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\title{Modular Forms, PS 9}
\author{Bryden Cais}
\begin{document}
\maketitle


\begin{exer}
    Suppose that $\ch$ is a primitive Dirichlet character modulo $q$ and that $(n,q)=d>1$.  Then we have
    \begin{align*}
        \sum_{a\mod q}\ch(a)e\left(\frac{an}{q}\right)&=\sum_{a\mod d}\sum_{b\mod q/d}\ch(a(q/d)+b)e\left(\frac{(a(q/d)+b)n}{q}\right)\\
        &=\sum_{a\mod d}\sum_{b\mod q/d}\ch(a(q/d)+b)e\left(\frac{bn}{q}\right)\\
        &=\sum_{b\mod q/d}e\left(\frac{bn}{q}\right)\sum_{\substack{j\mod q\\j\equiv b\mod q/d}}\ch(j).
    \end{align*}
    We claim that there exists some $c\equiv 1\mod q/d$ with $(c,q)=1$ such that $\ch(c)\neq 1$.  For suppose that
    $\ch(c)=1$ for all $(c,q)=1$ that satisfy $c\equiv 1\mod q/d$.  Then for any $b$ we have $\ch(cb)=\ch(c)\ch(b)=\ch(b)$
    for all $c\equiv 1\mod q/d$, or equivalently, $\ch(a)=\ch(b)$ for all $a\equiv b\mod q/d$ so that
    $\ch$ is then induced from a character modulo $q/d$.  Since $d>1$, this is a contradiction to our assumption that
    $\ch$ is primitive.  So let $c\equiv 1\mod q/d$ be such that $(c,q)=1$ and $\ch(c)\neq 1$.
    Then we have
    \begin{align*}
        \ch(c)\sum_{\substack{j\mod q\\j\equiv b\mod q/d}}\ch(j)&=\sum_{\substack{j\mod q\\j\equiv b\mod q/d}}\ch(cj)\\
        &=\sum_{\substack{j\mod q\\cj\equiv b\mod q/d}}\ch(cj)\\
        &=\sum_{\substack{j\mod q\\j\equiv b\mod q/d}}\ch(j)
    \end{align*}
    since $c\equiv 1\mod q/d$.  Since $\ch(c)\neq 1$ it follows that
    $$\sum_{\substack{j\mod q\\j\equiv b\mod q/d}}\ch(j)=0$$ for any $b$ and hence that
    $$\sum_{a\mod q}\ch(a)e\left(\frac{an}{q}\right)=0$$ whenever $(n,q)>1$.  This in conjunction with the result proved in lecture shows that
    $$\overline{\ch}(n)=\frac{1}{G(\ch)}\sum_{a\mod q}\ch(a)e\left(\frac{an}{q}\right)$$
    for all $n$.
\end{exer}

