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\title{Modular Forms, PS 11}
\author{Bryden Cais}
\begin{document}
\maketitle


\begin{exer}
    Let $\{f_j\}$ be a basis of normalized Hecke eigenforms for the space $\Sn$ and write
    $$f_j(z)=\sum_{n=1}^{\infty} \lambda_j(n)e(nz)$$ so that
    $$T_{n,\ch}(f_j)=\lambda_j(n) f_j$$ for {\em all} $n$.  Recall that we have the formula
    $$\lambda_j(m)\lambda_j(n)=\sum_{d|(m,n)} \chi(d)d^{k-1}\lambda_j\left(\frac{mn}{d^2}\right),$$
    a special case of which is
    \begin{align}
    \lambda_j(p^{r-1})\lambda_j(p)=\begin{cases}\lambda_j(p^r)+\chi(p)p^{k-1}\lambda_j(p^{r-2}) & \text{if } (p,N)=1\\
    \lambda_j(p^r) & \text{if } p|N\end{cases}\label{mul}
    \end{align}
    since $\chi$ is a Dirichlet character modulo $N$.
    Write
    $$\lambda_j(p)=\a_j(p)+\b_j(p)$$ with
    $$\a_j(p)\b_j(p)=\chi(p)p^{k-1}.$$
    We now show by induction that for $(p,N)=1$ we have
    \begin{align}
        \lambda_j(p^n)=\sum_{i=0}^n \a_j(p)^{n-i}\b_j(p)^i. \label{formula1}
    \end{align}
    The cases $n=1,0$ follow by definition.  Suppose that (\ref{formula1}) holds for $n<r$.  Applying formula (\ref{mul})
    yields
    \begin{align*}
        \lambda_j(p^{r})&=\lambda_j(p^{r-1})\lambda_j(p)-\chi(p)p^{k-1}\lambda_j(p^{r-2})\\
        &=\lambda_j(p)\sum_{i=0}^{r-1} \a_j(p)^{r-1-i}\b_j(p)^i-\chi(p)p^{k-1}\sum_{i=0}^{r-2} \a_j(p)^{r-2-i}\b_j(p)^i\\
        &=\sum_{i=0}^{r-1} \left(\a_j(p)^{r-i}\b_j(p)^i+\a_j(p)^{r-1-i}\b_j(p)^{i+1}\right)-\sum_{i=0}^{r-2} \a_j(p)^{r-1-i}\b_j(p)^{i+1}\\
        &=\left(\a_j(p)^{r}+\a_j(p)^{r-1}\b_j(p)\right)+\sum_{i=1}^{r-1}\a_j(p)^{r-1-i}\b_j(p)^{i+1}\\
        &=\left(\a_j(p)^{r}+\a_j(p)^{r-1}\b_j(p)\right)+\sum_{i=2}^{r}\a_j(p)^{r-i}\b_j(p)^{i}\\
        &=\sum_{i=0}^{r}\a_j(p)^{r-i}\b_j(p)^{i}
    \end{align*}
    by our induction hypothesis.
    Therefore, if $(p,N)=1$ we have
    \begin{align*}
        \left(1-\frac{\lambda_j(p)}{p^s}+\frac{\chi(p)p^{k-1}}{p^{2s}}\right)^{-1}&=
        \left(1-\frac{\a_j(p)+\b_j(p)}{p^s}+\frac{\a_j(p)\b_j(p)}{p^{2s}}\right)^{-1}\\
        &=\left(1-\frac{\a_j(p)}{p^s}\right)^{-1}\left(1-\frac{\b_j(p)}{p^s}\right)^{-1}\\
        &=\left(1+\frac{\a_j(p)}{p^s}+\frac{\a_j(p)^2}{p^{2s}}+\cdots\right)\left(1+\frac{\b_j(p)}{p^s}+\frac{\b_j(p)^2}{p^{2s}}+\cdots\right)\\
        &=\sum_{n=0}^{\infty} \left(\sum_{u+v=n} \a_j(p)^u\b_j(p)^v\right)p^{-ns}\\
        &=\sum_{n=0}^{\infty} \frac{\lambda_j(p^n)}{p^{ns}},
    \end{align*}
    while if $p|N$ then (\ref{mul}) tells us that
    $$\lambda_j(p^n)=\lambda_j(p)^n$$ so that in these cases we have
    \begin{align*}
        \sum_{n=0}^{\infty} \frac{\lambda_j(p^n)}{p^{ns}}&=\sum_{n=0}^{\infty} \left(\frac{\lambda_j(p)}{p^{s}}\right)^n\\
        &=\left(1-\frac{\lambda_j(p)}{p^s}\right)^{-1}.
    \end{align*}
\end{exer}

\begin{exer}
    Suppose that $f\in S_k(N,\chi)$ and let $K$ be defined by
    $$K(f(z))=\overline{f(-\bar{z})}.$$  Then we claim that
    if $g=Kf$ then $g\in S_k(N,\bar{\chi})$.  For let
    $$\gamma=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \G_0(N).$$
    Then we have
    \begin{align*}
        g|_{\gamma}&=(cz+d)^{-k}g\left(\frac{az+b}{cz+d}\right)\\
        &=(cz+d)^{-k}\overline{f\left(-\frac{a\bar{z}+b}{c\bar{z}+d}\right)}\\
        &=(cz+d)^{-k}\overline{f\left(\frac{a(-\bar{z})-b}{-c(-\bar{z})+d}\right)}\\
        &=(cz+d)^{-k}\overline{\chi(a)(-c(-\bar{z})+d)^{k}f\left(-\bar{z}\right)}\\
        &=(cz+d)^{-k}(-c(-z)+d)^{k}\overline{\chi(a)}\overline{f\left(-\bar{z}\right)}\\
        &=\overline{\chi(a)}g(z),\\
    \end{align*}
    where we have used the fact that $\bigl(\begin{smallmatrix}\phantom{-}a & -b\\ -c & \phantom{-}d\end{smallmatrix}\bigr)\in \G_0(N)$ (obviously).
    Clearly, then, $g\in S_k(N,\bar{\chi})$, as claimed.
\end{exer}



\end{document}
