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\section{Notations}

We will consider the group $SL_2(\R)$. Elements of this group will be usually denoted by $\gamma$, and matrix elements by $a$, $b$, $c$, $d$:
\[
\gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\R).
\]
This groups acts on the upper half plane $\HH$.

The group $SL_2(\R)$ naturally acts on the linear space of homogenious polynomials of degree $m$ in two variables. This is the same as the space of polynomials in one variable $X$ of degree not greater than $m$. We denote this space $V^m$. For
\[
p\in V^m, \; p(X) = p_0 + p_1 X + \dots + p_m X^m,
\]
we write the right action of the group on $p$ as follows:
\[
(p \gamma)(X) = (p |_{-m} \gamma)(X) = p(\gamma X) (cX + d)^m, \qquad \gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\R).
\]
The corresponding action on the left is
\[
(\gamma p)(X) = (p \gamma^{-1})(X) = p(\gamma^{-1} X) (-cX + a)^m.
\]

The space $V^m$ possess an invariant scalar product, which is given by 
\[
(\sum_{i=0}^m p_i X^i, \sum_{i=0}^m p_i' X^i) = \sum_{i=0}^m \frac {(-1)^i p_i p_{m-i}'}{\binom{m}{i}}.
\]

Let $S$ be a discrete subset of the upper half plane $\HH$ and $f(\z)$ be a function on $\HH-S$ with values in $\C$ and $w\in \Z$. Define differential operators
\begin{gather*}
\delta_w f = \frac{\partial f}{\partial z} + \frac{w}{\z - \zc} f,\\
\delta_w^- f = (\z - \zc)^2 \frac{\partial f}{\partial \zc},\\
\Delta f = (\z - \zc)^2 \frac {\partial}{\partial \z} \frac {\partial}{\partial \zc} f + w (\z - \zc) \frac {\partial}{\partial \zc}.
\end{gather*}
We think about $w$ as the weight, attached to the function $f$. The weight will be always clear from the context, so we will omit the subscript $w$. 
We will follow the following agreement: the operator $\delta$ increases weight by $2$, the operator $\delta^-$ decreases weight by $2$ and the operator $\Delta$ leaves weight untouched. Taking into account this agreement the following identities can be proved:
\begin{gather*}
\delta^- \delta - \delta \delta^- = w,\\
\delta \delta^- = \Delta,\\
\delta^- \delta = \Delta + w.\\
\end{gather*}

Let the group $SL_2(\R)$ act on functions of weight $w$ by the usual formula:
\[
(f|_w \gamma)(\z) = f(\gamma \z) (c\z + d)^{-w}.
\]
This is a right action. We also define the corresponding left action
\[
(\gamma f)(\z) = (f|_w \gamma^{-1})(\z) = f(\gamma^{-1} \z) (-c\z + a)^{-w}.
\]
Note that this action commutes with the operators $\delta$, $\delta^-$, $\Delta$, it maps functions defined on $\HH-S$ to functions defined on $\HH-\gamma S$.

It is also convenient to modify the complex conjugation for functions with weight to make it commuting with the action of the group. For $f$ of weight $w$ we put
\[
f^*(\z) = (\z - \zc)^w \overline{f(\z)}.
\]
Assign to $f^*$ weight $-w$. We can check that 
\begin{gather*}
f^{**} = (-1)^w f, \\
\delta (f^*) = (\delta^- f)^*, \\
\delta^- (f^*) = (\delta f)^*, \\
\Delta (f^*) = ((\Delta + w) f)^*, \\
\gamma (f^*) = (\gamma f)^*.
\end{gather*}
We remark, that for the weight $0$ the operator $*$ is the usual complex conjugation, the operator $\delta$ is the usual $\frac{\partial}{\partial z}$, and the operator $\Delta$ is the usual Laplace operator for the hyperbolic metric $-y^2(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2})$.

