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\author{Anton Mellit}
\title{Higher green's functions for modular forms}
\begin{document}
\maketitle

\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 $\HH$.

\section{Representations of $SL_2(\R)$}

The group $SL_2(\R)$ naturally acts on the following linear spaces:
\begin{itemize}
\item $V=\C^2$~--- the space of column vectors of length 2.
\item $V_m$~--- symmetric $m$-th power of $V$.
\item $V^m$~--- the space of homogenious polynomials of degree $m$ on $V$.
\end{itemize}
Note, that $V_m$ and $V^m$ are mutually dual, $V_0=V^0=\C$. Of course, $V_m$ is isomorphic to $V^m$ as a representation of $SL_2(\R)$, but we would like to represent elements of this spaces in different ways, also these two spaces have different natural $\Z$- structures~--- that is why we distinguish them. 

We explain the way we view elements of $V_m$, $V^m$. Start with $V^m$. Let $e_1$, $e_2$ be the natural basis on $V$. Let $e^1$, $e^2$ be the dual basis. Every homogenious polynomial on $V$ is a polynomial in $e^1$, $e^2$. If we substitute $e^1$ by a new variable $X$ and $e^2$ by $1$ we obtain a non-homogenious polynomial in one variable of degree less or equal $m$. We represent elements of $V^m$ as polynomials in the variable $X$ of degree less or equal $m$, i.e.
\[
p\in V^m, \; p(X) = p_0 + p_1 X + \dots + p_m X^m.
\]
The group acts on $V^m$ on the right by
\[
(p \gamma)(X) = (p |_{-m} \gamma)(X) = p(\gamma X) (cX + d)^m, \qquad \gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\R).
\]
There is a corresponding action on the left
\[
(\gamma p)(X) = (p \gamma^{-1})(X) = p(\gamma^{-1} X) (-cX + a)^m.
\]
We may also consider the corresponding homogenious polynomials in two variables
\[
p(X,Y) = p \left( \frac {X}{Y} \right) Y^m = p_0 Y^m + p_1 Y^{m-1} X + \dots + p_m X^m,
\]
the action of $SL_2(\R)$ is then given by
\[
(\gamma p)(X,Y) = p(dX - bY, -cX + aY).
\]

Now turn to $V_m$. Any $v\in V_m$ defines a functional on $V^m$. Suppose its value on $p$ is 
\[
v_0 p_0 + v_1 p_1 + \dots v_m p_m,
\]
then we represent $v$ as a raw vector $(v_0, v_1, v_2, \dots, v_m)$. For any two numbers $x, y \in \C$ we form a vector
\[
v_{x, y} = (y^m, y^{m-1} x, y^{m-2} x^2, \dots, x^m) \in V_m.
\]
Then for any $p\in V^m$
\[
(v_{x, y}, p) = p\left(\frac{x}{y}\right) y^m = p(x, y),
\]
hence
\[
\gamma v_{x, y} = v_{\gamma (x, y)} = v_{a x + b y, c x + d y}.
\]
In this way the action of $SL_2(\R)$ is given on a subset of $V_m$ which spans $V_m$.

The isomorphism between $V_m$ and $V^m$ can be given as follows. Define an invariant pairing between elements of the form $v_{x,y} \in V_m$
\[
(v_{x,y}, v_{x', y'}) = (x y' - x' y)^m = \sum_{i=0}^m (-1)^i \binom{m}{i} x^{m-i} y^i x'^i y'^{m-i}.
\]
Since it is a homogenious polynomial of degree $m$ in each pair of variables this induces an equivariant linear map from $V_m$ to $V^m$
\[
(v_0, v_1, \dots, v_m) \mapsto \sum_{i=0}^m (-1)^i \binom{m}{i} v_{m-i} X^i.
\]
This is an isomorphism of representations. It induces invariant pairings on $V^m$
\[
(\sum_{i=0}^m p_i X^i, \sum_{i=0}^m p_i' X^i) = \sum_{i=0}^m \frac {(-1)^i p_i p_{m-i}'}{\binom{m}{i}}
\]
and on $V_m$
\[
((v_i), (v_i')) = \sum_{i=0}^m (-1)^i \binom{m}{i} v_i v_{m-i}'.
\]

\section{Operations on functions with weight}

\subsection{Differential operators}
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*}
Df = \frac{1}{2\pi i} \frac{\partial f}{\partial \z},\\
\delta_w f = Df + \frac{1}{2\pi i} \frac{w}{\z - \zc} f,\\
\delta_w^- f = 2 \pi i (\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 $w$. For example, if we have a function which satisfies
\[
f(\gamma \z) = f(\z) (c \z + d)^w,
\]
for $\gamma$ from some subgroup of $SL_2(\R)$ then we will assign to $f$ weight $w$. Another example would be an element of a vector valued function which satisfies some transformation property. 

We will follow the following agreement: the operators $D$ and $\delta$ increase 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*}

\subsection{Complex conjugation}
Define operator $\eps$ as follows:
\[
(\eps f)(\z) = (2 \pi i)^w (\z - \zc)^w \overline{f(\z)}.
\]
Assign to $\eps f$ weight $-w$. We can check that 
\begin{gather}
\eps \eps f = f \\
\delta^- \eps f = -\eps \delta f \\
\delta \eps f = -\eps \delta^- f \\
\Delta \eps f = \eps (\Delta + w) f
\end{gather}

\subsection{Action of $SL_2(\R)$ on functions}
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 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 operators $\delta$, $\delta^-$, $\Delta$, $\eps$ it maps functions defined on $\HH-S$ to functions defined on $\HH-\gamma S$.

