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\title {Constructing Simultaneous Hecke Eigenforms} 

\author {Thomas R. Shemanske, Stephanie A. Treneer, Lynne H. Walling}

\date{\today}
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\evenhead{Thomas R. Shemanske, Stephanie A. Treneer, Lynne H. Walling}     %Running Heads on Even Pages
\oddhead{Constructing simultaneous Hecke Eigenforms}       %Running Heads on Odd Pages

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\subjclass {Primary 11F25; Secondary 11F11}

\keywords{Diagonalizing Hecke Operators, simultaneous eigenfunctions}

\begin{abstract}
  It is well known that newforms of integral weight are simultaneous
  eigenforms for all the Hecke operators, and that the converse is not
  true.  In this paper, we give a characterization of all simultaneous
  Hecke eigenforms associated to a given newform, and provide several
  applications.  These include determining the number of linearly
  independent simultaneous eigenforms in a fixed space which
  correspond to a given newform, and characterizing several situations
  in which the full space of cusp forms is spanned by a basis
  consisting of such eigenforms.  Part of our results can be seen as a
  generalization of results of Choie-Kohnen \cite{Choie-Kohnen} who
  considered diagonalization of ``bad'' Hecke operators on spaces with
  square free level and trivial character.  Of independent interest,
  but used herein, is a lower bound for the dimension of the space of
  newforms with arbitrary character.

\end{abstract}


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\section{Introduction}\label{sec:introduction}
For $N$ a positive integer, $\psi$ a Dirichlet character defined
modulo $N$, and $k \ge 2$ an integer, we let $S_k(N,\psi)$ denote the
space of cusp forms of weight $k$ for $\Gamma_0(N)$ with character
$\psi$, and $S_k^+(N,\psi)$ the subspace generated by the newforms.
For a prime $p$, we let $T_p$ (or $T_p^N$) denote the $p$th Hecke
operator for forms on $S_k(N,\psi)$. We use this notation for the
primes which divide the level as well, so for example if $q\slash N$,
our Hecke operator $T_q^N$ is the same as the operator $U_q$ in the
notation of \cite{Li-Newforms}.

It is well-known that $S_k(N,\psi)$ has a basis consisting of
simultaneous eigenforms for the algebra of Hecke operators generated
by $\{T_p^N \mid (p,N) = 1\}$, and via multiplicity-one that
$S_k^+(N,\psi)$ has a basis of simultaneous eigenforms for all the
Hecke operators.  Since $S_k^+(N,\psi)$ is generally a proper subspace
of $S_k(N,\psi)$, it is a natural question to consider the extent to
which the full space of cusp forms has a basis of simultaneous
eigenforms for all the Hecke operators.  Choie and Kohnen
\cite{Choie-Kohnen} considered the question of diagonalizing ``bad''
Hecke operators (that is, $T_q^N$ where $q \mid N$), and gave an upper
bound for the number of primes $q$ for which $T_q^N$ could not be
diagonalized on $S_k(N,\psi)$ where $N$ is square-free, $q \mid N$ and
$\psi$ is trivial.  An alternate perspective on that question is to
determine conditions under which simultaneous Hecke eigenforms are
newforms.  One result along these lines is Li's \cite{Li-Newforms}
Theorem 9: if $f\in S_k(N,\psi)$ is a simultaneous eigenform for all
Hecke operators $T_p^N$, and $f$ is also an eigenform for the operator
$KW_N$ (where $K$ is the conjugation operator and $W_N$ is the Fricke
involution), then $f$ is a newform.

In this paper we address the question broadly, in particular giving a
characterization of all simultaneous Hecke eigenforms associated to a
given newform for arbitrary level and character.  For a given newform
$h \in S_k(N_0,\psi)$, we first determine
(Theorem~\ref{thm:classify-for-T_q} and
Proposition~\ref{prop:sufficient-conditions}) the exact structure and
the eigenvalues of each form $f \in S_k(N,\psi)$ which is
Hecke-equivalent to $h$ and also an eigenfunction for $T_q^N$.  In
section 3, we address the diagonalizability of $T_q^N$ on a given
space of cusp forms, characterizing several situations in which the
full space of cusp forms is spanned by a basis consisting of such
eigenforms, as well as those situations when it is not
(Theorems~\ref{thm:Choie-Kohnen-recast} and
\ref{thm:cant-diagonalize}) .  To establish the later result we derive
a lower bound (Theorem~\ref{thm:newform-dimension}) for the dimension
of the space of newforms, $S_k^+(N,\psi)$; dimension formulas for the
space of newforms with trivial character are given by Martin
\cite{Martin}.  In Theorem~\ref{thm:Choie-Kohnen-generalized}, we
generalize the results of Choie-Kohnen producing an upper bound for
the number of primes $q$ for which $T_q^N$ fails to diagonalize.
Section 4 considers simultaneous Hecke eigenforms, and section 5 has
several examples delineating cases in which bases of simultaneous
eigenforms do or do not exist.





\section{Characterizing Hecke Eigenforms at primes dividing the level}


Throughout, we make the convention that all Dirichlet characters will
be considered as defined modulo their conductor, so that when
considering a modular form in $S_k(N,\psi)$, $\psi(d) \ne 0$ iff $d$
is relatively prime to the conductor.  In particular, there may well
be primes $q \mid N$ for which $\psi(q) \ne 0$.  Of course for any
prime $q \nmid N$, $\psi(q) \ne 0$.  The convention is necessary to
allow a uniform handling of all subspaces $S_k(N_0,\psi)$ where
$\cond(\psi) \mid N_0 \mid N$.


Let $h \in S_k(N_0,\psi)$ be a newform (always assumed nonzero), $N$
an integer divisible by $N_0$, and $f\in S_k(N,\psi)$ a nonzero
simultaneous Hecke eigenform having the same eigenvalues as $h$ for
all Hecke operators $T_\ell^N$, $\ell$ a prime with $(\ell,N) = 1$.
The eigenvalues of $h$ are given by $h \slash T_\ell^{N_0} =
\lambda_\ell h$ for all primes $\ell$, and we note that $T_\ell^N =
T_\ell^{N_0}$ when $(\ell, N) = 1$.  Moreover suppose that $f$ is also
a nonzero eigenform for $T_q^N$ where $q$ is a fixed prime dividing
$N$, and put $f\slash T_q^N = \kappa_q f$.  It is well-known
\cite{Atkin-Lehner, Li-Newforms}, that $f$ has the form
\[f=\sum_{d \mid \frac{N}{N_0}} \alpha_d\, h \slash B_d,\] where $B_d$
is the shift operator of \cite{Li-Newforms}, and the $\alpha_d$ are
complex scalars.


\begin{theorem}\label{thm:classify-for-T_q}
  Let the notation be as above.  Then assuming $q \mid N$ and $d \mid
  N/N_0$, we have:

  \begin{compactenum}
  \item If $q \mid N_0$ then $\alpha_d = 0$ if $q^2 \mid d$.
    \begin{compactenum}
    \item If $q \nmid N/N_0$, then $\kappa_q = \lambda_q$, and
      (vacuously) $\alpha_d = 0$ for $q\mid d$.

    \item If $q\mid N/N_0$, then $\alpha_{d} = (\kappa_q -
      \lambda_q)\alpha_{d/q}$ if $q \parallel d$. If $\kappa_q \ne 0$,
      then $\kappa_q = \lambda_q$, and $\lambda_q = 0$ implies
      $\kappa_q = 0$.
    \end{compactenum}

  \item If $q \nmid N_0$, then $\alpha_d = 0$ if $q^3 \mid d$.
    \begin{compactenum}
    \item If $\kappa_q \ne 0$, then $\alpha_d = 0$ if $q^2 \mid d$,
      $\alpha_{d} = (\kappa_q - \lambda_q)\alpha_{d/q}$ if
      $q \parallel d$, and $ \kappa_q = \frac12\left(\lambda_q \pm
        \sqrt{\lambda_q^2 - 4 \psi(q) q^{k-1}}\right) \ne \lambda_q$.
    \item If $\kappa_q = 0$, then $q^2 \mid N/N_0$, $\alpha_{d} =
      \psi(q)q^{k-1}\alpha_{d/q^2}$ if $q^2 \parallel d$, and
      $\alpha_{d} = -\lambda_q\alpha_{d/q}$ if $q \parallel d$.
    \end{compactenum}

  \end{compactenum}

 
 
\end{theorem}


\begin{proof}
  As above, we assume that $f=\sum_{d \mid \frac{N}{N_0}} \alpha_d\, h
  \slash B_d.$ We separate the argument by cases.
  \begin{itemize}
  \item \textbf{Case:} $q\mid N_0$, $q\nmid N/N_0$.

    Since $q \mid N_0$, $T_q^N = T_q^{dN_0}$ for any $d$.  Also note
    that since $q\nmid N/N_0$, any divisor $d \mid N/N_0$ satisfies
    $(d,q) = 1$, so that the shift and Hecke operators commute: $B_d
    \slash T_q^{dN_0} = T_q^{N_0}~\slash~B_d$.  Thus
    \[\kappa_q f = f \slash T_q^N = \sum_{d \mid \frac{N}{N_0}}
    \alpha_d\, h \slash B_d \slash T_q^{dN_0} = \sum_{d \mid
      \frac{N}{N_0}}\alpha_d\, h\slash T_q^{N_0} \slash B_d =
    \lambda_q\sum_{d \mid \frac{N}{N_0}}\alpha_d\, h\slash B_d
    =\lambda_q f,
    \]
    and so we have $\kappa_q = \lambda_q$.  Since $q \nmid N/N_0$, it
    is vacuously true that $\alpha_d = 0$ for $q\mid d$.

