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\newtheorem{exer}{Exercise}
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\title{Modular Forms, PS 8}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
Let $a/b\in \P(\Q)$ with $(a,b)=1$ and $1/0$ representing $\infty.$  Since $(a,b)=1$, we can find $x,y\in\Z$ so that
$xa+yb=1$.  Without loss of generality, we may assume that $x$ is odd: if it were even then $b$ would have to be
odd, in which case $(x+b)a+(y-a)b=1$ also and $x+b$ is odd.  It is also clear that we must have $(x,y)=1$, and
since $x$ is odd, $(x,4y)=1$.  There therefore exist integers $k,z$ so that $xz-4yk=1$.  Then
$$\gamma=\begin{pmatrix}x & y \\ 4k & z\end{pmatrix}\in \G_0(4)$$ and
$$\gamma\frac{a}{b}=\frac{1}{s}$$ for some integer $s$.  Now let $j\in \Z$ be such that $-2\leq s+4j< 2$.
then
$$\alpha_j=\begin{pmatrix}1 & 0\\ 4j & 1\end{pmatrix}\in\G_0(4)$$
and
$$\alpha_j\frac{1}{s}=\frac{1}{4j+s}.$$  Since $\bigl(\begin{smallmatrix}-1 & \phantom{-}0 \\ \phantom{-}0 & -1\end{smallmatrix}\bigr)\in \G_0(4),$
It follows that every $a/b\in\P(\Q)$ is equivalent under $\G_0(4)$ to one of
$$\infty=\frac{1}{0},\ 1,\ \frac{1}{2}.$$  It is not difficult to see that these three rationals are inequivalent modulo $\G_0(4)$ by
observing that the denominator of $g(p/q)$ is congruent to $q$ modulo $4$ for any $g\in \G_0(4)$ and $p/q\in \P(\Q)$.  Therefore, a set of representatives
for the cusps of $\G_0(4)/\H$ is $\{\infty,1,1/2\}$.  It is easy to see that
\begin{align*}
    \gamma_{1}&=\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}\\
        \gamma_{1/2}&=\begin{pmatrix}1 & 0\\2 & 1\end{pmatrix}\\
\end{align*}
satisfy $\gamma_{a}\infty=a$.

Now recall that
$$\T(z):=\sum_{n\in\Z}e(n^2z)$$ obeys
\begin{align*}
    \T(z+2)&=\T(z)\\
    \T(-1/z)&=\sqrt{\frac{z}{i}}\T(z).
\end{align*}
Moreover, we have
\begin{align*}
    \T(z+1)&=\sum_{n\in\Z} (-1)^n e(n^2z)\\
    &=2\sum_{n\in\Z}e((2n)^2 z)-\sum_{n\in\Z}e(n^2z)\\
    &=2\T(4z)-\T(z).