\section{Eigenfunctions of the laplacian}\label{eigenvalues}
If $f$ has weight $w$ and is an eigenfunction of $\Delta$ with eigenvalue $x$, i.e. $\Delta f = xf$, then 
\begin{gather*}
\Delta \delta f = (x + w) \delta f, \\
\Delta \delta^- f = (x - w + 2) \delta^- f, \\
\Delta \eps f = (x + w) \eps f.
\end{gather*}
So, operators $\delta$, $\delta^-$ and $\eps$ preserve the property to be an eigenfunction of the laplacian. We can compose a number from the weight and the eigenvalue which does not change under applications of these operators:
\[
x_0 = x - \frac {w (w-2) }{4}.
\]
Let us fix some $k\in \Z$, $k>0$ and consider functions with fixed 
\[
x_0 = k (1-k).
\]
Denote by $F_w$ the space of functions with weight $w$ and given value of $x_0$, i.e. satisfying
\begin{gather*}
\delta \delta^- f = \Delta f = (k (1-k) + \frac {w (w-2) }{4}) f = (k - \frac {w}{2}) (1-k-\frac{w}{2}) f,\\
\delta^- \delta f = (k + \frac {w}{2}) (1-k+\frac{w}{2}) f.
\end{gather*}
\begin{lem}
The following properties of $\delta$, $\delta^-$, $\eps$ are true:
\begin{enumerate}
\item The operator $\eps$ maps $F_w$ to $F_{-w}$.
\item The operator $\delta$ maps $F_w$ to $F_{w+2}$. It is invertible for all values of $w$, except, possibly, $2k - 2$ and $-2k$.
\item The operator $\delta^-$ maps $F_w$ to $F_{w-2}$. It is invertible for all values of $w$, except, possibly, $2k$ and $2-2k$.
\end{enumerate}
\end{lem}
In fact, choosing an element in $F_w$ for $2-2k \leq w \leq 2k-2$ determines a one dimensional subspace in $F_{w'}$ for each $w'$ from the same interval and operators $\delta$, $\delta^-$ act between these one dimensional subspaces by non-zero maps.

Fix some $f\in F_{2-2k}$. Define a sequence of functions
\[
f_l = \frac {\delta^{k+l-1} f}{(k+l-1)!} , \qquad 1-k \leq l \leq k-1.
\]
This sequence satisfies:
\begin{enumerate}
\item $f_l \in F_{2l}$
\item $f_{1-k} = f$
\item $\delta f_{l} = (l+k) f_{l+1}$ for $1-k \leq l \leq k-2$
\item $\delta^- f_{l} = (l-k) f_{l-1}$ for $2-k \leq l \leq k-1$
\end{enumerate}
Note, that the lemma above implies that any element of the sequence completely determines the full sequence. We make several remarks:
\begin{enumerate}
\item $\eps f_{k-1}$ has weight $1-k$, so we can construct the corresponding sequence for $\eps f_{k-1}$. It is trivial to check that the resulting sequence is $(\eps f_{-l})$.
\item $\delta^- f$ is zero if and only if $f$ is holomorphic.
\item $\delta^{2k-1} f$ is holomorphic.
\end{enumerate}

\section{Functions with values in representations}\label{funcs_in_reps}
We are going to consider functions with values in $V_m$ or $V^m$. Obviously the notion of weight, operators $D$, $\delta$, $\delta^-$, $\Delta$ and their properties mentioned above extend to such functions. Let the complex conjugation act on $V_m$, $V^m$ in the obvious way. Then the operator $\eps$ can also be defined and the same properties are valid. The group acts on such functions by the rule:
\[
(\gamma f)(\z) = \gamma (f(\gamma^{-1} \z)) (-c\z+a)^{-w},
\]
and again this commutes with operators $D$, $\delta$, $\delta^-$, $\eps$.

For each $l$ such that $1-k \leq l \leq k-1$ define a function with values in $V^{2k-2}$
\[
Q_l(\z)(X) = (2 \pi i)^{k-1-l} \frac {(X - \z)^{k-1-l} (X - \zc)^{k-1+l}}{(\z - \zc)^{k-1+l}}.
\]
Assign to $Q_l$ weight $2 l$. One can check:
\begin{enumerate}
\item $\delta Q_{k-1} = 0$
\item $\delta^- Q_{1-k} = 0$
\item $\delta Q_{l} = (l+1-k) Q_{l+1}$ for $1-k \leq l \leq k-2$
\item $\delta^- Q_{l} = (l+k-1) Q_{l-1}$ for $2-k \leq l \leq k-1$
\item $\eps Q_{l} = Q_{-l}$
\item $\gamma Q_{l} = Q_{l}$ for all $\gamma \in SL_2(\R)$
\end{enumerate}
It follows that $Q_l$ are eigenvalues for $\Delta$ with the same value $x_0 = k (1-k)$.
We have introduced an invariant bilinear form on $V^{2k-2}$. Let us compute its values on $Q_l$.
\begin{lem}
\[
(Q_i, Q_j) = \begin{cases} 
0, & \text{if $i\neq -j$} \\
(-1)^{i+k-1} \frac{1}{\binom{2k-2}{i+k-1}} (2\pi i)^{2k-2}, & \text{if $i= -j$} 
\end{cases}
\]
\end{lem}
\begin{proof}
We prove by induction on $i$ starting from $1-k$.
Since 
\[
Q_{1-k} = (2\pi i)^{2k-2} (X-\z)^{2k-2},
\]
for any polynomial $p\in V^{2k-2}$ we have
\[
(Q_{1-k}, p) = (2\pi i)^{2k-2} p(\z).
\]
Hence $(Q_{1-k}, Q_j)$ is not zero only for $j=k-1$ and in this case
\[
(Q_{1-k}, Q_{k-1}) = (2\pi i)^{2k-2}.
\]
If the statement is true for $i$ then for any $j$ since the weight of $(Q_i, Q_j)$ is $i+j$
\[
0 = \delta(Q_i, Q_j) = (\delta Q_i, Q_j) + (Q_i, \delta Q_j).
\]
Hence
\[
(Q_{i+1}, Q_j) = \frac{1}{i+1-k}(\delta Q_i, Q_j) = -\frac{1}{i+1-k}(Q_i, \delta Q_j)
\]
\[
= -\frac{j+1-k}{i+1-k}(Q_i, Q_{j+1}).
\]
We see, that if $i+j\neq -1$ this is zero. If $i+j=-1$ this equals exactly
\[
-(-1)^{i+k-1}\frac{-i-k}{(i+1-k)\binom{2k-2}{i+k-1}}(2\pi i)^{2k-2} = (-1)^{i+k} \frac{1}{\binom{2k-2}{i+k}}(2\pi i)^{2k-2}.
\]
\end{proof}