We list several formulae, which are convenient to use in computations. We assume that the constant function $1$ has weight $0$. Consider functions
\[
X-z,\qquad \frac{X-\zc}{z-\zc},
\]
which are thought as functions in $z$ of weights $-1$ and $1$ respectively. Then
\begin{gather*}
\delta 1 = \delta^- 1 = 0, \\
(X-z)^* = \frac{X-\zc}{z-\zc},\qquad \left(\frac{X-\zc}{z-\zc}\right)^* = -(X-z), \\
\delta (X-z) = -\frac{X-\zc}{z-\zc},\qquad \delta \frac{X-\zc}{z-\zc} = 0, \\
\delta^- (X-z) = 0,\qquad \delta^- \frac{X-\zc}{z-\zc} = X-z, \\
\delta (X-z)^a \left(\frac{X-\zc}{z-\zc}\right)^b = -a (X-z)^{a-1} \left(\frac{X-\zc}{z-\zc}\right)^{b+1}, \\
\delta^- (X-z)^a \left(\frac{X-\zc}{z-\zc}\right)^b = b (X-z)^{a+1} \left(\frac{X-\zc}{z-\zc}\right)^{b-1}, \\
\Delta (X-z)^a \left(\frac{X-\zc}{z-\zc}\right)^b = -b (a+1) (X-z)^a \left(\frac{X-\zc}{z-\zc}\right)^b.
\end{gather*}

\section{Eigenfunctions of the laplacian}\label{eigenvalues}
Let $S$ be a discrete subset of $\HH$.
For integers $k, w$ we denote by $F_{k, w}$ the space of functions on $\HH-S$ with weight $w$ satisfying
\[
\Delta f = (k (1-k) + \frac {w (w-2) }{4}) f.
\]
It is easy to check the following properties of the spaces $F_{k, w}$:
\begin{prop}
\begin{enumerate}
\item The space $F_{k, w}$ is invariant under the action of the group $SL_2(\R)$ (meaning, of course, that $\gamma\in SL_2(\R)$ changes $S$ to $\gamma S$).
\item The operator $*$ maps $F_{k, w}$ to $F_{k, -w}$.
\item The operator $\delta$ maps $F_{k, w}$ to $F_{k, w+2}$. It is invertible for all values of $w$, except, possibly, $2k - 2$ and $-2k$ with the inverse given by 
\[
\delta_w^{-1} = \frac4{(w+2k)(w-2k+2)} \delta_{w+2}^-.
\]
\item The operator $\delta^-$ maps $F_{k, w}$ to $F_{k, w-2}$. It is invertible for all values of $w$, except, possibly, $2k$ and $2-2k$ with the inverse given by
\[
(\delta_w^-)^{-1} = \frac4{(w-2k)(w+2k-2)} \delta_{w-2}.
\]
\end{enumerate}
\end{prop}
We will occasionally use negative powers of $\delta$ and $\delta^-$ when the argument belongs to the space $F_{k, w}$.

Next we state some basic facts
\begin{prop}\label{prop2_2}
For integer numbers $k$, $l$ the function 
\[
(X-z)^{k-l-1} \left(\frac{X-\zc}{z-\zc}\right)^{k+l-1}
\]
belongs to $F_{k, 2l}$ as a function of $z$.
\end{prop}
\begin{proof}
Use formulae listed in the end of the first chapter.
\end{proof}

\begin{prop}
For $f$, $g$ in $F_{k,0}$ and $1-k\leq l \leq k-1$ we have
\[
\delta^{-l} f \delta^l g = (\delta^-)^l f (\delta^-)^{-l}g.
\]
\end{prop}
\begin{proof}
Note, that
\[
(\delta^-)^l f = (-1)^l \frac{(k+l-1)!}{(k-l-1)!} \delta^{-l} f,
\]
and analogously
\[
\delta^l g = (-1)^l \frac{(k+l-1)!}{(k-l-1)!} (\delta^-)^{-l} g,
\]
which implies the required identity.
\end{proof}