\end{align*}
We therefore have
\begin{align*}
    \T(z)^8|_{\gamma_1}&=(z+1)^{-4}\T\left(\frac{z}{z+1}\right)^8\\
    &=(z+1)^{-4}\T\left(\frac{-1}{-1-\frac{1}{z}}\right)^8\\
    &=(z+1)^{-4}\left(\sqrt{\frac{-1-\frac{1}{z}}{i}}\right)^8\T\left(-1-\frac{1}{z}\right)^8\\
    &=(z+1)^{-4}\left(\sqrt{\frac{-1-\frac{1}{z}}{i}}\right)^8\left(2\T\left(-\frac{4}{z}\right)-\T\left(-\frac{1}{z}\right)\right)^8\\
    &=(z+1)^{-4}\left(\sqrt{\frac{-1-\frac{1}{z}}{i}}\right)^8\left(2\left(\sqrt{\frac{z}{4i}}\right)^8\T\left(\frac{z}{4}\right)-
    \left(\frac{z}{i}\right)^8\T\left(z\right)\right)^8\\
    &=(z+1)^{-4}\left(\sqrt{z+1}\right)^8\left(2\T\left(\frac{z}{4}\right)-\T(z)\right)^8\\
    &=\left(2\T\left(\frac{z}{4}\right)-\T(z)\right)^8
\end{align*}
Now since
$$\T(z)=1+2\sum_{n=1}^{\infty} e(n^2 z),$$ we see that the constant term in the fourier expansion of
$\left(2\T\left(\frac{z}{4}\right)-\T(z)\right)^8$ is $1$.  It follows that the value of $\T(z)^8$ at the
cusp $1$ is $1$.  Similarly, we find
\begin{align*}
    \T(z)^8|_{\gamma_{1/2}}&=(2z+1)^{-4}\T\left(\frac{z}{2z+1}\right)^8\\
    &=(2z+1)^{-4}\T\left(\frac{-1}{-2-\frac{1}{z}}\right)^8\\
    &=(2z+1)^{-4}\left(\sqrt{\frac{-2-\frac{1}{z}}{i}}\right)^8\T\left(-2-\frac{1}{z}\right)^8\\
    &=(2z+1)^{-4}\left(\sqrt{\frac{-2-\frac{1}{z}}{i}}\right)^8\T\left(-\frac{1}{z}\right)^8\\
    &=(2z+1)^{-4}\left(\sqrt{\frac{-2-\frac{1}{z}}{i}}\right)^8\left(\sqrt{\frac{z}{i}}{i}\right)^8\T\left(z\right)^8\\
    &=(2z+1)^{-4}\left(\sqrt{2z+1}\right)^8\T\left(z\right)^8\\
    &=\T\left(z\right)^8,
\end{align*}
so that the value of $\T(z)$ at the cusp $1/2$ is also $1$.
\end{exer}

\begin{exer}
    Let $\G$ be a congruence subgroup of $\SL{\Z}$ and write
    $$\SL{\Z}=\bigcup_{j=1}^n \a_j \G.$$  Suppose that $f\in M_0(\G)$ and let
    $$g(z)=\prod_{j=1}^n\left(f(\a_j^{-1}z)-f(z_0)\right).$$  We show that $g\in M_0(\SL{\Z})$.
    Observe that for $\gamma\in\SL{\Z}$ we have
    \begin{align*}
        g(z)|_{\gamma}=\prod_{j=1}^n\left(f(\a_j^{-1}\gamma z)-f(z_0)\right)\\
        &=\prod_{i=1}^n\left(f(\a_i^{-1} z)-f(z_0)\right)\\
        &=g(z)
    \end{align*}
    since as $\a_j$ runs over a complete set of representatives of $\G/\SL{\Z}$, so does
    $\gamma^{-1}\a_j$ for any $\gamma\in\SL{\Z}$.  That $g$ is holomorphic at the cusps follows
    directly from the fact that $f\in M_0{\G}$.  We therefore see that $g\in M_0(\SL{\Z})$.
\end{exer}

\begin{exer}
Using the same techniques as in exercise 1, it may be shown that the cusps of $\G_0(2)$ are represented by
$\infty,0$.  Moreover, the cusp $0$ has width 2 while $\infty$ has width 1.