Consider the following function:
\[
\wt f = \sum_{l=1-k}^{k-1} f_l Q_{-l}, \qquad \text{i.e.}
\]
\[
\wt f (\z)(X) = \sum_{l=1-k}^{k-1} (2 \pi i)^{k-1+l} f_l(\z) \frac {(X - \z)^{k-1+l} (X - \zc)^{k-1-l}}{(\z - \zc)^{k-1-l}},
\]
which is of weight $0$. Then we can apply our differential operators to $\wt f$ and obtain:
\begin{gather} \label{holimage}
\frac{1}{2\pi i} \frac{\partial \wt f}{\partial \z} = D \wt f = \delta \wt f = Q_{1-k} \delta f_{k-1} = (2 \pi i)^{2k-2} \frac {(X - \z)^{2k-2}}{(2 k - 2)!} \delta^{2 k - 1} f 
\\ \label{antihol}
2 \pi i (\z - \zc)^2 \frac{\partial \wt f}{\partial \zc} = \delta^- \wt f = Q_{k-1} \delta^- f_{1-k} = 2 \pi i (\z - \zc)^2 \left( \frac {X - \zc}{\z - \zc} \right)^{2k-2} \frac{\partial f}{\partial \zc}
\end{gather}
To prove this we have used Leibniz rule which holds for operators $\delta$, $\delta^-$.
For the operator $\eps$ we also have a simple formula:
\[
(\eps \wt f)(\z) = \wt{\eps f_{k-1}}(\z)
\]
\begin{lem} \label{ftilde}
The polynomial $f(\z)$ admits the following expression as a polynomial of $X-\z$:
\[
\wt f(\z)(X) = \sum_{n=0}^{2k-2} \frac{D^n f}{n!} (2 \pi i)^n (X-\z)^n,
\]
moreover
\[
\delta^{2k-1} f = D^{2k-1} f.
\]
\end{lem}
\begin{proof}
There exists a decomposition
\[
\wt f(\z)(X) = \sum_{n=0}^{2k-2} \frac{a_n}{n!} (2 \pi i)^n (X-\z)^n
\]
for some unknown functions $a_n$. Substituting $X=\z$ in the definition of $\wt f$ gives $a_0 = f$. Applying $D$ gives:
\[
\begin{split}
(D \wt f)(\z)(X) = \sum_{n=0}^{2k-3} \frac{D a_n - a_{n+1}}{n!} (2 \pi i)^n (X-\z)^n \\+ \frac{D a_{2k-2}}{(2k-2)!} (2 \pi i)^{2k-2} (X - \z)^{2k-2}.
\end{split}
\]
The identity proved above implies $a_{n+1} = D a_n$ and $\delta^{2k-1} f = D a_{2k-2}$
\end{proof}

\section{Holomorphic and antiholomorphic images}
Let $f$ be as in previous section. Since $\delta \wt f$ is holomorphic and the weight of $\wt f$ is zero
\[
\Delta \wt f = \delta^- \delta \wt f = \delta \delta^- \wt f = 0.
\]
This implies
\begin{lem}
\begin{enumerate}
\item $D \wt f = \delta \wt f$ is holomorphic. If $D \wt f = 0$, then $\wt f$ is antiholomorphic.
\item $\overline{D} \wt f = \overline{\delta \eps \wt f}$ is antiholomorphic. If $\overline{D} \wt f = 0$, then $\eps \wt f$ is antiholomorphic, i.e. $\wt f$ is holomorphic.
\item If both $D \wt f$ and $\overline{D} \wt f$ are $0$, then $\wt f$ does not depend on $\z$.
\end{enumerate}
\end{lem}
In the latter case $f(\z) = \wt f(\z)(\z)$ is a polynomial in $\z$ of degree not greater than $2k-2$. It is also true in the other way:
\begin{lem}\label{poly}
If $f(\z) = p(\z)$ is a polynomial of degree not greater than $2k-2$, then $f\in F_{2-2k}$ and $\wt f = p(X)$, thus $D \wt f = 0$ and $\overline{D} \wt f = 0$.
\end{lem}
\begin{proof}
The fact that $f \in F_{2-2k}$ holds since $\delta^- f$ = 0. To prove the rest use lemma \ref{ftilde} and the fact, that $f$ coincides with its Taylor expansion terminated at term of order $2k-2$.
\end{proof}