\begin{exer}
    Let $f\in S_k(1)$ be a Hecke eigenform, and let $\ch\mod q$ be a primitive Dirichlet character.
    Recall from lecture that if
    $$f=\sum_{n=1}^\infty a(n)e(nz)$$ then
    $$f_{\ch}=\sum_{n=1}^\infty a(n)\ch(n)e(nz)\in S_k(q^2,\ch^2).$$
    Moreover, we showed that
    $$T_{n,\ch^2}(f_{\ch})=\sum_{m=1}^{\infty}a_{n,\ch}(m)e(mz)$$
    where
    \begin{align*}
    a_{n,\ch}(m)&=\sum_{d|(m,n)}\ch^2(d)d^{k-1}\ch(mn/d^2)a(mn/d^2)\\
    &=\ch(mn)\sum_{d|(m,n)}d^{k-1}a(mn/d^2)\\
    &=\ch(mn)a(m)a(n)
    \end{align*}
    since $f$ is a Hecke eigenform for $T_n$ (normalized so as to have eigenvalue $a(n)$).
    But then we have
    $$T_{n,\ch^2}(f_{\ch})=a(n)\ch(n) f_{\ch}$$
    so that $f_{\ch}$ is a normalized eigenform for the operators $T_{n,\ch^2}$ (since the $n^{\text{th}}$ coefficient
    in the fourier expansion of $f_{\ch}$ is $a(n)\ch(n)$).  Recall that
    $$E_k(z)=1+\gamma_k\sum_{n=1}^{\infty}\sigma_{k-1}(n)e(nz)$$ (for some constant $\gamma_k$) is an eigenform for the $T_n$.
    We claim that
    $$E_{k,\ch}(z)=\ch(0)+\gamma_k\sum_{n=1}^{\infty}\ch(n)\sigma_{k-1}(n)e(nz)$$ is a Hecke eigenform for the
    $T_{n,\ch^2}$.  The proof is precisely the same as above, since we have not used that $f$ is a cusp form.
    Let us show that in general, $E_{k,\ch}$ is not a cusp form.  Suppose that $q$ is prime and that $\ch$ is primitive.  By exercise 1, we can write
    \begin{align*}
    E_{k,\ch}(z)&=C\sum_{a\mod q}\overline{\ch}(a)\sum_{n=1}^{\infty}\sigma_{k-1}(n)e\left(nz+n\frac{a}{q}\right)\\
    &=C^{\prime}\sum_{a\mod q}\overline{\ch}(a)(E_k(\gamma_a z)-1)\\
    &=C^{\prime}\sum_{a\mod q}\overline{\ch}(a)E_k(\gamma_a z)
    \end{align*}
    where $C,C^{\prime}$ are nonzero constants and
    $$\gamma_a=\begin{pmatrix}1 & a/q \\ 0 & 1\end{pmatrix}.$$
    Let
    $$\gamma=\begin{pmatrix}1 & x \\ q & y\end{pmatrix}$$ where $x,y$ are integers chosen so that $\gamma\in\SL{\Z}$.
    We have
    \begin{align*}
        E_{k,\ch}(z)|_{\gamma}&=C^{\prime}\sum_{a\mod q}\overline{\ch}(a)(qz+y)^{-k}E_k(\gamma_a\gamma z)
    \end{align*}
    Now observe that if $a\not\equiv -1\mod q$ then there exists $v_a$ such that $(a+1)v_a\equiv ay\mod q$ (since $q$ is prime).
    It follows that in these cases, we have a $v_a$ such that
    \begin{align*}
        \begin{pmatrix}1 & a/q \\0 & 1\end{pmatrix}\begin{pmatrix}1 & x \\q & y\end{pmatrix}&=
        \begin{pmatrix}a+1 & (ay-(a+1)v_a)/q+x \\q & y-v_a\end{pmatrix}\begin{pmatrix}1 & v_a/q \\0 & 1\end{pmatrix}
    \end{align*}
    and the matrix
    $$\begin{pmatrix}a+1 & (ay-(a+1)v_a)/q+x \\q & y-v_a\end{pmatrix}\in\SL{\Z}.$$
    We can therefore write
    \begin{align*}
        E_{k,\ch}(z)|_{\gamma}&=C^{\prime}\sum_{\substack{a\mod q\\a\not\equiv -1\mod q}}\overline{\ch}(a)\left(\frac{qz+y-v_a}{qz+y}\right)^{k}
        E_k(z+v_a/q)\\
        &+\overline{\ch}(-1)(qz+y)^{-k}E_k\left(\frac{qz+x+(q-1)y/q}{qz+y}\right).
    \end{align*}
    Since the value of $E_{k,\ch}$ at the cusp $1/q$ is $\lim_{z\rightarrow\infty}E_{k,\ch}(z)|_{\gamma},$ we see that this value is
    \begin{align*}
        &\lim_{z\rightarrow\infty}C^{\prime}\sum_{\substack{a\mod q\\a\not\equiv -1\mod q}}\overline{\ch}(a)\left(\frac{qz+y-v_a}{qz+y}\right)^{k}
        E_k(z+v_a/q)\\&+\lim_{z\rightarrow\infty}\overline{\ch}(-1)(qz+y)^{-k}E_k\left(\frac{qz+x+(q-1)y/q}{qz+y}\right)\\
        =&\sum_{\substack{a\mod q\\a\not\equiv -1\mod q}}\overline{\ch}(a)+\lim_{z\rightarrow\infty}\overline{\ch}(-1)(qz+y)^{-k}E_k\left(\frac{qz+x+(q-1)y/q}{qz+y}\right)
    \end{align*}
    since the value of $E_k(z)$ at $\infty$ is $1$.
    Recalling that
    $$E_k(z)=\sum_{\substack{(c,d)\in\Z^2\setminus\{0\}\\(c,d)=1}}(cz+d)^{-k}$$
    we see that

    \begin{align*}
    &\lim_{z\rightarrow\infty}\overline{\ch}(-1)(qz+y)^{-k}E_k\left(\frac{qz+x+(q-1)y/q}{qz+y}\right)=\\
    &\lim_{z\rightarrow\infty}\overline{\ch}(-1)\sum_{\substack{(c,d)\in\Z^2\setminus\{0\}\\(c,d)=1}}\left(c(qz+x+(q-1)y/q)+d(qz+y)\right)^{-k}=\\
    &\lim_{z\rightarrow\infty}2\overline{\ch}(-1)\left((qz+x+(q-1)y/q)-(qz+y)\right)^{-k}=\\
    &2\overline{\ch}(-1)\left(x-y/q)\right)^{-k}=\\
    &2\overline{\ch}(-1)\frac{-1}{q},
    \end{align*}
    since every term in the sum vanishes in the limit, except for the two terms corresponding to $(c,d)=(1,-1),\ (c,d)=(-1,1)$,
    and since $y-xq=1$.
    At long last, it follows that the value of $E_{k,\ch}(z)$ at the cusp $1/q$ is
    $$\sum_{\substack{a\mod q\\a\not\equiv -1\mod q}}\overline{\ch}(a)-\frac{2\ch(-1)}{q}=-\ch(-1)\left(1+\frac{2}{q}\right)\neq 0,$$
    and hence that $E_{k,\ch}(z)$ is not a cusp form.   
\end{exer}



\end{document}