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\title{Math 678, Lecture 2}
\author{Notes by Bryden Cais}
\date{Sept. 6, 2002}
\begin{document}
\maketitle

\section{Elliptic Functions}

        Modular functions naturally arise from the theory of elliptic functions.  An \textit{elliptic function} is a
        meromorphic function $f$ on $\C$ satisfying for all $z\in\C$.
        $$f(z+\omega_1)=f(z+\omega_2)=f(z)$$ for $\omega_1,\omega_2\in\C$ with $\Im(\omega_1/\omega_2)>0$ (so that the
        ratio is not real).  Thus if we let $\Lambda=\{m\omega_1+n\omega_2:(m,n)\in\Z^2\}$ be the lattice in $\C$
        spanned by $\omega_1,\omega_2$, we see that an elliptic function is just a meromorphic function on $\C/\Lambda$.

        \begin{exer}
            Discuss the case when $\omega_1/\omega_2\in \R$.
        \end{exer}

        We will construct an important example of a doubly periodic function by the process of ``averaging.''  The idea
        is o pick some suitable function $g$ that decays rapidly and to consider the sum
        $$f(z)=\sum_{\lambda\in\Lambda} g(z+\lambda).$$  Let us first derive some general properties of elliptic functions.

\section{General Properties of Elliptic Functions}
        Let $F$ be a \textit{fundamental parallelogram} for the elliptic function $f$ (that is, $F$ is the parallelogram
        with sides $\omega_1,\omega_2$).  It is not difficult to show that
        $$\frac{1}{2\pi i}\int_{F} \frac{f^{'}(z)}{f(z)}\,\mathrm{d}z=0$$
        sine the integrals along opposite sides of $F$ cancel.  This implies that any elliptic function has the same number
        of zeroes as poles, counted according to multiplicity.  (Another way to see this is to realize that $\C/\Lambda$ as
        a Riemann surface of genus 1).  Similarly, we see that no elliptic function can be holomorphic in $F$, for then it would
        be holomorphic on $\C$ and thus, by Louiville's theorem, a constant.  Finally, observe that
        $$\int_{F}f(z)\,\mathrm{d}z=0,$$
        so that the sum of the residues of any elliptic function is zero.  Thus, no elliptic function can have a single, simple pole.
        Moreover, if $f$ has only a single pole in $F$, we see that the residue there must be $0$.

\section{The Weierstrass $\wp$ Function}
        With these generalities in mind, consider the function $g(z)=z^{-k}$.  We then ask when $\sum g(z+\lambda)$ converges absolutely.
        Observe that in a circle of radius $R$ there are approximately $R^2$ lattice points.  More precisely, it is not hard to see that
        the number of lattice points inside a circle of radius $R$ is
        $$\frac{\pi}{A(F)}R^2+O(R),$$
        where $A(F)$ denotes the area of $F$.  In particular, the number of lattice points in the annulus $2^j\leq |\lambda|\leq 2^{j+1}$
        is bounded by $C 2^{2j}$ for some constant $C$.  Therefore, we see that
        $$\sum_{\lambda\in\Lambda}\frac{1}{(z+\lambda)^k}$$
        converges absolutely if and only if
        $$\sum_{j=0}^{\infty}\sum_{2^j\leq |\lambda|\leq 2^{j+1}}\frac{1}{\lambda^k}$$
        converges.  By our remarks above, this sum is bounded above by
        $$C\sum_{j=1}^{\infty} \frac{2^{2j}}{2^{jk}}=C\sum_{j=1}^{\infty} \frac{1}{2^{j(k-2)}},$$
        for some constant $C$.  Clearly, this converges for any $k>2$.

        As an example, we see that the sum
        $$f(z)=\sum_{\lambda\in\Lambda}\frac{1}{(z+\lambda)^3}$$
        converges absolutely so that $f(z)$ is an elliptic function.  Moreover, the sum
        $$\wp(z)=\frac{1}{z^2}+\sum_{\lambda\in\Lambda^{'}}\left\{\frac{1}{(z+\lambda)^2}-\frac{1}{\lambda^2}\right\}$$
        converges absolutely (as long as we do not rearrange anything inside the curly brackets) since each term
        behaves like $\lambda^{-3}$.  Here, $\Lambda^{'}=\Lambda\setminus\{0\}$.  Observe that $\wp(z)$ is clearly an even
        function of $z$ and that $\wp^{'}(z)=-2f(z)$.  This shows that for any $\lambda\in\Lambda,$ one has
        $$\wp(z+\lambda)=\wp(z)+C_{\lambda}$$ for some constant $C_{\lambda}$ depending only on $\lambda$.  Setting
        $z=-\lambda/2$ and using the fact that $\wp(z)$ is an even function shows that $C_{\lambda}=0$ for any $\lambda$,
        and hence that $\wp(z)$ is an elliptic function.