If a function has weight $2k$ and is holomorphic then it belongs to $F_{2k}$. Denote corresponding subspace of $F_{2k}$ by $F_{2k}^{hol}$. Denote the analogous subspace of $F_{2-2k}$ by $F_{2-2k}^{hol}$.
Denote by $\phi$ the map from $F_{2-2k}$ to $F_{2k}^{hol}$ which takes $f$ to 
\[
\phi(f)=D^{2k-1}f = \delta^{2k-1}f = \frac {(2k-2)! D \wt f}{(2\pi i)^{2k-2} (X-\z)^{2k-2}},
\]
see (\ref{holimage}). We call $\phi(f)$ the \emph{holomorphic image} of $f$.
\begin{lem}\label{locsurj}
The restriction of the map $\phi$ to $F_{2-2k}^{hol}$ is locally surjective.
\end{lem}
\begin{proof}
If $g \in F_{2k}^{hol}$ then we may take
\[
(2 \pi i)^{2k-2} \frac {(X - \z)^{2k-2}}{(2 k - 2)!} g,
\]
multiply by $2 \pi i d\z$, integrate it in some neighbourhood of a given point and obtain $h$, which is a function with values in $V^{2k-2}$. Put $f(\z) = h(\z)(\z)$ and assign to $f$ weight $2-2k$. Since $f$ is holomorphic $\delta^- f$ is zero, hence $f$ belongs to $F_{2-2k}$. Then both $h$ and $\wt f$ can be represented as polynomials in $X-\z$ as in lemma \ref{ftilde}. These polynomials have the same 'constant term' $f$ and both $D \wt f$, $D h$ are divisible by $(X-\z)^{2k-2}$. It follows, that $f = h$ (see the proof of lemma \ref{ftilde}), thus $\phi f = g$ in some neighbourhood of a given point.
\end{proof}

Since the kernel of the restriction of $\phi$ to $F_{2-2k}^{hol}$ is $V^{2k-2}$ as follows from the lemma \ref{poly} we obtain a locally exact sequence, i.e. an exact sequence of sheaves:
\[
0 \longrightarrow V^{2k-2} \longrightarrow F_{2-2k}^{hol} \xrightarrow{\;\,\phi\;\,} F_{2k}^{hol} \longrightarrow 0.
\]
Denote by $\overline{F_{2k}^{hol}}$ the space of functions whose complex conjugate belongs to $F_{2k}^{hol}$. Denote by $\phi'$ the map from $F_{2-2k}$ to $\overline{F_{2k}^{hol}}$ which sends $f$ to 
\[
\phi'(f)=\overline{D^{2k-1} \eps f_{k-1}} = \frac{(2k-2)! \overline{D} \wt f}{(2\pi i)^{2k-2} (X-\zc)^{2k-2}},
\]
see (\ref{holimage}). We call $\phi'(f)$ the \emph{antiholomorphic image} of $f$. Clearly
\[
\phi'(f) = \overline{\phi(\eps f_{k-1})}
\]
Using lemma \ref{locsurj} we conclude that $\phi'$ is also locally surjective. Its kernel is $F_{2-2k}^{hol}$. Therefore the following sequence is locally exact:
\[
0 \longrightarrow F_{2-2k}^{hol} \longrightarrow F_{2-2k} \xrightarrow{\;\,\phi'\;\,} \overline{F_{2k}^{hol}} \longrightarrow 0.
\]
Taking into account exactness of the above two sequences we conclude that the following sequence is locally exact:
\[
0 \longrightarrow V^{2k-2} \longrightarrow F_{2-2k} \xrightarrow{\phi+\phi'} F_{2k}^{hol} \oplus \overline{F_{2k}^{hol}}\longrightarrow 0.
\]
If we consider modular curve $X = \HH/\Gamma$ for some congruence subgroup $\Gamma \subset SL_2(\Z)$, or $X \subset \HH/\Gamma$ is an open subset then $V^{2k-2}$ gives a local system on $X$ and we get a long exact cohomology sequence:
\begin{multline}
0 \longrightarrow H^0(X, V^{2k-2}) \longrightarrow H^0(X, F_{2-2k}) \xrightarrow{\phi+\phi'} \\ H^0(X, F_{2k}^{hol}) \oplus \overline{H^0(X, F_{2k}^{hol})} \xrightarrow{\psi} H^1(X, V^{2k-2}) \longrightarrow \dots,
\end{multline}
$\psi$ is the boundary map.
Take $f\in H^0(X, F_{2-2k})$, i.e. a function from $F_{2-2k}$ which is invariant under $\Gamma$. Let $\phi(f) = g$ and $\phi'(f) = g'$. We know that $\psi(g) = -\psi(g')$. We see that the obstruction to represent $f$ as a sum of preimages of $g$ and $g'$ \emph{separately} is $\psi(g) \in H^1(X, V^{2k-2})$. %Note H^0(\Gamma

\section{Integrating}\label{integrating}
Consider a meromorphic modular form $f$ of weight $2k$ on $X = \HH / \Gamma$. Suppose the differential $f(\z) p(\z) d \z$ has zero residue for any polynomial $p\in V^{2k-2}$. Choose a basepoint $a \in \HH$. Integrating $f$ gives a cocycle with values in $V^{2k-2}$ in the following way:
\[
\sigma_a(\gamma) = (2 \pi i)^{2k-1} \int_{a}^{\gamma a} f(\z) (X - \z)^{2k-2} d \z.
\]
It satisfies 
\[
\sigma_a(\gamma \gamma') = \sigma_a(\gamma) + \gamma \sigma_a(\gamma').
\]
The integral of $f$ is a function with values in $V^{2k-2}$:
\[
I_a(x) = (2 \pi i)^{2k-1} \int_{a}^x f(\z) (X - \z)^{2k-2} d \z.
\]
This integral depends on the basepoint. The integrand $f(\z) (X - \z)^{2k-2} d\z$ is invariant w.r.t $\Gamma$. The function $I(x)$ is not, it has 'jumps' given by $\sigma$.
If we change basepoint from $a$ to $a'$ the integral and the cocycle change in the following way:
\[
I_a'(x) = I_a(x) + (2 \pi i)^{2k-1} \int_{a'}^a f(\z) (X - \z)^{2k-2} d \z,
\]
\begin{multline*}
\sigma_{a'}(\gamma) = \sigma_a(\gamma) + (2 \pi i)^{2k-1} \int_{a'}^a f(\z) (X - \z)^{2k-2} d \z -\\ (2 \pi i)^{2k-1} \int_{\gamma a'}^{\gamma a} f(\z) (X - \z)^{2k-2} d \z,
\end{multline*}
so if we introduce an element $v_{a a'}$ of $V^{2k-2}$ by 
\[
v_{a a'} = (2 \pi i)^{2k-1} \int_{a'}^a f(\z) (X - \z)^{2k-2} d \z,
\]
then
\[
I_a'(x) = I_a(x) + v_{a a'},
\]
\[
\sigma_{a'} = \sigma_a - \delta v_{a a'},
\]
where $\delta$ denotes the differential 
\[
(\delta v)(\gamma) = \gamma v - v.
\]
Suppose $\sigma$ is homologically trivial, i.e.
\[
\sigma_a = \delta v_a,
\]
for some $v_a \in V^{2k-2}$. Then it is natural to modify our definition of the integral as follows:
\[
I(x) = I_a(x) + v_a.
\]
We see that now the integral is defined up to addition of elements $v\in V^{2k-2}$ such that $\delta v = 0$. If $(V^{2k-2})^{\Gamma} = 0$, which holds if $k > 1$, then this gives a correct definition which does not depend on the choice of the base point. Summarizing