  \item \textbf{Case:} $q\mid N_0$, $q\mid N/N_0$.

    As in the previous case, we note that $q \mid N_0$ implies $T_q^N
    = T_q^{dN_0}$ for any $d$, and for any divisor $d \mid N/N_0$
    satisfying $(d,q) = 1$, the shift and Hecke operators commute:
    $B_d \slash T_q^{dN_0} = T_q^{N_0}~\slash~B_d$.  Finally we note
    that $B_qT_q^N=1$. With these observations we have
    \begin{align*}
      \kappa_q f = f \slash T_q^N &=\sum_{d\mid N/N_0} \alpha_d\, h\slash B_d\slash T_q^{dN_0}\\
      &=\sum_{d,\ q\nmid d} \alpha_d\, h\slash T_q^{N_0}\slash B_d
      +\sum_{d,\ q\mid d} \alpha_d\, h\slash B_{d/q}\\
      &=\sum_{d,\ q\nmid d} (\lambda_q\alpha_d+\alpha_{dq})h\slash B_d
      +\sum_{d,\ q^2\mid d} \alpha_d\, h\slash B_{d/q}.
    \end{align*}
    We now show the second summand does not appear.

    \begin{lemma*}
      $\alpha_d = 0$ if $q^2 \mid d$.
    \end{lemma*}

    \begin{proof}
      If $\kappa_q = 0$, then the linear independence of $\{h\slash
      B_d\}$ yields the result.  If $\kappa_q \ne 0$, let $M =\ds
      \max_{d\mid N/N_0}\{\ord_q(d) \mid \alpha_d \ne 0\}$.  Then we
      have
      \begin{align*}
        \kappa_q f &= \sum_{d\mid N/N_0}\kappa_q \alpha_d\, h\slash
        B_d =\sum_{i=0}^M \sum_{d\mid N/N_0,\ q^i \parallel d}
        \kappa_q \alpha_d h\slash B_d\\
        &=\sum_{d,\ q\nmid d} (\lambda_q\alpha_d+\alpha_{dq})h\slash
        B_d +\sum_{d,\ q^2\mid d} \alpha_d\, h\slash B_{d/q}.
      \end{align*}
      There is no issue if $M < 2$, so we assume $M \ge 2$.  In that
      case for a divisor $d$ with $\ord_q(d) = M$, we see that a term
      with $h\slash B_d$ occurs as a summand in $\kappa_q f$, but not
      in $\sum_{d,\ q^2\mid d} \alpha_d\, h\slash B_{d/q}$, so
      $\alpha_d = 0$, a contradiction.
    \end{proof}




    Applying this observation, the equation above becomes:

\begin{equation}\label{eqn:case2}
  \sum_{d\mid N/N_0,\ q^2 \nmid d} \kappa_q \alpha_d\,
  h\slash B_d = \kappa_q f = f \slash T_q^N =  \sum_{d\mid N/N_0,\ q\nmid d} (\lambda_q\alpha_d + \alpha_{dq}) h\slash
  B_d. 
\end{equation}

By the linear independence of the set $\{ h\slash B_d\}$, we deduce
from Equation~\ref{eqn:case2} that $\kappa_q \alpha_d = 0$ when
$q\parallel d$.  If $\kappa_q = 0$, then Equation~\ref{eqn:case2} is
zero, hence the coefficients of $h\slash B_d$ are all zero and we
conclude
\[\alpha_{d} = -\lambda_q \alpha_{d/q} = (\kappa_q - \lambda_q)
\alpha_{d/q} \qquad \textrm{if }q \parallel d.\] On the other hand, if
$\kappa_q \ne 0$, then $\alpha_d = 0$ when $q \parallel d$, so
Equation~\ref{eqn:case2} becomes

\[
\kappa_q f = \sum_{d\mid N/N_0,\ q \nmid d} \kappa_q \alpha_d\,
h\slash B_d = \sum_{d\mid N/N_0,\ q\nmid d} \lambda_q\alpha_d\,
h\slash B_d = \lambda_q f,
\]
and hence $\kappa_q = \lambda_q$.  It follows that $\lambda_q = 0$
implies $\kappa_q = 0$, and $ 0 =\alpha_{d} = (\kappa_q -
\lambda_q)\alpha_{d/q}$ if $q \parallel d$.


\item \textbf{Case:} $q\nmid N_0$, $q\mid N/N_0$.
 
  In this case $T_q^N = T_q^{N_0}-\psi(q)q^{k-1}B_q$, and we have
  \begin{align}\label{eqn:q-nmid-N_0}
    \kappa_q f&= \sum_{d\mid N/N_0}\kappa_q \alpha_d\, h\slash B_d =\sum_{d\mid N/N_0} \alpha_d\, h\slash B_d\slash T_q^N\notag\\
    &=\sum_{d,\ q\nmid d} \alpha_d\, h\slash T_q^N\slash B_d
    +\sum_{d,\ q\mid d} \alpha_d\, h\slash B_{d/q}\notag\\
    &=\sum_{d,\ q\nmid d}\alpha_d h\slash
    (T_q^{N_0}-\psi(q)q^{k-1}B_q)\slash B_d
    +\sum_{d,\ q\mid d} \alpha_d\, h\slash B_{d/q}\notag\\
    &=\sum_{d,\ q\nmid d} (\lambda_q\alpha_d+\alpha_{dq}) h\slash B_d
    +\sum_{d,\ q\parallel d}(\alpha_{dq}-\psi(q) q^{k-1}\alpha_{d/q})
    h\slash B_d +\sum_{d,\ q^3\mid d}\alpha_d\, h\slash B_{d/q}.
  \end{align}
  To simplify this expression, we show
  \begin{lemma*}
    $\alpha_d = 0$ if $q^3 \mid d$.
  \end{lemma*}


  \begin{proof}
    This is completely analogous to the previous lemma.  Let $M =\ds
    \max_{d\mid N/N_0}\{\ord_q(d) \mid \alpha_d \ne 0\}$. There is no
    issue if $M < 3$, so assume $M \ge 3$.  In that case for a divisor
    $d$ with $\ord_q(d) = M$, we see that a term with $h\slash B_d$
    occurs in $\kappa_q f$, but not $\sum_{d,\ q^3\mid d}\alpha_d\,
    h\slash B_{d/q}$, so $\alpha_d = 0$, a contradiction.
  \end{proof}

  To go further, we first suppose that $\kappa_q \ne 0$.  If
  $q \parallel N/N_0$, we have $\alpha_d = 0$ for $q^2 \mid d$ by
  convention.  Otherwise, let $d \mid N/N_0$ with $q^2 \mid d$.  The
  coefficient of $h\mid B_d$ in $\kappa_q f$ is $\kappa_q \alpha_d$
  while it is $\alpha_{dq}$ in $\sum_{d,\ q^3\mid d} \alpha_d\,
  h\slash B_{d/q}$.  By the lemma, $\alpha_{qd} = 0$, so we infer
  $\alpha_d =0 $.



  Applying these observations to the above expression for $\kappa_q f$
  yields
  \begin{equation}\label{eqn:case3}
    \kappa_q f = \sum_{d\mid N/N_0} \kappa_q \alpha_d h\slash B_d = \sum_{d\mid N/N_0,\ q\nmid d}(\lambda_q \alpha_d +
    \alpha_{dq}) h\slash B_d +  \sum_{d\mid N/N_0,\ q\nmid d}
    -\psi(q)q^{k-1}\alpha_d\, h\slash B_{dq}.
  \end{equation}

  Comparing coefficients of $h\slash B_d$ and $h\slash B_{dq}$ we
  obtain for $q \parallel d$:
  \begin{align*}
    \alpha_{d} &= (\kappa_q - \lambda_q)\alpha_{d/q}, \textrm{ and}\\
    \kappa_q\alpha_{d} &= -\psi(q)q^{k-1}\alpha_{d/q}.
  \end{align*}


  Substituting the expression for $\alpha_{d}$ from the first equation
  into the second yields the quadratic $(\kappa_q^2 -
  \lambda_q\kappa_q + \psi(q)q^{k-1})\alpha_{d/q} = 0$.  Note that
  $\alpha_{d/q} = 0$ for all $d$ with $q \parallel d$ would imply
  $\alpha_d = 0$ for all $d | N/N_0$, hence $f = 0$. Thus

  \[ \kappa_q = \frac12\left(\lambda_q \pm \sqrt{\lambda_q^2 - 4
      \psi(q) q^{k-1}}\right),
  \]
  and we note that $\kappa_q \ne \lambda_q$ since $\psi(q) \ne 0$.

