is the same and the growth condition is clear for weakly holomorphic modular forms as these have finitely many negative powers of $q$ in their Fourier expansion at every cusp. To show this, we recall a further identity which expresses $\Delta_k$ in terms of the raising and lowering operators
\[\Delta_k=-R_{k-2}L_k.\]
But it is clear from the definition of $L_k$ that this kills analytic functions.
\end{proof}
\section{Fourier Expansions}
Perhaps the most useful feature of classical modular forms is their Fourier expansion. As nonanalytic functions, harmonic Maass forms may not have $q$-series expansions, however they still have a canonical decomposition into two pieces, one of which is a $q$-series as for weakly holomorphic forms, and another Fourier expansion in terms of nonholomorphic functions. 
\begin{theorem}[Bruinier-Funke \cite{BruinierFunke}]\label{MaassFE}
If $f\in H_{2-k}(\Gamma_1(N))$ for $0<N\in\Z$ and $2-k>0$, then $f$ has a Fourier expansion
\begin{equation}
f(z)=\displaystyle\sum_{n\gg-\infty}c^+(n)q^n+\sum_{n<0}c^-(n)\Gamma(k-1,4\pi|y|n)q^n.
\end{equation}
\end{theorem}
Here $\Gamma(a;x):=\int_{x}^{\infty}e^{-t}t^{a-1}dt$ is the incomplete-gamma function. For $2-k\leq0$, the proof will show that the same result is true but for a shifted argument inside the incomplete-gamma function. However, for our restricted growth condition, we will soon see that there are no non-holomorphic harmonic Maass forms of such weight, as there are no cusp forms of nonpositive weight. 
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\indent
The first sum in the Fourier expansion of a harmonic Maass form is called the \emph{holomorphic part}, and is denoted $f^+$, and the second sum involving incomplete Gamma functions is called the \emph{non-holomorphic part} and denoted $f^-$. 
\begin{proof}
As the translation matrix $\bigl( \begin{smallmatrix}1&1\\0&1 \end{smallmatrix}\bigr)\in\Gamma_1(N)$ and any harmonic function is real-analytic, a harmonic Maass form $f$ has a Fourier expansion of the form
\[f(z)=\sum_{n\in\Z}c(n,y)e(nx),\]
where $y=\Im z$ and $e(x):=e^{2\pi i x}$. As $f$ is a harmonic function and functions are uniquely determined by their Fourier series, we have that for all $n$,
\[\Delta_k c(n,y)e(nx)=0.\]
This is a second order differential equation, so to find the solution space we only need to write down two independent solutions. For $n=0$, the equation is
\[-y^2\frac{\partial^2}{\partial y^2}c(0,y)-ky\frac{\partial}{\partial y}c(0,y),\]
and it is easily checked that two independent solutions are given by $1$ and $y^{1-k}$.
For $n\neq0$, we set $c(n,y)=b(2\pi ny)$, and we find that $b(w)$ satisfies
\begin{equation}\frac{\partial^2}{\partial w^2}b(w)+\frac kw\left(\frac{\partial}{\partial w}b(w)+b(w)\right)=0.
\end{equation}
One can check that $e^{-w}$ is a particular solution, and a second, independent, solution is given by 
\begin{equation}H(w)=e^{-w}\displaystyle\int_{-2w}^{\infty}e^{-t}t^{-k}dt.\end{equation}
This integral converges for $k<1$, but it can be analytically continued to $\C$ in a similar fashion as the $\Gamma$-function. We note that for $w<0$, which is the case we are interested in, that 
\[H(w)=\Gamma(1-k,-2w).\]
Thus, we see that 
\begin{equation}
c(n,y)=\begin{cases}c^+(0)+c^-(0)y^{1-k}&\text{ if }n=0\\ c^+(n)e^{-2pi y}+c^-(n)H(2\pi n y)e(nx).\end{cases}\end{equation}
Finally, we need to show that the correct Fourier coefficients vanish as per our theorem, namely that $c^+(n)=0$ for all but finitely many $n$ and that $c^-(n)=0$ for $n\geq0$. For the first condition, note that the growth condition (3) in our definition of Maass forms implies there can only be finitely many negative powers of $q$. In the non-holomorphic piece, we remark that by realizing $H(w)$ as a specialization of a Whittaker function, standard formulas on special functions imply the asymptotic
\[H(w)\sim(-2w)^{-k}e^w\]
as $w\rightarrow\infty$. Thus, the growth condition (3) again implies that the coefficients in the nonholomorphic part are all negative (although in the relaxed growth condition of Bruinier and Funke, there can be finitely many nonnegative terms).

