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\title{Modular Forms, PS 10}
\author{Bryden Cais}
\begin{document}
\maketitle


\begin{exer}
    Suppose that $f\in S_k(1)$ is a Hecke eigenform for all $T_n$.  We show that if $p|N$ then $f$ considered as an element of
    $S_k(N,\ch_0)$ is {\em not} an eigenform for $T_{p,\ch}$.  Recall from class that we showed that for $p|n$ one has
    $$T_{p,\ch_0}=\frac{1}{p}\sum_{b\mod p} \begin{pmatrix}1 & b \\ & p\end{pmatrix}$$ while
    $$T_p=\frac{1}{p}\sum_{b\mod p} \begin{pmatrix}1 & b \\ & p\end{pmatrix}+p^{k-1}\begin{pmatrix}p & \\ & 1\end{pmatrix}.$$
    Suppose that $f$ is an eigenform for $T_{p\ch_0}$.  Then we have
    $$\frac{1}{p}\sum_{b\mod p} f\left(\frac{z+b}{p}\right)=\mu_p f(z)$$ and
    $$\frac{1}{p}\sum_{b\mod p} f\left(\frac{z+b}{p}\right)+p^{k-1}f(pz)=\lambda_p f(z).$$
    We may assume that $\lambda_p\neq \mu_p$ for otherwise it is clear that $f=0$ identically.
    However, subtracting these two equations yields
    \begin{align*}
        f(z)=\frac{p^{k-1}}{\lambda_p-\mu_p}f(pz).
    \end{align*}
    Applying this formula twice gives
    $$f(z)=\left(\frac{p^{k-1}}{\lambda_p-\mu_p}\right)^2f(p^2z),$$
    from which it follows that if we write
    $$f(z)=\sum_{n=1}^{\infty} a(n)e(nz)$$
    then $a(n)=0$ unless $p^2|n$.  In particular, $a(l)=0$ for {\em all} primes $l$.  Now since $f$ was supposed a normalized eigenform
    for $T_n$ for all $n$, it follows that we have the relation
    $$a(n)a(m)=\sum_{d|(m,n)} d^{k-1}a\left(\frac{mn}{d^2}\right)$$
    and thus that $a(n)=0$ for all $n$.  This gives $f=0$ identically, a contradiction.
\end{exer}

\begin{exer}
    Suppose that $f\in S_k(1)$ is a normalized eigenform for all $T_n$ and let $l|n$.  We show that $f(lz)$ as considered
    as an element of $S_k(N,\ch_0)$ is an eigenform for all $T_{n,\ch_0}$ with $(n,N)=1$ and eigenvalue $a(n)$.
    Recall that
    $$T_{n,\ch_0}(f(z))=\frac{1}{n}\sum_{ad=n}\ch_0(a)a^k\sum_{b\mod d} \begin{pmatrix}a & b\\ & d\end{pmatrix}f(z).$$
    It follows that
    \begin{align*}
        T_{n,\ch_0}(f(lz))&=\frac{1}{n}\sum_{ad=n}\ch_0(a)a^k\sum_{b\mod d} \begin{pmatrix}l & \\ & 1\end{pmatrix}\begin{pmatrix}a & b\\ & d\end{pmatrix}f(z)\\
        &=\frac{1}{n}\sum_{ad=n}\ch_0(a)a^k\sum_{b\mod d} \begin{pmatrix}a & bl\\ & d\end{pmatrix}\begin{pmatrix}l & \\ & 1\end{pmatrix}f(z).
    \end{align*}
    Now since $(n,N)=1$ and $l|N,$ we have $(l,N)=1$.  Thus, as $b$ runs over all classes modulo $d$ so does $lb$ (since $d|n$).  Moreover,
    since $a|n$ and $(n,N)=1$, we see that $\ch_0(a)=1$ for all $a|n$ since $\ch_0$ is the trivial character modulo $N$.  We then have
    \begin{align*}
        T_{n,\ch_0}(f(lz))&=\frac{1}{n}\sum_{ad=n}a^k\sum_{b\mod d} \begin{pmatrix}a & b\\ & d\end{pmatrix}\begin{pmatrix}l & \\ & 1\end{pmatrix}f(z)\\
        &=a(n)f(lz),
    \end{align*}
    since $f$ was assumed to be an eigenform for all $T_n$, normalized to have eigenvalue $a(n)$.
\end{exer}



\end{document}