  Finally, we assume $\kappa_q = 0$. Then all the coefficients of the
  $h\slash B_d$ in equation~\ref{eqn:q-nmid-N_0} are zero, yielding
  $\alpha_d\lambda_q + \alpha_{dq} = 0$ for $q\nmid d$, and
  $\alpha_{dq} = \psi(q)q^{k-1}\alpha_{d/q}$ for $q \parallel d$.
  Note that if $q \parallel N/N_0$, by convention we would have
  $\alpha_{dq} = 0$ in the last equation, leading to $\alpha_{d/q} =
  0$ and hence $\alpha_d = 0$ implying $f = 0$.  Thus $\kappa_q = 0$
  forces $q^2 \mid N/N_0$, which completes the proof.
\end{itemize}
\end{proof}


As above, let $h \in S_k(N_0,\psi)$ be a newform, and $N$ an integer
divisible by $N_0$.  Denote the \textit{class} of $h$ by
\[ [h] = \{f\in S_k(N,\psi) : f\slash T_p^N = h\slash T_p^N\textrm{
  for all primes } p\nmid N\}.
\]
By the theory of newforms, we know
\[ [h] = \bigoplus_{d\mid N/N_0} \la h \slash B_d\ra,
\]
that is, $f \in [h]$ if and only if $f = \sum_{d\mid N/N_0}\alpha_d\,
h\slash B_d$ for scalars $\alpha_d$. It is clear from the general
theory of newforms that any such $f$ is a simultaneous eigenform for
all Hecke operators $T_p^N$ for primes $p \nmid N$. In
Theorem~\ref{thm:classify-for-T_q}, we have given necessary conditions
on the coefficients $\alpha_d$ for $f$ to be an eigenform for $T_q^N$
for a prime $q\mid N$ and eigenvalue $\kappa_q$.  However, the
necessary conditions are also sufficient.


\begin{prop}\label{prop:sufficient-conditions}
  Let $h \in S_k(N_0,\psi)$ be a newform, $N$ an integer divisible by
  $N_0$, and $q$ a prime dividing $N$.  Set $h \slash T_q^{N_0} =
  \lambda_q h$, and fix $\kappa_q$ and constants $\alpha_d$ for $d\mid
  N/N_0$ according to the following scheme (any unconstrained
  constants are arbitrary):
  \begin{compactitem}
  \item $q \mid N_0$ and $q \nmid N/N_0$: Let $\kappa_q = \lambda_q$.
  \item $q \mid N_0$ and $q \mid N/N_0$: Let $\kappa_q = \lambda_q$ or
    0, and put $\alpha_d = 0$ if $q^2 \mid d$, and $\alpha_{d} =
    (\kappa_q - \lambda_q)\alpha_{d/q}$ if $q \parallel d$.
  \item $q \nmid N_0$: Set $\alpha_d = 0$ if $q^3 \mid d$.
    \begin{compactitem}
    \item Let $\kappa_q = \frac12\left(\lambda_q \pm \sqrt{\lambda_q^2
          - 4 \psi(q) q^{k-1}}\right)$, and note $\kappa_q \ne
      0,\lambda_q$. For $q^2\mid d$ put $\alpha_d = 0$; for
      $q \parallel d$, put $\alpha_d = (\kappa_q - \lambda_q)
      \alpha_{d/q}$.
    \item Moreover, if $q^2 \mid N$, we can also let $\kappa_q = 0$,
      and for $q^2 \parallel d$, put $\alpha_{d} =
      \psi(q)q^{k-1}\alpha_{d/q^2}$ and for $q \parallel d$, put
      $\alpha_{d} = -\lambda_q \alpha_{d/q} = (\kappa_q - \lambda_q)
      \alpha_{d/q}$.

    \end{compactitem}
  \end{compactitem}

  \smallskip

  Then $f = \sum_{d\mid N/N_0}\alpha_d\, h\slash B_d$ is an eigenform
  for $T_q^N$ with eigenvalue $\kappa_q$.
\end{prop}

\begin{proof}
  The proposition follows immediately from the computations already
  present in Theorem~\ref{thm:classify-for-T_q}.
\end{proof}
\section{Comparison to Choie-Kohnen}

As in the previous section, let $h \in S_k(N_0,\psi)$ be a newform,
and $N$ an integer divisible by $N_0$, and denote the class of $h$ by
$[h]$.  From \cite{Atkin-Lehner, Li-Newforms} we know that if
$S_k^+(N_0,\psi)$ denotes the space generated by newforms of level
$N_0$,
\begin{align*}
  S_k(N,\psi) &= \bigoplus_{\cond(\psi) \mid N_0 \mid N}\
  \bigoplus_{d\mid
    N/N_0} \left(S_k^+(N_0,\psi) \slash B_d\right)\\
  &=\bigoplus_{\cond(\psi) \mid N_0 \mid N}\ \bigoplus_{h}\ [h],
\end{align*}
where the last sum is over normalized newforms $h \in
S_k^+(N_0,\psi)$.


\begin{lemma}\label{lemma:SE-on-classes}
  $T_q^N$ is diagonalizable on $S_k(N,\psi)$ if and only if there is a
  basis of $S_k(N,\psi)$ consisting of simultaneous eigenforms for
  $T_q^N$ as well as for all $T_p^N$, $p$ a prime with $p \nmid N$.
  Moreover, for each $N_0 \mid N$ and each normalized newform $h\in
  S_k(N_0,\psi)$, $T_q^N$ is diagonalizable on $S_k(N, \psi)$ if and
  only if it is diagonalizable on each class $[h]$.
\end{lemma}

\begin{proof}
  For both statements, only the forward direction requires proof.  If
  $T_q^N$ is diagonalizable on $S_k(N,\psi)$, then $S_k(N,\psi) =
  \bigoplus_i E_i$ where the $E_i$ are the eigenspaces corresponding
  to the distinct eigenvalues of $T_q^N$.  For a prime $p \nmid N$,
  the Hecke operators $T_p^N$ and $T_q^N$ commute so each eigenspace
  is invariant under all the $T_p^N$, $p \nmid N$.  Since Hecke theory
  tells us that $S_k(N,\psi)$ admits a basis of simultaneous
  eigenforms for all the $T_p^N$, and each $E_i$ is invariant under
  this collection of operators, each $E_i$ also admits such a basis,
  $\B_i$, every element of which is also (by definition) an eigenform
  for $T_q^N$.

  Now consider the second statement. Every element of the basis $\B_i$
  belongs to a unique class $[h]$ of some newform $h \in
  S_k(N_0,\psi)$ with $N_0 \mid N$.  We collect the elements of the
  $\B_i$ which belong to a given class $[h]$. Since $S_k(N,\psi)$ is
  the direct sum of such classes and all the $\B_i$ taken together
  span $S_k(N,\psi)$, we see that $T_q^N$ is diagonalizable on each
  class $[h]$.
\end{proof}


Below we reverse the process of the lemma, starting with the class of
a newform $[h]$, and investigate how to decompose the class $[h]$ into
subspaces, extracting the various eigenspaces of $T_q^N$ for $q \mid
N$, and give conditions under which $T_q^N$ can be diagonalized on
$[h]$.  We then use these results to generalize those of Choie and
Kohnen \cite{Choie-Kohnen}. We also apply these results in section 4
to determine when there exist simultaneous eigenforms for all the
Hecke operators, and determine the number of such eigenforms which are
linearly independent.


For a prime $q\mid N$ and $h \in S_k(N_0,\psi)$ a newform,
Theorem~\ref{thm:classify-for-T_q} implies that $[h]$ contains at most
three eigenspaces for $T_q^N$.  With $f\slash T_q^N = \kappa_q f$, we
have
\begin{equation}\label{eqn:possible-eigenvalues}
  \kappa_q =
  \begin{cases}
    \lambda_q &\textrm{ when } q\nmid N/N_0, \\
    0, \lambda_q &\textrm{ when } q \mid N/N_0,\ q\mid N_0, \\
    \frac12 \left(\lambda_q \pm \sqrt{\lambda_q^2 -
        4\psi(q)q^{k-1}}\right) &\textrm{ when } q \parallel N/N_0,\ q
    \nmid N_0,\textrm{ and }\\
    0, \frac12 \left(\lambda_q \pm \sqrt{\lambda_q^2 -
        4\psi(q)q^{k-1}}\right) &\textrm{ when } q^2 \mid N/N_0,\ q
    \nmid N_0.
  \end{cases}
\end{equation}


When $q \nmid N/N_0$, we have observed that every element of $[h]$ is
an eigenform for $T_q^N$ having eigenvalue $\lambda_q$, so $T_q^N$
diagonalizes on $[h]$. Thus we restrict our attention to the case
where $q\mid N/N_0$.  Write $N/N_0 = q^\mu M_0$, with $q \nmid M_0$.
For $d_0 \mid M_0$, put $U_{d_0} = \bigoplus_{i=0}^\mu \la h\slash
B_{d_0q^i}\ra$ where $\la h\slash B_d\ra$ denotes the $\C$-linear span
of $h\slash B_d$.  Using that $[h] = \bigoplus_{d_0 \mid M_0}
U_{d_0}$, Theorem~\ref{thm:classify-for-T_q} shows that every
eigenform $f \in [h]$ with $f \slash T_q^n = \kappa_q f$ has the form
$f = \sum_{d_0 \mid M_0} f_{d_0}$ with $f_{d_0} \in U_{d_0}$, and
Proposition~\ref{prop:sufficient-conditions} shows that each $f_{d_0}$
also satisfies $f_{d_0} \slash T_q^N = \kappa_q f_{d_0}$.  Thus
$T_q^N$ diagonalizes on $[h]$ if and only if it diagonalizes on each
$U_{d_0}$.  Further, Theorem~\ref{thm:classify-for-T_q} and
Proposition~\ref{prop:sufficient-conditions} also show that each
subspace $U_{d_0}$ contains precisely $m$ linearly independent
eigenforms for $T_q^N$ where $m$ is the number of distinct eigenvalues
$\kappa_q$ given in Equation~\ref{eqn:possible-eigenvalues}.  Since
the dimension of $U_{d_0} = \mu + 1$, $T_q^N$ diagonalizes on $[h]$ if
and only if $m = \mu + 1$.  Note that since $m \le 3$, $T_q^N$
diagonalizes on $[h]$ only if $\mu \le 2$.  Moreover when $\mu = 2$
and $q\mid N_0$, we see from above that there are most $m = 2 < 3 =
\mu+1$ distinct eigenvalues, so once again $T_q^N$ cannot diagonalize
in this case.