\end{proof}
\section{The $\xi$-operator and ``Shadows''}
Our next main result concerns the Fourier expansion of the nonholomorphic part, which shows that it is essentially the Fourier expansion of a cusp form of ``dual weight''. To describe this, we require the $\xi$-operator
\begin{equation}
\xi_k:=2iy^k\overline{\frac{\partial}{\partial\overline z}}
\end{equation}
We note that a harmonic Maass form is killed by $\xi_k$ if and only if $f\in M_k^!$. Remarkably, the $\xi$-operator sends Maass forms to modular forms of dual weight. 
\begin{theorem} We have that 
\[\xi_{2-k}\: H_{2-k}\twoheadrightarrow S_k
\]
\end{theorem}
\begin{proof}
We begin by computing the action of $L_k$ on the Fourier expansion of a harmonic Maass form.
\begin{lemma}
Let $f\in H_k$ have a Fourier expansion as in \ref{MaassFE}. Then 
\[L_kf=L_kf^-=\sum_{n<0}c^-(n)(-4\pi n)^{1-k}e(n\overline z).\]
\end{lemma}
\begin{proof}
The first equality which states that $L_kf^+=0$ is clear. For the second fact, we plug in the Fourier expansion. Note that 
\[\frac{\partial}{\partial\overline z}=\frac12\cdot\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right),\]
so that 
\[\frac{\partial}{\partial\overline z}\Gamma(k-1,4\pi|n|y)q^n=\frac i2\frac{\partial}{\partial y}\displaystyle\int_{4\pi|n|y}^{\infty}e^{-t}t^{-k}dt.\]
Using the Fundamental Theorem of Calculus and plugging in gives the result.
\end{proof}
To prove that $\xi_kf$ is a cusp form, we first note that it is a holomorphic function on $\H$ as $\xi_k=y^{k-2}\overline{L_k}$ and the powers of $y$ cancel to give an ordinary $q$-series. Specifically, we find that
\begin{equation}\label{ShadowComp}\xi_{2-k}f^-(z)=-(4\pi)^{k-1}\sum_{n\geq1}\overline{c^-(-n)}n^{k-1}q^n.\end{equation} The modular transformation properties are routine to check, so we leave them
to the reader. To show, surjectivity of the map $\xi_{2-k}$ onto $S_k$, Bruinier and Funke use a geometric argument, or alternatively the theory of Poincar\'e series can be used to construct preimages.

\end{proof}
Following Zagier, we call the image $\xi_k f$ the \emph{shadow} of $f$. 
The computation in (\ref{ShadowComp}) allows us to realize the nonholomorphic part of $f$ as a period integral of its shadow. Namely, a simple change of variables $\tau\rightarrow\tau-z$ shows that we have the following integral identity.
\begin{equation}\displaystyle\int_{-\overline z}^{i\infty}\frac{e^{2\pi i n\tau}}{(-i(\tau+z))^{2-k}}d\tau=i(2\pi n)^{1-k}\Gamma(k-1,4\pi ny)q^{-n}.\end{equation}
Plugging the Fourier expansion for $f^-$ into this formula, we find
\begin{equation}f^-(z)=-\int_{-\overline z}^{i\infty}\frac{\xi_{2-k}f}{(-i(\tau+z))^{2-k}}d\tau.\end{equation}
This realization played an important role in Zwegers' work on Ramanujan's mock theta functions \cite{Zwegers1,Zwegers2}, where we will see that the shadows of the mock theta functions are weight $3/2$ unary theta series. 