Recall that $\D(z)\in S_{12}(\SL{\Z})$ has no zeroes in $\H$ and a simple $0$ at $\infty$.  Moreover, suppose that
$$\gamma=\begin{pmatrix}a & b\\c & d\end{pmatrix}\in \G_0(2).$$  Then
$$\D(2z)|_{\gamma}=(cz+d)^{-12}\D\left(\frac{a(2z)+2b}{(c/2)(2z)+d}\right)=\D(2z).$$  It follows that
$$\D(z)\D(2z)\in S_{24}(\G_0(2)).$$  Moreover, since
$$\D(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24},$$ we see that $\D(z)\D(2z)=q^3+\cdots$ and therefore has
a zero of order $3$ at infinity.  Let
$$\gamma=\begin{pmatrix}0 & -1\\ 1 & \phantom{-}0\end{pmatrix}\in \SL{\Z}.$$
It is clear that $\gamma$ interchanges $0$ and $\infty$.  Moreover, we have
\begin{align*}
\D(z)\D(2z)|_{\gamma}&=(z)^{-24}\D(-2/z)\D(-1/z)\\&=z^{-24}(z/2)^{12}\D(z/2)z^{12}\D(z)\\&=2^{-12}\D(z/2)\D(z)\\&=2^{-12}q^{3/2}+\cdots,\end{align*}
so that, since $0$ has width 2, $\D(z)\D(2z)$ has a zero of order $3$ at $0$.  Since $\D(z)$ is never $0$ on $\H$,
we can define a function $f(z):=(\Delta(z)\Delta(2z))^{1/3}$ that is holomorphic on $\H$.  We claim that $f\in S_{8}(\G_0(2)).$
Indeed, recall that
$$\D(z)=\eta(z)^{24}$$ and that
\begin{align*}
    \eta(z+1)&=e^{\pi/12}\eta(z)\\
    \eta(-1/z)&=e^{-i\pi/4}\sqrt{z}\eta(z).
\end{align*}
Moreover, the group $\G_0(2)$ is generated by the transformations $z\mapsto z+1$ and
$$z\mapsto \frac{z-1}{2z-1}=\frac{-1}{-2-\frac{1}{z-1}}.$$
We have
\begin{align*}
    f(z+1)&=(\eta(z+1)\eta(2z+2))^8\\
    &=(e^{\pi i/12}\eta(z)e^{2\pi i/12}\eta(2z))^8\\
    &=(\eta(z)\eta(2z))^8\\
    &=f(z),
\end{align*}
and
\begin{align*}
    f\left(\frac{-1}{-2-\frac{1}{z-1}}\right)&=\left(\eta\left(\frac{-1}{-2-\frac{1}{z-1}}\right)\eta\left(\frac{-1}{-1-\frac{1}{2z-2}}\right)\right)^8\\
&=\left(e^{-i\pi/4}\sqrt{-2-\frac{1}{z-1}}\eta\left(-2-\frac{1}{z-1}\right)e^{-i\pi/4}\sqrt{-1-\frac{1}{2z-2}}\eta\left(-1-\frac{1}{2z-2}\right)\right)^8\\
&=\left(e^{-i\pi/4-2\pi i/12}\sqrt{-2-\frac{1}{z-1}}\eta\left(\frac{-1}{z-1}\right)e^{-i\pi/4-i\pi/12}\sqrt{-1-\frac{1}{2z-2}}\eta\left(\frac{-1}{2z-2}\right)\right)^8\\
&=\left(e^{-9i\pi/12}\sqrt{-2-\frac{1}{z-1}}e^{-i\pi/4}\sqrt{z-1}\eta\left(z-1\right)\sqrt{-1-\frac{1}{2z-2}}e^{-i\pi/4}\sqrt{2z-2}\eta\left(2z-2\right)\right)^8\\
&=\left(e^{-15i\pi/12}\sqrt{-2z+1}e^{-i\pi/12}\eta\left(z\right)\sqrt{-2z+1}e^{-2i\pi/12}\eta\left(2z\right)\right)^8\\
&=\left(e^{-18i\pi/12}(-2z+1)\eta\left(z\right)\eta\left(2z\right)\right)^8\\
&=(2z-1)^8 f(z).
\end{align*}
It follows that $f\in M_8(\G_0(2))$.  Since we have determined that $f^3$ has a zero or order 3 at each of the cusps $0,\infty$,
we must have that $f$ has a simple zero at each cusp.  Therefore $f\in S_8(\G_0(2))$.