We quantify the above observations a bit further. Still assuming
$q\mid N/N_0$, if $q\mid N_0$, there are two distinct eigenvalues
precisely when $\lambda_q \ne 0$; by Theorem 3 of \cite{Li-Newforms}
this occurs iff $q \parallel N_0$ or $\ord_q(\cond(\psi)) =
\ord_q(N_0)$.  If $q \nmid N_0$, there are two independent eigenforms
for $T_q^N$ (with non-zero eigenvalues $\kappa_q$) precisely when
$\lambda_q^2 \ne 4\psi(q)q^{k-1}$, that is when $\lambda_q$ fails to
achieve the Deligne bound.  There is an additional independent
eigenform with eigenvalue $\kappa_q = 0$ if and only if $\mu \ge 2$.
For later convenience we denote by $\fQ$ the set of primes $q \mid
N/N_0$ (just characterized) yielding a maximal number of distinct
eigenvalues $\kappa_q$, and tabulate their number.


\renewcommand{\arraystretch}{1.5}
\begin{center}
  \label{table:eigenvalues}
  \begin{tabular}{|c|c||c|c|}
    \hline
    \multicolumn{2}{|c||}{ $\mu = \ord_q(N/N_0) \ge 1$; $q \in \fQ$ provided:} & {Number of distinct eigenvalues $\kappa_q$} \\\hline
    $q\mid N_0$& $\ord_q(\cond(\psi)) = \ord_q(N_0)$ or
    $q \parallel N_0$&2 \\\hline
    $q\nmid N_0$&$\lambda_q^2 \ne 4 \psi(q)q^{k-1}$&$\min(3, \mu+1)$\\\hline
  \end{tabular}%\newline

\begin{tabular}{|c|c||c|}
  \hline
  \multicolumn{2}{|c||}{  $\mu = \ord_q(N/N_0) \ge 1$; $q \notin \fQ$ provided:} & {Number of distinct eigenvalues $\kappa_q$} \\\hline
  $q\mid N_0$& $q^2 \mid  N_0$ and $q \mid N_0/\cond(\psi)$&1 \\\hline
  $q\nmid N_0$&$\lambda_q^2 = 4 \psi(q)q^{k-1}$&$\min(2,\mu)$\\\hline

\end{tabular}%\newline
\end{center}





With this in hand, we now generalize the first part of Choie and
Kohnen's theorem \cite{Choie-Kohnen} characterizing when ``bad'' Hecke
operators can be diagonalized.

\begin{theorem}\label{thm:Choie-Kohnen}
  For a prime $q \mid N$, the Hecke operator $T_q^N$ is diagonalizable
  on $S_k(N,\psi)$ only if $S_k(N,\psi)$ contains no newform of level
  $N_0$ with $q^3 \mid N/N_0$, or with $q^2\mid N/N_0$ and $q \mid
  N_0$.  Assuming this condition, $T_q^N$ is diagonalizable if and
  only if for each $N_0$ with $\cond(\psi) \mid N_0 \mid N$ and each
  newform $h \in S_k(N_0,\psi)$ with $h \slash T_q^{N_0} = \lambda_q
  h$, either $q \nmid N/N_0$ or $q \in \fQ$.
\end{theorem}

\begin{proof} We know that
  \begin{equation*}
    S_k(N,\psi) =\bigoplus_{\cond(\psi) \mid N_0 \mid N}\ \bigoplus_{h}\ [h],
  \end{equation*}
  where the sum is over normalized newforms $h \in S_k(N_0,\psi)$.  By
  Lemma~\ref{lemma:SE-on-classes}, it suffices to determine when
  $T_q^N$ is diagonalizable on each class $[h]$. Given a newform $h
  \in S_k(N_0,\psi)$, we have seen from the discussion preceding the
  theorem that $T_q^N$ is diagonalizable on $[h]$ only if $\mu =
  \ord_q(N/N_0) \le 2$ and if $\mu = 2$, $q \nmid N_0$.  Thus the
  given conditions are necessary. Moreover, if $q \nmid N/N_0$, every
  element of $[h]$ is an eigenform for $T_q^N$, so we restrict our
  attention to the case $q \mid N/N_0$.

  Consider a newform $h \in S_k(N_0,\psi)$.  As before, write $N/N_0 =
  q^\mu M_0$, with $q \nmid M_0$, and recall we are assuming $\mu=1$
  or $\mu = 2$ and $q\nmid N_0$.  For $d_0 \mid M_0$, put $U_{d_0} =
  \bigoplus_{i=0}^\mu \la h\slash B_{d_0q^i}\ra$. We have observed
  above since $[h] = \bigoplus_{d_0 \mid M_0} U_{d_0}$, that $T_q^N$
  diagonalizes on $[h]$ if and only if it diagonalizes on each
  $U_{d_0}$, and that $T_q^N$ diagonalizes on $U_{d_0}$ if and only if
  $\dim U_{d_0}$ is equal to the number of distinct eigenvalues
  $\kappa_q$.  From the tables above it is clear that the dimension
  ($\mu+1$) equals the number of distinct eigenvalues if and only if
  $q \in \fQ$.
\end{proof}

We summarize the above results in a more compact formulation.

\begin{theorem}\label{thm:Choie-Kohnen-recast} Let $q$ be prime, and let $\psi$ be a
  Dirichlet character with conductor $\mathfrak{f}=q^\nu M_0$, with
  $\nu\geq 0$ and $q\nmid M_0$.  Let $M$ be an integer with $M_0\mid
  M$ and $q\nmid M$. If $s\leq 2$, then $T_q$ is diagonalizable on
  $S_k(q^{\nu+s}M,\psi)$ if and only if one of the following is true:

  \begin{compactenum}
  \item $s=0$,

  \item $s=1$ and $\nu\geq 1$,

  \item $s>0$, $\nu=0$, and $S_k(q^{\nu+s}M,\psi)$ contains no newform
    $h$ of level $N_0$ with $q\nmid N_0$, $T_q^{N_0}h=\lambda_qh$, and
    $\lambda_q^2=4\psi(q)q^{k-1}$, or

  \item $s=2$, $\nu\geq 1$, and $S_k(q^{\nu+s}M,\psi)$ contains no
    newform of level $N_0$ with $\mathrm{ord}_q(N_0)=\nu\mbox{ or
    }\nu+1$.

  \end{compactenum}
\end{theorem}

\begin{proof} Set $N=q^{\nu+s}M$.  We first interpret
  Theorem~\ref{thm:Choie-Kohnen} in this setting.  Since $s\leq 2$,
  $q^3\nmid N/\mathfrak{f}$, so $S_k(N,\psi)$ contains no newform of
  level $N_0$ with $q^3\mid N/N_0$.  The only case in which
  $\mathfrak{f}\mid N_0\mid N$ with $q\mid N_0$ and $q^2\mid N/N_0$
  occurs when $s=2$ and $\mathrm{ord}_q(N_0)=\nu\geq 1$. In this case,
  if $S_k(N,\psi)$ contains a newform of level $N_0$ then $T_q^N$ is
  not diagonalizable.  Otherwise, $T_q^N$ is diagonalizable if and
  only if for each $\mathfrak{f}\mid N_0\mid N$ and each newform $h\in
  S_k(N_0,\psi)$, either $q\nmid N/N_0$ or $q\in\mathfrak{Q}_{N_0,h}$.

  First suppose that one of conditions (1) - (4) holds.  If (1) holds
  then $q\nmid N/\mathfrak{f}$, so $q\nmid N/N_0$ for all
  $\mathfrak{f}\mid N_0\mid N$, hence $T_q^N$ is diagonalizable.  If
  (2) holds, then $\mathrm{ord}_q(N_0)=\nu\mbox{ or }\nu+1$ for each
  $\mathfrak{f}\mid N_0\mid N$.  In the first case,
  $\mathrm{ord}_q(N_0)=\mathrm{ord}_q(\mathfrak{f})$ so
  $q\in\mathfrak{Q}_{N_0,h}$ for each $h$ with such level.  In the
  second case, $q\nmid N/N_0$.  Therefore $T_q^N$ is diagonalizable.
  Now suppose that (3) holds.  Then $q\in\mathfrak{Q}_{N_0,h}$ for
  each newform $h$ of level $N_0$ with $q\nmid N_0$.  If
  $\mathrm{ord}_q(N_0)=s$ then $q\nmid N/N_0$.  Finally if $s=2$ and
  $q\mid\mid N_0$, then $q\in\mathfrak{Q}_{N_0,h}$ for each $h$ with
  such level.  Hence $T_q^N$ is diagonalizable. Lastly, suppose that
  (4) holds.  Then each newform $h$ contained in $S_k(N,\psi)$ has
  level $N_0$ with $\mathrm{ord}_q(N_0)=\nu+2$, so that $q\nmid
  N/N_0$.  Therefore $T_q^N$ is diagonalizable.