\begin{lem}
If the class of $f$ in $H^1(\Gamma, V^{2k-2})$ is zero and $k>1$, then $I(x)$ is a correctly defined $\Gamma$-equivariant meromorphic function with values in $V^{2k-2}$ satisfying
\[
D I = (2 \pi i)^{2k-2} f(\z) (X - \z)^{2k-2}.
\]
\end{lem}

To calculate $I(x)$ it is convenient to choose base point $a=x$. Then $I_a(x)=0$ and $I(x) = v_x$, where $v_x\in V^{2k-2}$ is a solution of the system of linear equations
\[
\gamma v_x - v_x = (2 \pi i)^{2k-1} \int_x^{\gamma x} f(\z) (X - \z)^{2k-2} d \z,
\]
$\gamma$ runs over a set of generators of $\Gamma$.

Let us allow $f(\z) p(\z) d \z$ to have residues which satisfy certain integrality property, we say that
\begin{defn}\label{integ_res}
The meromorphic modular form of weight $2k$ has integral residues if the differential form
\[
 (2 \pi i)^{2k-1} \binom{2k-2}{n} f(\z) \z^n d \z
\]
has integral residues for all $n$, $0\leq n \leq 2k-2$.
\end{defn}
\begin{rem}
The normalization can be modified to suit our needs.
\end{rem}
Let $V_{\Z}^{2k-2}\subset V^{2k-2}$ denote the abelian group of polynomials with integral coefficients. This group is stable under the action of $SL_2(\Z)$. We make the quotient 
\[
V_{\C/2\pi i}^{2k-2} = V^{2k-2} / V_{\Z}^{2k-2}.
\]
We can develop similar theory as above with coefficients in $V_{\C/2\pi i}^{2k-2}$. The only difference is that here for $k>1$ the group of invariants
\[
(V_{\C/2\pi i}^{2k-2})^\Gamma
\]
may be not trivial, but rather a finite torsion group. Denote its exponent by $e$. Then the integral may be not well defined, but the image of $e I(x)$ in $V_{\C/2\pi i}^{2k-2}$ is well defined. So, in this case the lemma holds:
\begin{lem}
If $f$ has integral residues, the class of $f$ in $H^1(\Gamma, V_{\C/2\pi i}^{2k-2})$ is zero and $k>1$, exponent of $(V_{\C/2\pi i}^{2k-2})^\Gamma$ is $e$ then $e I(x)$ is a correctly defined $\Gamma$-equivariant function with values in $V_{\C/2\pi i}^{2k-2}$ satisfying
\[
D e I = e (2 \pi i)^{2k-2} f(\z) (X - \z)^{2k-2}.
\]
Moreover, $I$ is holomorphic everywhere except poles of $f$.
\end{lem}

\section{Green's functions}
Consider Green's function $G(\z_1, \z_2)=G_{2k}^{\Gamma}(\z_1, \z_2)$ for $\Gamma \subset SL_2(\Z)$~--- congruence subgroup, $k$~--- positive integer greater than $1$. It is a real-valued function, defined on the set
\[
\{(\z_1, \z_2)| \z_i\in \HH, \z_1\neq \gamma\z_2,\; \text{for all}\; \gamma\in \Gamma\}
\]
satisfying the following properties:
\begin{enumerate}
\item $G(\gamma_1 \z_1, \gamma_2 \z_2) = G(\z_1, \z_2)$ for all $\gamma_1, \gamma_2 \in \Gamma$.
\item $\Delta_i G = k(1-k) G$, where $\Delta_i$ denotes Laplace operator with respect to $\z_i$.
\item When $\z_1$ tends to $\z_2$ $G$ has asymptotics $\log|\z_1-\z_2|^2$.
\item When $\z_1$ tends to infinity $G$ tends to $0$.
\end{enumerate}
There exists the following formula for $G$:
\[
G(\z_1, \z_2) = -2\sum_{\gamma\in\Gamma} Q_{k-1}(1-2\frac{|\z_1-\gamma \z_2|^2}{(\z_1-\zc_1)(\gamma\z_2-\gamma\zc_2)}),
\]
where $Q_{k-1}$ is the $k-1$-th Legendre function. 
Denote 
\[
t(\z_1, \z_2) = 1-2\frac{|\z_1-\z_2|^2}{(\z_1-\zc_1)(\z_2-\zc_2)}.
\]
It satisfies
\[
t(\z_1, \z_2)-1 = 2\frac{(\z_1-\z_2)(\zc_2-\zc_1)}{(\z_1-\zc_1)(\z_2-\zc_2)},
\]
\[
t(\z_1, \z_2)+1 = 2\frac{(\z_1-\zc_2)(\z_2-\zc_1)}{(\z_1-\zc_1)(\z_2-\zc_2)}.
\]
It follows, that
\[
G(\z_1, \z_2) = -2 Q_{k-1}(t(\z_1, \z_2)) + g(\z_1, \z_2),
\]
where $g$ is a function, smooth in the neighbourhood of the set $\{\z_1=\z_2\}$.
Denote 
\[
f(\z_1, \z_2) = -2 Q_{k-1}(t(\z_1, \z_2)).
\]
Note, that $f$ also satisfies properties of $G$ except the first property. Instead it satisfies 
\begin{enumerate}
\item[(i')] $G(\gamma \z_1, \gamma \z_2) = G(\z_1, \z_2)$ for all $\gamma\in SL_2(\R)$.
\end{enumerate}