\end{exer}

\begin{exer}
    In this exercise we determine the fourier expansion for the unrestricted Eisenstein series
    $$G_k(z;u,v;N):=\sum_{\substack{c\equiv u\pod N\\d\equiv v\pod N\\(c,d)\neq (0,0)}}\frac{1}{(cz+d)^k},$$ for $k\geq 3.$
    Recall that we have the formula
    $$\sum_{n\in \Z}\frac{1}{(w+n)^k}=\frac{1}{(k-1)!}(-2\pi i)^k\sum_{n=1}^{\infty} n^{k-1}q^n.$$
    We have
    \begin{align*}
        G_k(z;u,v;N)&=\sum_{\substack{c\equiv u\pod N\\d\equiv v\pod N\\(c,d)\neq (0,0)}}\frac{1}{(cz+d)^k}\\
        &=\sum_{j\in \Z}\sum_{l\in \Z}\frac{1}{((lN+u)z+(jN+v))^k}\\
        &=N^{-k}\sum_{j\in \Z}\sum_{l\in \Z}\frac{1}{(j+((lN+u)z+v)/N)^k}\\
        &=N^{-k}\sum_{j\in\Z}\frac{1}{(j+(uz+v)/N)^k}\\
        &+N^{-k}\sum_{j\in \Z}\sum_{l=1}^{\infty}\left(\frac{1}{(j+((lN+u)z+v)/N)^k}+(-1)^k\frac{1}{(j+((lN-u)z-v)/N)^k}\right)\\
        &=\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{n=1}^{\infty}n^{k-1}e\left(\frac{uz+v}{N}n\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{n=1}^{\infty}\sum_{l=1}^{\infty}n^{k-1}\left(e\left(\frac{(lN+u)z+v}{N}n\right)+(-1)^k e\left(\frac{(lN-u)z-v}{N}n\right)\right)\\
        &=\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}\sum_{n=1}^{\infty}(nN+r)^{k-1}e\left(\frac{uz+v}{N}(nN+r)\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}\sum_{n=1}^{\infty}\sum_{l=1}^{\infty}(nN+r)^{k-1}e\left(\frac{(lN+u)z+v}{N}(nN+r)\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}\sum_{n=1}^{\infty}\sum_{l=1}^{\infty}(nN+r)^{k-1}e\left(\frac{(lN-u)z-v}{N}(nN+r)\right)\\
        &=\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(vr/N)\sum_{n=1}^{\infty}(nN+r)^{k-1}e\left(\frac{uz}{N}(nN+r)\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(vr/N)\sum_{n=1}^{\infty}\sum_{l=1}^{\infty}(nN+r)^{k-1}e\left(\frac{(lN+u)(nN+r)z}{N}\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(-vr/N)\sum_{n=1}^{\infty}\sum_{l=1}^{\infty}(nN+r)^{k-1}e\left(\frac{(lN-u)(nN+r)z}{N}\right)\\
    \end{align*}
    \begin{align*}
        &=\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(vr/N)\sum_{\substack{n\equiv r\pod N\\n>0}}\sum_{\substack{l\equiv u\pod N\\l>0}}^{\infty}n^{k-1}e\left(\frac{lnz}{N}\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(-vr/N)\sum_{\substack{n\equiv r\pod N\\n>0}}\sum_{\substack{l\equiv -u\pod N\\l>0}}^{\infty}n^{k-1}e\left(\frac{lnz}{N}\right)\\
        &=\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(vr/N)\sum_{\substack{n\equiv ru\pod N\\n>0}}\left(\sum_{\substack{d|n\\d\equiv r\pod N}}d^{k-1}\right)e\left(\frac{nz}{N}\right)\\
        &+\frac{(-2\pi i/N)^{k}}{(k-1)!}\sum_{r=0}^{N-1}e(-vr/N)\sum_{\substack{n\equiv -ru\pod N\\n>0}}\left(\sum_{\substack{d|n\\d\equiv r\pod N}}d^{k-1}\right)e\left(\frac{nz}{N}\right)\\
    \end{align*}
It seems that this should have a nicer form, but I can't seem to find it.  

\end{exer}



\end{document}