  Now suppose that none of (1) through (4) is true.  Then since (1) is
  false, $s=1\mbox{ or }2$.  If $s=1$, then since (2) is false,
  $\nu=0$.  Then since (3) is false, $S_k(N,\psi)$ must contain some
  newform $h$ of level $N_0$ with $q\nmid N_0$ and
  $T_q^{N_0}h=\lambda_qh$ with $\lambda_q^2=4\psi(q)q^{k-1}$.  Then
  $q\not\in\mathfrak{Q}_{N_0,h}$, so $T_q^N$ is not diagonalizable.
  Now suppose that $s=2$.  If $\nu=0$ then by the previous argument,
  $T_q^N$ is not diagonalizable.  If $\nu\geq 1$ then since (4) is
  false, $S_k(N,\psi)$ must contain some newform $h$ of level $N_0$
  with either $\mathrm{ord}_q(N_0)=\nu\mbox{ or }\nu+1$.  If
  $\mathrm{ord}_q(N_0)=\nu$, then $q\mid N_0$ and $q^2\mid N/N_0$ so
  $T_q^N$ is not diagonalizable.  If $\mathrm{ord}_q(N_0)=\nu+1$ then
  $\mathfrak{f}\mid N_0/q$ and $q^2\mid N_0$, so
  $q\not\in\mathfrak{Q}_{N_0,h}$, and hence $T_q^N$ is not
  diagonalizable.
\end{proof}


In the next result, we extend the work of Choie and Kohnen
\cite{Choie-Kohnen} (where they considered square-free level and
trivial character) by showing that if $k$ is even, $s=1\mbox{ or }2$
and $\nu=0$, then case (3) of Theorem~\ref{thm:Choie-Kohnen-recast}
holds for all but finitely many primes $q$.

\begin{theorem}\label{thm:Choie-Kohnen-generalized} Let $k$ be an even integer, and let $\psi$ be a
  Dirichlet character whose conductor $\mathfrak{f}$ divides $M$.
  Then $T_q$ is diagonalizable on both $S_k(qM,\psi)$ and
  $S_k(q^2M,\psi)$ for all primes $q\nmid M$ except for a finite
  number $r\leq C(M,k,\psi)$ of exceptions, where
  \[C(M,k,\psi):=\sum_{\mathrm{cond}(\psi)\mid M_0\mid
    M}\mathrm{dim}S_k^+(M_0,\psi)\sum_{\mu\geq
    1}\left[\frac{g_{M_0,k}}{2^\mu}\right],\] and
  \[g_{M_0,k}=\sum_{\chi \,\mathrm{mod}\,
    M_0}\mathrm{dim}S_k^+(M_0,\chi).\]
\end{theorem}

\begin{proof} By Theorem~\ref{thm:Choie-Kohnen-recast}, the only way
  that a given $T_q$ can fail to diagonalize on either $S_k(qM,\psi)$
  or $S_k(q^2M,\psi)$ is if there is a newform $h\in S_k(M_0,\psi)$
  for some $M_0$ with $\mathfrak{f}\mid M_0\mid M$ which has
  $T_q^{M_0}h=\lambda_qh$ with $\lambda_q^2=4\psi(q)q^{k-1}$.  Fix an
  $M_0$ with $\mathfrak{f}\mid M_0\mid M$ and a newform $h\in
  S_k(M_0,\psi)$ with eigenvalues $\lambda_n$.  Let $K_h$ be the
  number field (Proposition 2.8 of \cite{Shimura-hilbert},
  \cite{Shimura-hilbert-errata}) obtained by adjoining all the
  $\lambda_n$ and $\zeta_{2M}$, a primitive $2M$th root of unity, to
  $\mathbb{Q}$.  Then since $k$ is even, $\sqrt{q}\in K_h$ for each
  prime $q\nmid M$ such that $\lambda_q^2=4\psi(q)q^{k-1}$.  We call
  such a $q$ an exceptional prime for $h$.  Now if $p_1,p_2,\dots,p_s$
  are different primes, the degree of
  $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_s})/\mathbb{Q}$ is
  $2^s$.  Since
  \[\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{q}:q\mbox{ an exceptional
    prime for }h)\subseteq K_h\] and $K_h$ is a finite extension of
  $\mathbb{Q}$, there must be a finite number $r_h$ of exceptional
  primes for $h$.  In particular, $r_h\leq
  \mathrm{ord}_2([K_h:\mathbb{Q}])$.

  The group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on
  normalized eigenforms in $S_k(M_0,\psi)$ by sending $f=\sum
  a(n)q^n\in S_k(M_0,\psi)$ to $f^\sigma=\sum a(n)^\sigma q^n\in
  S_k(M_0,\psi^\sigma)$, for each
  $\sigma\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$
  (Proposition 2.6 of \cite{Shimura-hilbert},
  \cite{Shimura-hilbert-errata}).  Let $K_{M_0,k}=\prod K_h$ be the
  composite field where the product runs over all characters $\chi$
  modulo $M_0$ and all newforms $h\in S_k(M_0,\chi)$.  Since each
  automorphism of the Galois closure of $K_{M_0,k}/\mathbb{Q}$
  permutes these newforms, it can be considered as a subgroup of
  $S_{g_{M_0,k}}$, the symmetric group on $g_{M_0,k}$ elements, where
  \[g_{M_0,k}=\sum_{\chi \bmod M_0}\mathrm{dim}S_k^+(M_0,\chi).\] Then
  $[K_h:\mathbb{Q}]\mid [K_{M_0,k}:\mathbb{Q}]\mid g_{M_0,k}!$, so
  \[r_h\leq \mathrm{ord}_2(g_{M_0,k}!)=\sum_{\mu\geq
    1}\left[\frac{g_{M_0,k}}{2^\mu}\right].\] Now $T_q$ diagonalize on
  neither $S_k(qM,\psi)$ nor $S_k(q^2M,\psi)$ if $q$ is an exceptional
  prime for a single newform $h\in S_k(M_0,\psi)$ for any
  $\mathfrak{f}\mid M_0\mid M$.  Therefore we get an upper bound for
  $r$, the number of primes $q$ for which $T_q$ fails to diagonalize,
  by summing over all such newforms.  Then
  \[r\leq \sum_{\mathrm{cond}(\psi)\mid M_0\mid
    M}\mathrm{dim}S_k^+(M_0,\psi)\cdot r_h\leq
  \sum_{\mathrm{cond}(\psi)\mid M_0\mid
    M}\mathrm{dim}S_k^+(M_0,\psi)\left(\sum_{\mu\geq
      1}\left[\frac{g_{M_0,k}}{2^\mu}\right]\right).\]

\end{proof}

\begin{remark}
  One could obtain a more explicit, though considerably larger, upper
  bound. For example, $g_{M_0,k}\leq \mathrm{dim} S_k(\Gamma_1(M_0))$
  for which one could use the known dimension formulas.
\end{remark}


We conclude this investigation of diagonalization with the following
``negative'' result for levels divisible by a high power of $q$.

\begin{theorem}\label{thm:cant-diagonalize} Let $q$ be prime, and let $\psi$ be a Dirichlet
  character with conductor $\mathfrak{f}=q^\nu M_0$, with $\nu\geq 0$
  and $q\nmid M_0$. Let $M$ be an integer with $M_0\mid M$ and $q\nmid
  M$, and let $s\geq 3$ be an integer. Except possibly for finitely
  many $k\geq 2$ with $\psi(-1)=(-1)^k$ and finitely many $q$, $T_q$
  is not diagonalizable on $S_k(q^{\nu+s}M,\psi)$.
\end{theorem}

\begin{proof} Let $N=q^{\nu+s}M$ and $N_0=q^{\nu+2}M$.  For each
  $s\geq 3$, we have $\mathfrak{f}\mid N_0\mid N$.  Further, $q\mid
  N/N_0$, $q^2\mid N_0$ and $\mathfrak{f}\mid N_0/q$.  Hence if
  $S_k(N,\psi)$ contains a newform $h$ of level $N_0$, then
  $q\not\in\mathfrak{Q}_{N_0,h}$, so $T_q^N$ is not diagonalizable.
  But by Theorem~\ref{thm:newform-dimension}, for all but finitely
  many $k\geq 2$ with $\psi(-1)=(-1)^k$ and finitely many $q$,
  $\mbox{dim }S_k^+(N_0,\psi)\geq 1$, and hence $T_q$ is not
  diagonalizable on $S_k(N,\psi)$.
\end{proof}






\section{Simultaneous Hecke Eigenforms}

We now turn to the question of characterizing simultaneous Hecke
eigenforms in $S_k(N,\psi)$ for all Hecke operators $T_\ell^N$, $\ell$
a prime.  From the previous section and the theory of newforms, for a
given simultaneous eigenform $f \in S_k(N,\psi)$, the only primes
which need careful analysis are primes $q \mid N/N_0$ where $N_0$ is
the level of the associated newform.  We make this explicit.


\begin{theorem}\label{thm:characterizeSE}
  Let $f\in S_k(N,\psi)$ be a nonzero simultaneous eigenform for all
  the Hecke operators $T_\ell^N$, $\ell$ a prime, and put $f \slash
  T_q^N = \kappa_q f$ for each prime $q \mid N$.  Associated to $f$ is
  a newform $h\in S_k(N_0,\psi)$ (with $\cond(\psi) \mid N_0 \mid N$)
  such that $f=\sum_{d\mid N/N_0}\alpha_d\, h\slash B_d$.  Then
  $\alpha_1 \ne 0$, and normalizing with $\alpha_1 = 1$, we have that
  $\alpha_d = \prod_{q \mid d} \alpha_{q^{\mu_q}}$, where $\mu_q =
  \ord_q(d)$.  Further, we have $\alpha_{q^e} = 0$ for $e \ge 3$, and
  \[\alpha_q = (\kappa_q - \lambda_q) \textrm{ and }\alpha_{q^2} =
  \begin{cases}
    0& q \mid N_0,\\
    0& q\nmid N_0, \kappa_q \ne 0,\\
    \psi(q) q^{k-1}& q\nmid N_0, \kappa_q = 0,
  \end{cases}
  \]
\end{theorem}

\begin{proof}
  This is immediate from Theorem~\ref{thm:classify-for-T_q}, which
  also indicates the possible eigenvalues $\kappa_q$.
\end{proof}

\begin{remark}
  The converse to the above theorem is also true. Starting with a
  newform $h$, and choosing the $\kappa_q$ and $\alpha_d$ as in the
  theorem, Proposition~\ref{prop:sufficient-conditions} guarantees
  that $f=\sum_{d\mid M}\alpha_d\, h\slash B_d$ is a simultaneous
  eigenform for all $T_q^N$ with $q \mid N$, and hence for all
  $T_\ell^N$, $\ell$ a prime.
\end{remark}