Let us introduce the following operation. For two functions $f$, $g$ from $F_{k,0}$ we put
\[
f * g = \sum_{l=1-k}^{k-1} \delta^{-l} f \delta^l g,
\]
which has weight $0$.
Note, that the previous proposition implies
\[
f*g = \sum_{l=1-k}^{k-1} (-1)^l (\delta^-)^{-l} f (\delta^-)^l g,
\]
so
\[
\overline{f*g} = \overline{f}*\overline{g}.
\]
It is also easy to see, that 
\[
\frac{\partial}{\partial z} (f*g) = (-1)^{k-1}(\delta^{1-k}f \delta^k g + \delta^k f \delta^{1-k} g).
\]
Consider a function $Q_z(X)^{k-1}$. By Proposition \ref{prop2_2} 
\[
    Q_z(X)^{k-1} \in F_{k, 0}, \qquad \text{as a function of $z$.}
\]
One can compute, that for $1-k \leq l \leq k-1$
\[
\delta^l Q_z(X)^{k-1} = (-1)^l \frac{(k-1)!}{(k-l-1)!} (X-z)^{k-1-l} \left(\frac{X-\zc}{z-\zc}\right)^{k-1+l}.
\]
For $f\in F_{k, 0}$ we denote
\begin{multline*}
\wt f = (-1)^{k-1} \binom{2k-2}{k-1} f * Q_z(X)^{k-1}  
\\= (-1)^{k-1} \sum_{l=1-k}^{k-1} \frac{(2k-2)!}{(k+l-1)!(k-1)!} (X-z)^{k-1+l} \left(\frac{X-\zc}{z-\zc}\right)^{k-1-l} \delta^l f.
\end{multline*}
It is easy to check, that
\[
\wt{\gamma f} = \gamma \wt f,\qquad \text{for $\gamma\in SL_2(\R)$,}
\]
where $\gamma$ acts on $\wt f$ by the simultaneous action on $z$ in weight $0$ and $X$ in weight $2-2k$. Also
\[
\overline{\wt f} = (-1)^{k-1} \wt{\overline f}.
\]
One can compute the scalar product of $\delta^i Q_z(X)^{k-1}$ and $\delta^j Q_z(X)^{k-1}$ as follows:
\begin{lem}
\[
(\delta^i Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1}) = \begin{cases} 
0, & \text{if $i\neq -j$} \\
(-1)^{k-1-i} \binom{2k-2}{k-1}^{-1} & \text{if $i=-j$.}
\end{cases}
\]
\end{lem}
\begin{proof}
We prove by the induction on $i$ starting from $1-k$.
Since 
\[
\delta^{1-k} Q_z(X)^{k-1} = (-1)^{k-1} \frac{(k-1)!}{(2k-2)!} (X-\z)^{2k-2},
\]
for any polynomial $p\in V_{2k-2}$ we have
\[
(\delta^{1-k} Q_z(X)^{k-1}, p) = (-1)^{k-1} \frac{(k-1)!}{(2k-2)!} p(\z).
\]
Hence $(\delta^{1-k} Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1})$ is not zero only for $j=k-1$ and in this case
\[
(\delta^{1-k} Q_z(X)^{k-1}, \delta^{k-1} Q_z(X)^{k-1}) = \binom{2k-2}{k-1}^{-1}.
\]
If the statement is true for $i$ then for any $j$, taking into account, that the weight of $(\delta^i Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1})$ is $2i+2j$
\begin{multline*}
0 = \delta(\delta^i Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1}) 
\\= (\delta^{i+1} Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1}) + (\delta^i Q_z(X)^{k-1}, \delta^{j+1} Q_z(X)^{k-1}).
\end{multline*}
Hence
\[
(\delta^{i+1} Q_z(X)^{k-1}, \delta^j Q_z(X)^{k-1}) = -(\delta^i Q_z(X)^{k-1}, \delta^{j+1} Q_z(X)^{k-1}).
\]
We see, that if $i+j\neq -1$ this is zero. If $i+j=-1$ this equals exactly
\[
-(-1)^{k-1-i} \binom{2k-2}{k-1}^{-1}.
\]
\end{proof}

Therefore the original function $f$ can be recovered as
\[
f = (\wt f, Q_z(X)^{k-1}).
\]

Note also that $\Delta \wt f=0$. This is true because 
\[
\delta^- \delta^k f = (\Delta+2 k - 2) \delta^{k-1} f = 0.
\]

Let us summarize.
\begin{thm}
Let $f\in F_{k, 0}$ for $k\geq 1$. Then the function $\wt f$ satisfies the following properties:
\begin{eqnarray*}
\wt f \in F_{0,0}\otimes V_{2k-2},\\
(\wt f, Q_z(X)^{k-1}) = f,\\
\frac{\partial \wt f}{\partial z} = (X-z)^{2k-2}\frac{(-1)^{k-1}\delta^k f}{(k-1)!},\\
\frac{\partial \wt f}{\partial \zc} = (X-\zc)^{2k-2}\frac{\bar{\delta}^k f}{(k-1)!}.
\end{eqnarray*}
\end{thm}

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