\section{The Laurent Expansion of $\wp(z)$}
        Let us now compute the laurent expansion of $\wp(z).$  One readily computes
        \begin{align*}
            \frac{1}{(z+\lambda)^2}&=\frac{1}{\lambda^2}\frac{1}{(1+(z/\lambda))^2}\\
            &=\frac{1}{\lambda^2}\left(1-2\frac{z}{\lambda}+3\frac{z^2}{\lambda^2}-\cdots\right),\\
            \intertext{so that}
            \wp(z)&=\frac{1}{z^2}+\sum_{\lambda\in\Lambda^{'}}\sum_{n=1}^{\infty}(-1)^n(n+1)\frac{z^n}{\lambda^{n+2}}\\
            &=\frac{1}{z^2}+\sum_{n=1}^{\infty}(-1)^n(n+1)G_{n+2}(\Lambda)z^n,\\
            \intertext{where}
            G_{n}(\Lambda)&:=\sum_{\lambda\in\Lambda^{'}}\frac{1}{\lambda^n},
        \end{align*}
        and the change in order of summation above is
        justified by absolute convergence.  Now observe that if $\lambda\in\Lambda^{'}$ then also $-\lambda\in\Lambda^{'}$
        so that $G_{n}(\Lambda)=0$ if $n$ is odd.  Thus we have
        $$\wp(z)=\frac{1}{z^2}+\sum_{n=1}^{\infty}(2n+1)G_{2n+2}(\Lambda)z^{2n}.$$

\section{The Differential Equation Satisfied by $\wp(z)$}
        With the laurent expansion of $\wp(z)$ we can now compute
        \begin{align*}
            (\wp^{'}(z))^2&=\frac{4}{z^6}+\frac{c_1}{z^2}+c_2z^2+\cdots\\
            (\wp(z))^3&=\frac{1}{z^6}+\frac{b_1}{z^2}+b_2z^2+\cdots,
        \end{align*}
        so that there exists a constant $d$ such that
        $$(\wp^{'}(z))^2-4(\wp(z))^3-d\wp(z)=z^2+\cdots$$
        But the left and side is obviously an elliptic function.  On the other hand, the only possible pole is at $z=0$ and we see
        that the left hand side is analytic there.  Thus it is a holomorphic elliptic function and therefore a constant.  This will
        turn out to have profound consequences.

\section{Eisenstein Series}

        The series $G_{n}(\Lambda)$ are known as Eisenstein Series.  It is clear that they converge absolutely for
        $n\geq 3$ and that $G_{n}(\Lambda)=0$ if $n$ is odd.  Observe that $G_n{\Lambda}$ depends only on the lattice
        $\lambda$ and not any chosen basis for $\lambda$.  Let
        $$\gamma=\begin{pmatrix}a & b\\ c& d\end{pmatrix}\in\SL{\Z}.$$  Then it is clear that $\gamma$ takes a basis
        of $\Lambda$ to itself where we have
        \begin{align*}
            \omega_1&\mapsto a\omega_1+b\omega_2\\
            \omega_2&\mapsto c\omega_1+d\omega_2.
        \end{align*}
        This gives an action of $\SL{\Z}$ on the $G_n(\Lambda)$ under which $G_{n}(\Lambda)$ is invariant.  Observe that
        we can require, without loss of generality, that $z=\omega_1/\omega_2\in\H.$  Then the action of $\SL{\Z}$
        on $\Lambda$ induces the usual fractional linear action on $\H$, that is,
        $$\gamma z=\frac{az+b}{cz+d}.$$
        Write
        \begin{align*}
            G_k(\Lambda)&=\sum_{(m,n)\in\Z^2\setminus\{0\}}\frac{1}{(m\omega_1+n\omega_2)^k}\\
            &=\frac{1}{\omega_2^k}\sum_{(m,n)\in\Z^2\setminus\{0\}}\frac{1}{(m(\omega_1/\omega_2)+n)^k}.
        \end{align*}
        Therefore we define
        $$G_{k}(z)=\sum_{(m,n)\in\Z^2\setminus\{0\}}\frac{1}{(mz+n)^k}.$$
        By our discussion above, we readily see that
        $$G_k(\gamma z)=(cz+d)^{k}G_k(z)$$
        for $\gamma\in\SL{\Z}$ as above.

\section{Modular Forms}

\begin{defn}
A \textit{Modular Form} of weight $k$ for $\Gamma=\SL{\Z}$ is a holomorphic function $f$ on $\H$ such that
\begin{enumerate}
    \item   $f(\gamma z)=(cz+d)^k f(z)$ for any $\gamma=\left(\begin{smallmatrix}a & b\\ c& d\end{smallmatrix}\right)\in\SL{\Z}$
    \item   $f(z)$ is holomorphic at the cusps of $\H/\Gamma$.
\end{enumerate}
\end{defn}

Item 2 requires explanation.  Observe that any modular form is periodic with period 1 (by item 1) and therefore has a fourier expansion,
or, equivalently, a laurent expansion in the variable $q=e(z)$.  Write
$$f(z)=\sum_{n=M}^{\infty}a_nq^n,$$
where $M$ is possibly negative.  Observe that $q:\H\longrightarrow D\setminus \{0\}$ where $D$ is the unit disc.  We say that
$f$ is holomorphic at the cusps of $\H/\Gamma$ if the function $\sum_{n=M}^{\infty}a_nq^n$ is holomorphic at $q=0$ (corresponding
to $z=\infty\in\H$).

\begin{defn}
    A \textit{cusp form} for $\SL{\Z}$ is a modular form that is 0 at the cusps; i.e. a modular form $f(z)=\sum_{n=0}^{\infty}a_nq^n$
    with $a_0=0$.
\end{defn}

\end{document}