Now we wish to count the number of linearly independent simultaneous
Hecke eigenforms that are associated to a given newform.

\begin{theorem}\label{thm:countSE}
  Let $h\in S_k(N_0,\psi)$ be a newform and let $N$ be an integer such
  that $N_0\slash N$. For all primes $q \slash N$, put $h\slash
  T_q^{N_0} = \lambda_q h$. The number of linearly independent
  simultaneous eigenforms $f\in S_k(N,\psi)$ which are eigenforms for
  all $\{T_\ell^N\}$, $\ell$ a prime and which have the same
  eigenvalues as $h$ under all $T_p$, $p\nmid N$ is $2^{|A|} 3^{|B|}$,
  where $A$ and $B$ are sets of primes dividing $N/N_0$ satisfying
  \[
  q\in B = B(N, N_0, h) \iff q\nmid N_0, q^2 \mid N/N_0, \lambda_q^2
  \ne 4 \psi(q) q^{k-1},
  \]
  and
  \[
  q \in A = A(N, N_0, h) \iff
  \begin{cases}
    q\mid N_0 \textrm{ and } \lambda_q \ne 0,
    \textrm{ or}\\
    q\nmid N_0, q \parallel N/N_0, \lambda_q^2 \ne 4 \psi(q)
    q^{k-1}, \textrm{ or}\\
    q\nmid N_0, q^2 \mid N/N_0, \lambda_q^2 = 4 \psi(q) q^{k-1}.
  \end{cases}
  \]
\end{theorem}

\begin{remark}
  By Theorem 3 of \cite{Li-Newforms}, the first condition stated to
  define $A$ ($q\mid N_0$ and $\lambda_q \ne 0$) is equivalent to
  $q \parallel N_0$, or $q^2 \mid N_0$ and $\ord_q(\cond(\psi)) =
  \ord_q(N_0)$.
\end{remark}


\begin{proof}
  Theorem~\ref{thm:characterizeSE} indicates the shape of every
  simultaneous eigenform $f$ of level $N$ associated to the newform
  $h$: $f=\sum_{d\mid N/N_0}\alpha_d\, h\slash B_d$, where without
  loss, $\alpha_1 = 1$, and $\alpha_d$ is completely determined as the
  product of $\alpha_{q^e}$ where $e = \ord_q(d)$.  We see all such
  values $\alpha_{q^e}$ are uniquely determined except for the value
  of $\alpha_q = \kappa_q - \lambda_q$ which has as many distinct
  values as distinct eigenvalues $\kappa_q$. It is now a simple matter
  using Theorem~\ref{thm:classify-for-T_q} to verify that the sets $A$
  and $B$ characterize those cases in which $\kappa_q$ can have 2 or 3
  distinct eigenvalues.
\end{proof}


\section{Examples}

Theorem~\ref{thm:countSE} tells how to compute the number of
simultaneous eigenforms in $S_k(N,\psi)$ associated to a newform $h
\in S_k(N_0,\psi)$ with $h\slash T_q^{N_0} = \lambda_q h$ in terms of
the sets $A$ and $B$.  Knowledge of the eigenvalue $\lambda_q$ for $q
\nmid N_0$ can often be problematic, but there are cases in which it
is easy to calculate explicitly the sets $A$ and $B$.  We characterize
one particularly useful situation, and give some examples.

Let $N_0\mid N$ with $N$ and $N_0$ having exactly the same prime
divisors.  Then $B= B(N,N_0,h) = \emptyset$, and
\begin{equation}
  \label{eqn:fQ}
  A = A(N,N_0,h) = \{ q\mid N/N_0 :  \ord_q(N_0) = 1 
  \textrm{ or } \ord_q(\cond(\psi))\}.
\end{equation}


\begin{example}
  Let $\psi$ be a character with square-free conductor $D$, and let
  $N$ be an integer with $D \mid N \mid D^2$.  Then $S_k(N,\psi)$ has
  a basis consisting of simultaneous eigenforms for all Hecke
  operators.
\end{example}
\begin{proof}
  Let $N_0$ be such that $D \mid N_0 \mid N$, and consider a newform
  $h \in S_k(N_0,\psi)$.  To compute $A$, let $q \mid N/N_0$ be a
  prime.  Observe that $q \mid N/N_0$ implies $q\mid D$.  Since $N
  \mid D^2$, $1 \le \ord_q(N_0) \le \ord_q(N) \le 2$, and if
  $\ord_q(N_0) = 2$, then $q \nmid N/N_0$.  Thus $q \mid N/N_0$
  implies $\ord_q(N_0) = 1$ and hence $q \in A$.  Thus $2^{| A|} =
  \sigma_0(N/N_0)$ (since $N/N_0$ is square-free), where $\sigma_0(m)$
  is the number of positive divisors of $m$. Since
  \[
  S_k(N,\psi) = \bigoplus_{D\mid N_0\mid N}\ \bigoplus_{d\mid N/N_0}
  S_k^+(N_0,\psi) \slash B_d \cong \bigoplus_{D\mid N_0\mid N}\
  \sigma_0(N/N_0)\ S_k^+(N_0,\psi),
  \]
  with the isomorphism as modules for the Hecke algebra generated by
  $T_p^N$ for all primes $p \nmid N$, the result is clear.  In the
  isomorphism we use the convention that for a space $S$, $mS =
  \bigoplus_{i=1}^m S$.
\end{proof}



As a second example, we consider a situation in which the conductor of
the character $\psi$ can be large.

\begin{example}
  Let $q$ be an odd prime and $\psi$ a character of conductor $q^\nu$,
  $\nu \ge 1$.  Then $S_k(q^{\nu+\mu}, \psi)$, $\mu \ge 0$, has a
  basis of simultaneous eigenforms for all the Hecke operators when
  $\mu = 0, 1$; for $\mu \ge 3$ it has such a basis only for finitely
  many $k$ and $q$.
\end{example}

\begin{proof}
  The only issue concerns the diagonalizability of the operator
  $T_q^N$ where $N = q^{\nu+\mu}$. The case of $\mu = 0, 1$ is
  addressed by Theorem~\ref{thm:Choie-Kohnen-recast}, while the case
  of $\mu \ge 3$ is addressed by Theorem~\ref{thm:cant-diagonalize}.
\end{proof}

\begin{example}
  Somewhat complementary to the previous example, we consider the case
  of $S_k(q^2,1)$ providing instances when $T_q^N$ can be
  diagonalized.  To that end we consider normalized newforms $h$ of
  level 1, $q$ and $q^2$ and whether $T_q^N$ can be diagonalized on
  $[h]$.  For level $q^2$ the answer is affirmative from the theory of
  newforms.  For each newform $h \in S_k(q,1)$, Equation~\ref{eqn:fQ}
  yields $A = \{q\}$ providing the requisite two linearly independent
  simultaneous eigenforms, $h$ and $h - \lambda_q h$.  Now consider $h
  \in S_k(1,1)$.  When $k < 12$ or $k=14$, $S_k(1,1) = 0$, and there
  are no classes to consider.  On the other hand, when $S_k(1,1) \ne
  0$, the situation is more subtle.  $T_q^N$ will diagonalize on $[h]$
  provided that $\lambda_q^2 \ne 4q^{k-1}$.  As an example, consider
  weight 12. There are three simultaneous eigenforms in $S_{12}(q^2,
  1)$ equivalent to $\Delta = (2\pi)^{12}\sum_{n=1}^\infty \tau(n)
  e^{2 \pi i n \tau}\in S_{12}(1,1)$ with Fourier coefficients given
  by the Ramanujan $\tau$-function.  Choose a prime $q$ for which
  $\lambda^2_q = \tau^2(q) \ne 4 q^{11}$; tables of values of the
  $\tau$-function abound.

  Using Theorem~\ref{thm:characterizeSE}, we produce three linearly
  independent simultaneous eigenforms $f = h + \alpha_q h\slash B_q +
  \alpha_{q^2} h \slash B_{q^2}$ satisfying $f \slash T_q^N = \kappa_q
  f$ where $\kappa_q = 0, \frac12 \left(\lambda_q \pm
    \sqrt{\lambda_q^2 - 4\psi(q)q^{k-1}}\right)$ (all distinct).  Thus
  $T_q$ diagonalizes on $S_k(q^2, 1)$.