Since the function $G$ is in the space $F_0$ with respect to each of the arguments in the sence of the section \ref{eigenvalues} we can calculate holomorphic and antiholomorphic images with respect to some of the variables, or with respect to both. The same applies to the function $f$. Denote by $G_h$, $G_a$, $f_h$, $f_a$ corresponding holomorphic and antiholomorphic images of $G$ and $f$ w.r.t. $\z_1$. Denote by $G_{hh}$, $G_{ha}$, $G_{ah}$, $G_{aa}$ holomorphic image w.r.t. $\z_2$ of $G_h$, antiholomorphic image w.r.t. $\z_2$ of $G_h$, e.t.c. The same for $f$.
\begin{lem}
For $f_h$ and $f_a$ we have the following formulas:
\[
f_h(\z_1, \z_2)=(k-1)!^2 (-1)^{k-1} (2 \pi i)^{-k} \frac{(\z_2-\zc_2)^k}{(\z_1 - \z_2)^k (\z_1 - \zc_2)^k},
\]
\[
f_a(\z_1, \z_2)=(k-1)!^2 (-1)^{k-1} (2 \pi i)^{-k} \frac{(\z_2-\zc_2)^k}{(\zc_1 - \z_2)^k (\zc_1 - \zc_2)^k}.
\]
\end{lem}
\begin{proof}
The function $f_h$ satisfies the following properties:
\begin{itemize}
\item $f_h$ is holomorphic w.r.t $\z_1$
\item $f_h(\gamma \z_1, \gamma \z_2) = f_h(\z_1, \z_2) (c\z_1 + d)^{2k}$ for all $\gamma\in SL_2(\R)$.
\end{itemize}
Hence the function
\[
\frac{f_h(\z_1, \z_2) (\z_1 - \z_2)^k (\z_1 - \zc_2)^k} {(\z_2-\zc_2)^k}
\]
is invariant under the diagonal action of $SL_2(\R)$ and is holomorphic in $\z_1$ on its domain of definition. Hence if we fix $\z_2$ the function as a function of $\z_1$ must be constant along circles centered in $\z_2$. This is impossible for non-constant holomorphic functions. Therefore
\[
f_h(\z_1, \z_2) = c_1 \frac{(\z_2-\zc_2)^k}{(\z_1 - \z_2)^k (\z_1 - \zc_2)^k} 
\]
for some constant $c_1\in\C$.
To find the constant recall that 
\[
f_h = (k-1)!\delta_1^k f,
\]
where $\delta_1$ denotes the operator $\delta$ applied w.r.t. the argument $\z_1$. Since when $\z_1$ approaches $\z_2$
\[
f(\z_1, \z_2) \sim \log|\z_1-\z_2|^2,
\]
modulo terms of smaller growth.
and the function $f$ is a sum of products of logarithms and rational functions in $\z_1$, $\z_2$, $\zc_1$, $\zc_2$ we can apply $\delta_1$ to both sides and obtain
\[
f_h(\z_1, \z_2) \sim (k-1)!^2 (-1)^{k-1} (2 \pi i)^{-k}(\z_1-\z_2)^{-k},
\]
which implies that 
\[
c_1 = (k-1)!^2 (-1)^{k-1} (2 \pi i)^{-k}.
\]
Since the function $f$ is real its antiholomorphic image is simply the complex conjugate of the holomorphic image, which implies the second formula.
\end{proof}
Put $k'=1-k$ then
\[
f_h(\z_1, \z_2) = (k-1)!^2 (-1)^{k-1} (2\pi i)^{k'-1} \frac{(\z_1-\z_2)^{k'-1}(\z_1-\zc_2)^{k'-1}}{(\z_2-\zc_2)^{k'-1}}
\]
\[
=(k-1)!^2 (-1)^{k-1} Q_{k'\,0}(\z_2)(\z_1),
\]
where $Q_{k'\,l}$ is defined in the same way as polynomials $Q_l$ in section \ref{funcs_in_reps}, with $k'$ instead of $k$ - it is not a polynomial, but a rational function in $X$. However, most of the properties remain true, in particular
\[
\delta^k Q_{k'\,0} = (1-k')(2-k')\dots(k-k') Q_{k'\,k} = \frac{(2k-1)!}{(k-1)!} (2\pi i)^{-2k} (X-\z)^{-2k},
\]
which implies
\[
f_{hh}(\z_1, \z_2) = (k-1)!^2 (2k-1)! (-1)^{k-1} (2\pi i)^{-2k} (\z_1-\z_2)^{-2k},
\]
\[
f_{ha}(\z_1, \z_2) = (k-1)!^2 (2k-1)! (-1)^{k-1} (2\pi i)^{-2k} (\z_1-\zc_2)^{-2k},
\]
\[
f_{ah}(\z_1, \z_2) = (k-1)!^2 (2k-1)! (-1)^{k-1} (2\pi i)^{-2k} (\zc_1-\z_2)^{-2k},
\]
\[
f_{aa}(\z_1, \z_2) = (k-1)!^2 (2k-1)! (-1)^{k-1} (2\pi i)^{-2k} (\zc_1-\zc_2)^{-2k}.
\]
For $G_{**}$ we can conclude, that 
\begin{lem}
\begin{itemize}
\item The function $G_{hh}$ is a meromorphic cusp form of weight $2k$ in both variables with pole along diagonal, such that
\[
G_{hh}(\z_1-\z_2) - (k-1)!^2 (2k-1)! (-1)^{k-1} (2\pi i)^{-2k}(\z_1-\z_2)^{-2k}
\]
is holomorphic in the neighbourhood of $\{\z_1=\z_2\}$.
\item The function $G_{ha}$ is a holomorphic cusp form of weight $2k$ in first variable and $\overline{G_{ha}}$ is a holomorphic cusp form of weight $2k$ in second variable.
\item $G_{aa}$, $G_{ah}$ are complex conjugates of $G_{hh}$, $G_{ha}$ correspondingly.
\end{itemize}
\end{lem}