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\indent
\section{The Bruinier-Funke Pairing}
We conclude this section with a proof of a useful fact by Bruinier and Funke that harmonic Maass forms which are not holomorphic modular forms have an exponential singularity at some cusp. 
\begin{theorem}[Bruinier-Funke \cite{BruinierFunke}]\label{PrincipalPart}
Let $f$ be a harmonic Maass form of weight $k$ such that $f^-\neq0$. Then $f$ has a nonzero principal part at some cusp. 
\end{theorem}
\begin{proof}
The proof relies on an important paring defined by Bruinier and Funke. We first recall the Petersson inner product $(\cdot,\cdot)\:M_k\times S_k\rightarrow\C$ defined by 
\begin{equation}\label{InnerProductDefn}
(f,g):=\int_{\Gamma\backslash\H}f(z)\overline{g(z)}y^k\frac{dx\,dy}{y^2}.
\end{equation}
We then define the a pairing $\{\cdot,\cdot\}\:M_k\times H_{2-k}\rightarrow\C$ using the $\xi$-operator:
\begin{equation}\label{BFPairingDefn}
\{g,f\}:=(g,\xi_{2-k}).
\end{equation}
The idea is to consider $\{\xi_{2-k}f,f\}=(\xi_{2-k}f,\xi_{2-k}f)$, which is non-zero as $f^-\neq0$ implies that $\xi_{2-k}f$ is a non-zero cusp form. The result then follows from an alternative expression for $\{\cdot,\cdot\}$ in terms of the Fourier expansions of $f$ and $g$. 
\begin{proposition}
If $h$ runs over all the cusps, and the $n^{\text{th}}$ Fourier coefficient of $g$ at a cusp $h$ is $b(h,n)$, and the $n^{\text{th}}$ Fourier coefficient of $f^+$ at a cusp $h$ is $a^+(h,n)$, we have that
\[\{g,f\}=\sum_h\sum_{n\leq0}a^+(h,n)b(h,-n).\]
\end{proposition}
\begin{proof}
We consider the usual fundamental domain $\mathcal{F}$ for $\Gamma=\operatorname{SL}_2(\Z)$, and we also consider the truncated fundamental domain:
\[\mathcal{F}_T:=\{z\in\mathcal{F}\colon\Im z\leq T\}.\]
By a simple calculation, using Stokes' theorem we find that 
\[\int_{\mathcal{F}_T}g(z)L_kf(z)\frac{dx\, dy}{y^2}=-\int_{\partial\mathcal{F}_T}g(z)f(z)dz.\]
By the usual relation between $L_k$ and $\xi_k$, we see that the right-hand side is just $\int_{\mathcal{F}_T}g(z)\overline{\xi_{2-k}f(x)}y^k\frac{dx\, dy}{y^2}$. For the right-hand side, as $f$ and $g$ have dual weights, we have that $g(z)f(z)dz$ is an invariant $1$-form, so that its integral on all the pieces of $\mathcal{F}_T$ except the top line from $-\frac12+iT$ to $\frac12+iT$ cancel out. Hence, the right-hand side is equal to 
\[\int_{-\frac12}^{\frac12}g(x+iT)f(x+iT)dx.\]
By our growth assumption in our definition of harmonic Maass forms, and as for a $q$ series such an integral just picks out the constant term, we have that this integral is equal to
\[\sum_h\sum_na^+(h,n)b(h,-n)+O(e^{-CT}).\]
Taking the limit $T\rightarrow\infty$ gives the result.

\end{proof}


\end{proof}