\end{example}


\begin{example}
  The results of section 2 also can provide a multiplicity-one theorem
  in the following narrow context. Let $N_0 \ge 2$ be an integer, and
  consider any newform $h \in S_k(N_0^2, 1)$. Let $N = N_0^2 N_1$
  where the only primes dividing $N_1$ divide $N_0$.  Let $f \in
  S_k(N,1)$ be an simultaneous eigenform for all Hecke operators $T_p$
  for $p \nmid N$ which is equivalent to $h$.  Then $f = h$.

  \begin{proof}
    This is immediate from
    Proposition~\ref{prop:sufficient-conditions}, since for each $q
    \mid N$, the $q$th eigenvalue of $h$ ($\lambda_q)$ is zero forcing
    $f= \sum \alpha_d h \mid B_d = h$.
  \end{proof}

\end{example}

\begin{remark}

  Another point worth noting concerns the interpretation when the sets
  $A$ and $B$ are empty.  In such a case, Theorem~\ref{thm:countSE}
  implies there is a unique simultaneous eigenform $f$ in
  $S_k(N,\psi)$ associated to a newform $h \in S_k(N_0,\psi)$.  It is
  not necessarily the case that the simultaneous eigenform is the
  newform $h$.  For example, choose a prime $q \parallel N/N_0$ with
  $q \nmid N_0$.  Then $B = \emptyset$ and $q \notin A$ means that
  $\lambda_q^2 = 4\psi(q) q^{k-1} \ne 0$ and hence by
  Theorem~\ref{thm:characterizeSE}, $f = \sum_{d \mid N/N_0} \alpha_d
  h\slash B_d$ with $\alpha_q = -\lambda_q/2 \ne 0$.
   

 


\end{remark}


\section{Dimensions of spaces of newforms}

To justify the last part of Theorem~\ref{thm:Choie-Kohnen-recast}, we
compute a lower bound for the dimension of the space of newforms
$S_k^+(q^{\nu + r}, \psi)$ where $q$ is a prime, $r\ge 2$, and $\psi$
a character with conductor $q^\nu$, $\nu \ge 0$.  We make implicit use
of the trace formula for Hecke operators as given in \cite{Ross}, in
particular Ross's formula for the dimension of the space of cusp
forms.  For trivial character, one can find a formula for the
dimension of the space of newforms in \cite{Martin}.


\begin{theorem}\label{thm:newform-dimension}
  Let $q$ be a prime, and $\psi$ be a Dirichlet character of conductor
  $\f = q^\nu M_0$, $q \nmid M_0$, $\nu \ge 0$.  Let $r$ be an
  integer, $r \ge 2$, and let $M$ be any integer divisible by $M_0$
  ($q\nmid M$); we further require that $\ord_2(M/M_0) \ne 1$. Then
  except for finitely many values of $k \ge 2$ with $\psi(-1) =
  (-1)^k$ and finitely many values of $q$, we have the dimension of
  the space of newforms $S_k^+(q^{\nu+r}M,\psi)$ is positive.
\end{theorem}

\begin{proof}
  We consider the Hecke algebra generated by all operators $T_p$ with
  $p \nmid qM$, and recall our shorthand of writing $mS$ for
  $\bigoplus_{i=1}^m S$ in any isomorphism of modules for the Hecke
  algebra.  Let $N = q^{\nu+r}M$.  Then

\begin{align*}
  S_k(N,\psi) &= \bigoplus_{\f\, \mid N_0\mid N}\ \bigoplus_{d \mid
    N/N_0} S_k^+(N_0, \psi) \slash B_d\\
  &\cong \bigoplus_{\f\, \mid N_0\mid N}\ \sigma_0(N/N_0) S_k^+(N_0,
  \psi)\\
  &= \bigoplus_{d\mid N/\f} \sigma_0(N/d\f) S_k^+(d\f, \psi).
\end{align*}
We adapt the notation of Martin \cite{Martin}, who gives a formula for
the dimension of the space of newforms with trivial character, and put
\[
g_0(d) = \dim S_k(d\f, \psi), \qquad g_0^+ = \dim S^+_k(d\f, \psi).
\]
The above decomposition yields the following relations of dimensions:
\begin{equation}
  \label{eqn:g_0-equals}
  g_0(N/\f) = \sum_{d\mid N/\f} g_0^+(d)\, \sigma_0(N/d\f) = (g_0^+ *
  \sigma_0) (N/\f),
\end{equation}


\noindent where * is the standard Dirichlet convolution of arithmetic
functions.  While $g_0^+(n)$ is not a multiplicative function,
$\sigma_0(n)$ is, and in complete analogy with \cite{Martin}, we let
$\lambda$ be the Dirichlet inverse of $\sigma_0$: $\lambda = \mu *
\mu$ ($\mu$ the Mobius function).  Thus $\lambda$ is a multiplicative
function with values (for $p$ a prime) $\lambda(p) = -2$,
$\lambda(p^2) = 1$, and $\lambda(p^j) = 0$ for $j \ge 3$.  Taking the
Dirichlet convolution of both sides of Equation~\ref{eqn:g_0-equals}
yields
\begin{equation}
  \label{eqn:g^+-equals}
  \dim S_k^+(N,\psi) = g_0^+(N/\f) = \sum_{d\mid N/\f} g_0(d)\, \lambda(N/d\f) = (g_0 *
  \lambda) (N/\f).
\end{equation}

The goal is to use the above expression to produce a formula for the
dimension of the space of newforms as a function of the prime $q$ and
weight $k$.  Using a parametrized version of the notation from
\cite{Ross}, we obtain a formula for the dimension of the space of
cusp forms with arbitrary character:
\begin{equation}
  \label{eqn:dimension-cusp}
  g_0(d) = -s_0(d\f) - s_1(d\f) + \delta + m(d\f) - p(d\f),
\end{equation}
where letting $\omega(n)$ be the number of prime divisors of an
integer $n$, we have
\begin{align*}
  |s_0(d\f)| &\le 2^{\omega(d\f)-2},\\
  |s_1(d\f)| &\le \frac13 2^{\omega(d\f)},\\
  \delta &=
  \begin{cases}
    1&\textrm{if $k=2$ and $\psi=1$},\\
    0&\textrm{otherwise,}
  \end{cases}\\
  m(d\f) &= \frac{(k-1)}{12}\, d\f\, \prod_{\ell \mid d\f} (1 +
  1/\ell)
  \textrm{  ($\ell$ a prime), and}\\
  p(d\f) &= \frac12 \prod_{\ell \mid d\f} par(\ell)\textrm{ ($\ell$ a
    prime),}
\end{align*}
where $par(\ell)$ are the parabolic terms as computed in Theorem 1 of
\cite{Ross}.

Now we distribute the convolution of $\lambda$ through the summands
defining $g_0$ and obtain
\begin{align}
  \label{eqn:dimen-new-split}
  g_0^+(N/\f) &= \sum_{d\mid N/\f} g_0(d)\,
  \lambda(N/d\f)\notag\\
  &= -\sum_{d\mid N/\f}s_0(d\f)\,\lambda(N/d\f) -\sum_{d\mid
    N/\f}s_1(d\f)\,\lambda(N/d\f)+ \delta \sum_{d\mid
    N/\f}\lambda(N/d\f)\notag\\
  &\hspace{1.5in}+ \sum_{d\mid N/\f}m(d\f)\,\lambda(N/d\f)-
  \sum_{d\mid N/\f} p(d\f)\,\lambda(N/d\f).
\end{align}

If we put $N/\f = q^rM_1$ with $q\nmid M_1$, then since $r\ge 2$,
\begin{equation*}
  \sum_{d\mid N/\f}\lambda(N/d\f) = \sum_{j=0}^r \lambda(q^j)
  \sum_{d_1 \mid M_1} \lambda(M_1/d_1) = 0,
\end{equation*}
so
\begin{align}
  \label{eqn:dimen-new-split-1}
  g_0^+(N/\f) &= \sum_{d\mid N/\f} g_0(d)\,
  \lambda(N/d\f)\notag\\
  &= -\sum_{d\mid N/\f}(s_0(d\f) + s_1(d\f))\, \lambda(N/d\f) +
  \sum_{d\mid N/\f}\left(m(d\f) - p(d\f) \right)\,\lambda(N/d\f),
\end{align}
hence
\begin{align}
  \label{eqn:dimen-new-split-2}
  g_0^+(N/\f) &= | g_0^+(N/\f)| \ge \notag\\
  &\ |\sum_{d\mid N/\f}\left(m(d\f) - p(d\f) \right)\,\lambda(N/d\f)|
  - |\sum_{d\mid N/\f}(s_0(d\f) + s_1(d\f))\, \lambda(N/d\f) |.
\end{align}
We shall show that the second term is bounded and the first goes to
infinity as $q$ or $k$ do which will establish our result.

We consider the second term:
\begin{equation}
  \label{eqn:s-terms}
  |\sum_{d\mid N/\f}(s_0(d\f) + s_1(d\f))\, \lambda(N/d\f) | \le 
  \sum_{d\mid N/\f}\left(|s_0(d\f)| +| s_1(d\f)|\right)\,
  |\lambda(N/d\f)| \le \sum_{d\mid N/\f} 2^{\omega(d\f)}|\lambda(N/d\f)|,
\end{equation}
since from above we have that
\begin{equation*}
  |s_0(d\f)| +| s_1(d\f)| \le 2^{\omega(d\f) - 2} + \frac13
  2^{\omega(d\f)} < 2^{\omega(d\f)}.
\end{equation*}
Writing $N/\f = q^r M_1$ with $q \nmid M_1$ and recalling that $\f =
q^\nu M_0$, we have
\begin{align*}
  \sum_{d\mid N/\f} 2^{\omega(d\f)}|\lambda(N/d\f)|&= \sum_{j=0}^r
  |\lambda(q^{r-j})| \sum_{d_1 \mid M_1}
  2^{\omega(q^{j+\nu} d_1 M_0)}|\lambda(M_1/d_1)|\notag\\
  &\le \sum_{j=0}^r |\lambda(q^{r-j})| \sum_{d_1 \mid M_1}
  2^{1+\omega(M))}|\lambda(M_1/d_1)|\notag\\
  &\le 2^{3 + \omega(M)}\sum_{d_1 \mid M_1} |\lambda(M_1/d_1)|,
\end{align*}
which is a constant depending only on $M$ and independent from $k$ and
$q$.  Thus it remains only to show that


\begin{equation*}
  |\sum_{d\mid N/\f}\left(m(d\f) - p(d\f) \right)\,\lambda(N/d\f)| \to \infty,
\end{equation*}
as $q$ or $k$ go to infinity.  What we show is that
\begin{equation*}
  \sum_{d\mid N/\f}\left(m(d\f) - p(d\f) \right)\,\lambda(N/d\f) =
  \frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q),
\end{equation*}
with constants $A, B$ depending only on $M$, $A > 0$, and $\mathcal M,
\mathcal P$ functions of $q$ with the expression having the desired
limits as $q$ or $k$ go to infinity.