Since the function $G$ is in the space $F_0$ with respect to each of the arguments in the sence of the section \ref{eigenvalues} we can apply construction of section \ref{eigenvalues} to $G$ and get a sequence of functions $G_{ij}(\z_1, \z_2)$ for $1-k \leq i,j \leq k-1$, which satisfies
\begin{enumerate}
\item $G_{ij}$ is in $F_{2i}$ w.r.t. $\z_1$ and $F_{2j}$ w.r.t. $\z_2$
\item $G_{0 0} = G$
\item $\delta_1 G_{ij} = (i+k) G_{i+1\,j}$ for $1-k \leq i \leq k-2$
\item $\delta_2 G_{ij} = (j+k) G_{i\,j+1}$ for $1-k \leq j \leq k-2$
\item $\delta_1^- G_{ij} = (i-k) G_{i-1\,j}$ for $2-k \leq i \leq k-1$
\item $\delta_2^- G_{ij} = (j-k) G_{i\,j-1}$ for $2-k \leq j \leq k-1$
\end{enumerate}
Here we denote by $\delta_i$, $\delta_i^-$ corresponding operators w.r.t. $\z_i$. We can construct a function with values in polynomials in two valiables $X_1$, $X_2$ as in section \ref{funcs_in_reps}:
\[
\wt G(\z_1, \z_2) = \sum_{i,j=1-k}^{k-1} G_{ij}(\z_1, \z_2) Q_{-i}(\z_1)(X_1) Q_{-j}(\z_2)(X_2).
\]
The function $G_h$ has weight $0$ w.r.t. $\z_2$, so if we make a sequence $G_{hl}$ from $G_h$ with respect to variable $\z_2$ we obtain
\[
D_1 \wt G(\z_1, \z_2) = \frac{(2\pi i)^{2k-2}(X_1-\z_1)^{2k-2}}{(2k-2)!} \sum_{l=1-k}^{k-1} G_{hl}(\z_1, \z_2) Q_{-l}(\z_2)(X_2)
\]
\[
=\frac{(2\pi i)^{2k-2}(X_1-\z_1)^{2k-2}}{(2k-2)!} \wt G_h(\z_1, \z_2)(X_2).
\]
Differentiating w.r.t. $\z_2$ gives
\[
D_2 D_1 \wt G(\z_1, \z_2) = \frac{(2\pi i)^{4k-4}(X_1-\z_1)^{2k-2}(X_2-\z_2)^{2k-2}}{(2k-2)!^2}G_{hh}(\z_1, \z_2).
\]
Note, that functions $G_{**}$ are holomorphic (antiholomorphic) modular forms, so we can actually hope to express them in terms of, say, Eisenstein series and $j$-invariant. In order to recover $G$ we need to integrate $G_{**}$, and this can be impossible in the class of modular forms if certain cohomology groups are not trivial. Let us make an additional assumption:
\begin{assumption}
Suppose, there are no cusp forms of weight $k$ for $\Gamma$.
\end{assumption}
One simplification comes immediately:
\begin{prop}
Functions $G_{ah}$ and $G_{ha}$ are identically zero.
\end{prop}
Recall, that for the function $\wt{G_h}$ this implies:
\begin{prop}
Function $\wt{G_h}$ is holomorphic in its domain of definition.
\end{prop}
Final step for recovering of $\wt G$ would be to integrate $\wt{G_h}$. Let us study residue of the following differential at the point $\z_1=\z_2$:
\[
w_G = (X_1-\z_1)^{2k-2} \wt{G_h}(\z_1, \z_2)(X_2) 2\pi i d \z_1.
\]
Clearly it is the same as the residue of 
\[
w_f = (X_1-\z_1)^{2k-2} \wt{f_h}(\z_1, \z_2)(X_2) 2\pi i d \z_1.
\]
The latter can be calculated explicitly. Since
\[
f_{hl}(\z_1, \z_2) = (k-1)!^2 (-1)^{k-1} Q_{k'\,l}(\z_2)(\z_1),
\]
in particular,
\[
f_{h\,1-k}(\z_1, \z_2) = (k-1)!^2 (-1)^{k-1} (2\pi i)^{-1} (\z_1-\z_2)^{-1}\left(\frac{\z_2-\zc_2}{\z_1-\zc_2}\right)^{2k-1}
\]
we see that the Laurent expansion of $f_{h\,1-k}$ in the variable $\z_1-\z_2$ (treating $\z_2$ as a parameter) is
\[
f_{h\,1-k}(\z_1, \z_2) = (k-1)!^2 (-1)^{k-1} (2\pi i)^{-1} (\z_1-\z_2)^{-1} + O(1).
\]
Hence
\[
D_2^n f_{h\,1-k}(\z_1, \z_2) = (k-1)!^2 (-1)^{k-1} n! (2\pi i)^{-1-n} (\z_1-\z_2)^{-1-n} + O(1).
\]
We use the lemma \ref{ftilde} to compute the Laurent expansion of $\wt f_h$:
\[
\wt f_h(\z_1, \z_2)(X_2) = (k-1)!^2 (-1)^{k-1} (2\pi i)^{-1} \sum_{n=0}^{2k-2} (X_2 - \z_2)^n (\z_1-\z_2)^{1-n} + O(1).
\]
Since 
\[
\res (X_1-\z_1)^{2k-2}(\z_1-\z_2)^{1-n} d\z_1 = (-1)^n\binom{2k-2}{n}(X_1-\z_2)^{2k-2-n}
\]
we can compute the residue term by term:
\[
\res w_f = (k-1)!^2 (-1)^{k-1} \sum_{n=0}^{2k-2} (-1)^n\binom{2k-2}{n} (X_2 - \z_2)^n (X_1-\z_2)^{2k-2-n}
\]
\[
= (k-1)!^2 (-1)^{k-1} (X_1-X_2)^{2k-2}.
\]
Hence
\begin{prop}
The residue at $\z_1=\z_2$ of the differential 
\[
w_G = (X_1-\z_1)^{2k-2} \wt{G_h}(\z_1, \z_2)(X_2) 2\pi i d \z_1
\]
equals
\[
(k-1)!^2 (-1)^{k-1} (X_1-X_2)^{2k-2}.
\]
\end{prop}
Let $p\in V^{2k-2}$ be a polynomial with integer coefficients.
Since $w_G$ is a polynomial in $X_2$ we can consider value of the bilinear form on $w_G$ and $p$, which will be denoted $(w_G, p)_{X_2}$. It is again a differential. The residue at $\z_1=\z_2$ is
\[
\res (w_G, p)_{X_2} = (k-1)!^2 (-1)^{k-1} p(X_1).
\]
Therefore
\begin{prop}
The following form has integral residues w.r.t. variable $\z_1$ (see definition \ref{integ_res}):
\[
\frac{1}{(2\pi i)^{2k-2}(k-1)!^2} (\wt{G_h}(\z_1, \z_2), p)_{X_2}.
\]
\end{prop}
Therefore using results from section \ref{integrating} we obtain a function $I(\z_1,\z_2, p)$ with values in $V^{2k-2}_{\C/2\pi i}$ which satisfies
\[
D_1 I(\z_1,\z_2,p) = e \frac{1}{(k-1)!^2} (X_1-\z_1)^{2k-2} (\wt{G_h}(\z_1, \z_2), p)_{X_2},
\]
$e$ is a positive integer, defined in section \ref{integrating}. This $I$ is holomorphic in $\z_1$, $\z_2$ for $\z_1\neq\gamma \z_2$ for all $\gamma\in\Gamma$. Let $q$ be another polynomial with integer coefficients. Let $K$ be the least common multiple of numbers $\binom{2k-2}{i}$. Then the pairing with $Kq$ is defined for elements of $V^{2k-2}_{\C/2\pi i}$ and takes values in $\C/2\pi i$. We introduce a function
\[
\hat G(\z_1, \z_2, q, p) = \frac{(I(\z_1, \z_2, p), K q)_{X_1}}{Ke}
\]
Note, that $\hat G$ takes values in $\C/\frac{1}{K e}2\pi i$ and
\[
D_1 \hat G(\z_1, \z_2, q, p) = \frac{1}{(k-1)!^2} q(\z_1) (\wt{G_h}(\z_1, \z_2)(X_2), p)_{X_2},
\]
which implies that
\[
((\wt G(\z_1, \z_2)(X_1, X_2),q)_{X_1},p)_{X_2} = \frac{(k-1)!^2 (2\pi i)^{2k-2}}{(2k-2)!} 2 \Re \hat G(\z_1, \z_2, q, p),
\]
here $\Re$ denotes the real part of a complex number. Let $x_1$, $x_2$ be two CM points on $\HH$. Let
\[
a_i x_i^2 + b_i x_i + c_i = 0
\]
be corresponding minimal equations for $x_i$ with integral coefficients. Let 
\[
p_i(x) = (a_i x^2 + b_i x + c_i)^{k-1}
\]
be two polynomials  with integer coefficients. We can express the value of $G$ on $x_1$, $x_2$ as follows:
\[
G(x_1, x_2) = (2\pi i)^{4-4k}\binom{2k-2}{k-1}^2((\wt G(x_1, x_2), Q_0(x_1))_{X_1}, Q_0(x_2))_{X_2},
\]
since $Q_l$ are orthogonal. 
\[
Q_0(x_i) = (2\pi i)^{k-1} \frac{(X-x_i)^{k-1}(X-\bar x_i)^{k-1}}{(x_i-\bar x_i)^{k-1}} = (2\pi i)^{k-1} \frac{p_i(X)}{D_i^{\frac{k-1}2}},
\]
where $D_i = b_i^2-4 a_i c_i$~--- the discriminant.
Therefore
\[
G(x_1, x_2) = (2\pi i)^{2-2k}\binom{2k-2}{k-1}^2 (D_1 D_2)^{\frac{1-k}2} ((\wt G(x_1, x_2), p_1)_{X_1}, p_2)_{X_2}
\]
\[
=\binom{2k-2}{k-1} (D_1 D_2)^{\frac{1-k}2} 2 \Re\hat G(x_1, x_2, p_1, p_2).
\]
\end{document}
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