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\title{Math 678, Lecture 1}
\author{Notes by Bryden Cais}
\date{Sept. 4, 2002}
\begin{document}
\maketitle
\section{Modular Functions}
    Observe that we have a discrete action of $\Z$ on $\R$ by translations:
    $$z\mapsto z+a.$$  We can ask for the $\R$ vector space of smooth functions invariant under this action,
    that is, the space of smooth fnctions on $\R/\Z$.  Fourier analysis tells us that this is spanned
    by the functions
    $$\left\{e(nz):n\in\Z\right\},$$
    where $e(z)=\exp(2\pi i z)$.  The situation becomes much more interesting when we consider the following
    generalization: Let $\H=\left\{z\in\C:\Im(z)>0\right\}$ be the usual upper half plane.  Then we have
    an action of $\SL{\Z}$ on $\H$ via fractional linear transformation:
    $$\begin{pmatrix}a&b\\c&d\end{pmatrix}z=\frac{az+b}{cz+d}.$$
    (The right way to think about this action is as the action inherited from the natural action of
    $\SL(\Z)$ on $\C^2\setminus\{0\}/\C^*$).  We then want to study the holomorphic functions on $\H$ invariant
    under the action of $\SL{\Z}.$  This turns out to be the wrong generalization: it isn't hard to show that
    $\H/\SL{\Z}$ has the structure of a compact Riemann surface (as long as we add the point at infinity).  It is
    a fact that any holomorphic function on a compact Riemann surface must be constant.  So we relax the conditions
    on the functions under consideration to holomorphic functions $f$ on $\H$ such that
    $$f(\gamma z)=j(\gamma,z)f(z),$$
    for some simple function $j$ with no poles in $\H$.  Clearly, if we are to have a genuine action of $\SL{\Z}$ on
    these functions, we must have
    \begin{align}
        f(\gamma_1\gamma_2 z)&=f(\gamma_1(\gamma_2 z)),\nonumber\\
        \intertext{which implies that we must have}
        j(\gamma_1\gamma_2, z)&=j(\gamma_1,\gamma_2z)j(\gamma_2,z)\label{jprop}.
    \end{align}
    The simplest function $j$ that one can imagine is the function $j(\gamma,z)=(cz+d)^k$, where
        $$\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$  It is not difficult to verify that this
        function satisfies (\ref{jprop}).  (Notice that it is sufficient to check this for $k=1$ since
        $z\mapsto z^k$ is a homomorphism.)

        We will therefore define (at least for now) a \textit{modular function} to be a holomorphic function $f$ on $\H$ satisfying
        $$f(\gamma z)=(cz+d)^k f(z)$$
        for all $\gamma=\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\in \SL{\Z}$ and $z\in\H.$
        Observe that since
        $$\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\SL{\Z},$$
        any such $f$ must have $k$ even, and is periodic with period 1, so it has a fourier series expansion.

        One of the main threads running throughout the course will be the connection between $L$ series
        and modular functions.  To motivate this connection, consider the Riemann zeta function defined by
        $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}.$$  It is not hard to see that this series defines
        a holomorphic function of $s$ for $\Re(s)>1$.  Moreover, $\zeta(s)$ satisfies a functional equation:
        \begin{eqnarray}
            \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{s-1/2}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).
        \label{fneq}
        \end{eqnarray}
        The standard way to prove this is to take mellin transforms of both sides of the identity
        \begin{eqnarray}
            \sum_{n\in\Z}e(-n^2/z)=\sqrt{\frac{z}{i}}\sum_{n\in\Z}e(n^2 z),
        \label{theta}
        \end{eqnarray}
        which in turn is proved using the Poisson summation formula.  This states that for any smooth function $f$
        with compact support we have
        $$\sum_{n\in\Z}f(n)=\sum_{n\in\Z}\hat{f}(z)$$

        \begin{exer}
            Prove the Poisson summation formula.
        \end{exer}

\section{Divisor Sums and Sums of Squares}

        Modular functions are ubiquitous in modern number theory.  Define
        $$\sigma_{k-1}(n)=\begin{cases}\frac{(k-1)!\zeta(k)}{(2\pi i)^k}\quad &\text{if $n=0$}\\
        \sum_{d|n}d^{k-1}\quad &\text{otherwise}\end{cases}.$$
        In 1916, Ramanujan stated the remarkable result
        \begin{eqnarray*}
            \sum_{a+b=m}\sigma_{r-1}(a)\sigma_{s-1}(b)=\frac{(r-1)!(s-1)!\z(r)\z(s)}{(r+s-1)!\z(r+s)}\sigma(r+s-1)+\varepsilon,
        \end{eqnarray*}
        where $\varepsilon$ is an error term.  It turns out that for $m<12,$ we have $\varepsilon=0$ (which will turn out to be
        because the space of cusp forms of dimension $m<12$ is just $\{0\}$).  This identity is an analytic way of expressing
        very important algebraic information about the structure of modular forms of a given weight.

        Another number theoretic context in which modular forms naturally arise is that of the problem of the number of
        ways of representing a positive integer as the sum of $k$ squares.  Let $r_k(n)$ denote the number of ways that
        $n$ can be represented as the sum of $k$ squares.  The formulae
        \begin{align*}
            r_2(n)&=\sum_{d|n}\left(\frac{-4}{d}\right)\\
            r_4(n)&=8(\sum_{d|n} d-\sum_{4d|n} 4d)
        \end{align*}
        are classical.  It turns out that $r_k(n)$ san similarly be expressed as such a divisor sum, though as $k$ increases
        things tend to get much more complicated.  This result comes straight out of the theory of modular functions.

\section{The $\tau$ function}

        A beautiful example of the interplay between modular functions and number theory arises when one considers the
        function
        $$\Delta(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}=\sum_{n=1}^{\infty}\tau(n)q^n,$$
        where $q=e(z)$.  One can show that $\Delta$ is a modular function (of weight 12).  The numbers $\tau(n)$
        are obviously integers.  Moreover, they enjoy remarkable number theoretic properties.  Namely,
        we have
        \begin{align*}
            \tau(m)\tau(n)&=\tau(mn)\\
            \intertext{whenever $(m,n)=1$, and}
            |\tau(p)|&\leq 2p^{11/2}
        \end{align*}
        for all primes $p$.  Both of these results are deep (especially the latter, which was only proved recently by Deligne) and
        come from the theory of modular forms.

\end{document}