We first consider the ``Mass'' term: $\ds \sum_{d\mid N/\f}m(d\f)
\,\lambda(N/d\f)$.  For an integer $n$, let $m_0(n) = n
\prod_{\ell\mid n} (1 + 1/\ell)$ where the product is over all primes
$\ell$ dividing $n$.  Then $m(n) = \frac{k-1}{12} m_0(n)$, and $m_0$
is a multiplicative function.  Thus once again writing $N/\f = q^r
M_1$ with $q \nmid M_1$ and recalling that $\f = q^\nu M_0$,
\begin{align*}
  \sum_{d\mid N/\f}m(d\f)\,\lambda(N/d\f) &=
  \frac{k-1}{12}\sum_{j=0}^r m_0(q^{\nu+j})\lambda(q^{r-j}) \sum_{d_1
    \mid M_1}
  m_0(d_1M_0)\lambda(M_1/d_1)\\
  &= \frac{k-1}{12} (m_0(q^{\nu+r}) -2m_0(q^{\nu+r-1}) +
  m_0(q^{\nu+r-2})) \sum_{d_1 \mid M_1}
  m_0(d_1M_0)\lambda(M_1/d_1)\\
  &= \frac{k-1}{12} \mathcal M(q) \sum_{d_1 \mid M_1}
  m_0(d_1M_0)\lambda(M_1/d_1),
\end{align*}
where
\begin{equation*}
  \mathcal M(q) = \begin{cases}
    q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2& \textrm{if }\nu + r - 2 > 0, \\
    q^2 -q -1 = (q-1)^2 + q-2& \textrm{if } \nu=0, r=2.
  \end{cases}
\end{equation*}
We now wish to show that $ \sum_{d_1 \mid M_1}
m_0(d_1M_0)\lambda(M_1/d_1)$ is positive.  Since both $m_0$ and
$\lambda$ are multiplicative, it suffices to show this when $M_1 =
p^e$, $M_0 = p^f$ are prime powers ($e+f \ge 1$).
\begin{align*}
  \sum_{d_1 \mid M_1} m_0(d_1M_0)\lambda(M_1/d_1) &=
  \sum_{j=0}^{e} m_0(p^{f+j})\lambda(p^{e-j})\\
  &=\begin{cases}
    m_0(p^f)&\textrm{if } e = 0\\
    -2m_0(p^f) + m_0(p^{f+1})& \textrm{if } e = 1\\
    m_0(p^{e-2+f}) - 2m_0(p^{e-1+f}) + m_0(p^{e+f})& \textrm{if }
    e\ge 2\\
  \end{cases}\\
  &=
  \begin{cases}
    p^f + p^{f-1}&\textrm{if } e = 0\\
    p^{f+1}-p^f -2p^{f-1}& \textrm{if } e = 1\\
    p^2 - p -1 & \textrm{if }e= 2, f=0\\
    p^{e+f-3}(p+1)(p-1)^2&  \textrm{if }e+f\ge 3,\\
  \end{cases}
\end{align*}
where we understand $p^{-1} = 0$, and this sum is trivially checked to
be positive for all primes $p \ge 2$. Note the case with $e=1$ (when
$p=2$) is precluded by the theorem's hypothesis $\ord_2(M/M_0) =
\ord_2(M_1) \ne 1$.



Finally we turn to the parabolic terms: $\ds \sum_{d\mid N/\f}p(d\f)
\,\lambda(N/d\f)$.  For an integer $n$, let $p_0(n) = \prod_{\ell\mid
  n} par(\ell)$ where the product is over all primes $\ell$ dividing
$n$, and $par(\ell)$ is defined as in Theorem 1 of \cite{Ross}.  Then
$p(n) = (1/2) p_0(n)$, and $p_0$ is a multiplicative function, $p_0(1)
= 1$.  Once again writing $N/\f = q^r M_1$ with $q \nmid M_1$ and
recalling that $\f = q^\nu M_0$,
\begin{align*}
  \sum_{d\mid N/\f}p(d\f)\,\lambda(N/d\f) &= \frac{1}{2}\sum_{j=0}^r
  p_0(q^{\nu+j})\lambda(q^{r-j}) \sum_{d_1 \mid M_1}
  p_0(d_1M_0)\lambda(M_1/d_1)\\
  &= \frac{1}{2} (p_0(q^{\nu+r}) -2p_0(q^{\nu+r-1}) +
  p_0(q^{\nu+r-2})) \sum_{d_1 \mid M_1}
  p_0(d_1M_0)\lambda(M_1/d_1)\\
  &= \frac{1}{2} \mathcal P(q) \sum_{d_1 \mid M_1}
  p_0(d_1M_0)\lambda(M_1/d_1),
\end{align*}
where $ \mathcal P(q) = (p_0(q^{\nu+r}) -2p_0(q^{\nu+r-1}) +
p_0(q^{\nu+r-2}))$.

Thus it remains only to show that
\begin{equation*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| \to \infty,
  \textrm{as $k$ or $q$ } \to \infty
\end{equation*}
with constants $A, B$ depending only on $M = M_0M_1$, $A > 0$.  The
case for $k \to \infty$ is clear, so we focus on this expression as a
function of $q$.

To compute $p_0(q^{\nu+j})$, we set a bit of notation.  Let $\mu_j =
\lfloor\frac{\nu+j}{2}\rfloor$.  From Theorem 1 of \cite{Ross}, we
have
\begin{equation*}
  p_0(q^{\nu+j}) = par(q) = 
  \begin{cases}
    2q^j&\textrm{if } \nu \ge \mu_j + 1,\\
    (q^{\mu_j} + q^{\mu_j - 1})& \textrm{if } \nu \le \mu_j,\ \nu+j \textrm{ even},\\
    2q^{\mu_j}& \textrm{if } \nu \le \mu_j,\ \nu+j \textrm{ odd}.\\
  \end{cases}
\end{equation*}
Since we need to compute $ \mathcal P(q) = (p_0(q^{\nu+r})
-2p_0(q^{\nu+r-1}) + p_0(q^{\nu+r-2}))$ as part of $|\frac{k-1}{12} A
\mathcal M(q) - B\mathcal P(q)|$, we need to break the argument into
cases.

\textbf{Case 0:} Special case $\nu = 0$, $r = 2$.

$|\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| =|\frac{k-1}{12} A
(q^2 - q - 1) - B (q-2)|\to \infty $ as $q \to \infty$.

Henceforth we can assume $\nu + r \ge 3$, so $\mathcal M(q)=
q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2$.

\textbf{Case 1:} $\nu \ge \mu_r + 1$.  Then $\nu \ge \mu_r + 1 \ge
\mu_{r-1} + 1 \ge \mu_{r-2} + 1$.  Note that this case cannot occur
unless $\nu \ge 3$.  Then $\mathcal P(q) = 2q^{r-2}(q-1)^2$, so
\begin{align*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| &=|\frac{k-1}{12}
  A
  q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2) - B(2q^{r-2}(q-1)^2)|\\
  &=q^{r-2}(q-1)^2[|\frac{k-1}{12} A q^\nu (1+\frac{1}{q}) - B|] \to
  \infty \textrm{ as } q \to \infty.
\end{align*}


\textbf{Case 2:} $\nu \le \mu_{r-2}$.  Then $\nu \le \mu_{r-2} \le
\mu_{r-1} \le \mu_r$.

When $\nu+r$ is even, $\mu_r = (\nu+r)/2 \ge 2$, and we have
\begin{align*}
  \mathcal P(q) &= (q^{\mu_r} + q^{\mu_r - 1}) - 4q^{\mu_{r-1}} +
  q^{\mu_{r-2}} + q^{\mu_{r-2}-1}\\
  &= (q^{\mu_r} + q^{\mu_r - 1}) -4 q^{\mu_r-1} + q^{\mu_r -1} +
  q^{\mu_r - 2}\\
  &= q^{\mu_r -2}(q -1)^2,
\end{align*}
so
\begin{align*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| &=|\frac{k-1}{12}
  A
  q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2) - B( q^{\mu_r -2}(q -1)^2)|\\
  &=q^{\mu_r-2}(q-1)^2[|\frac{k-1}{12} A q^{\mu_r} (1+\frac{1}{q}) -
  B|] \to \infty \textrm{ as } q \to \infty.
\end{align*}

When $\nu+r$ is odd, $\mu_r = \mu_{r-1} = \mu_{r-2}+1$, so
\begin{equation*}
  \mathcal P(q) = 2q^{\mu_r} - 2(q^{\mu_{r-1}} + q^{\mu_{r-1}-1}) +
  2q^{\mu_{r-2}} = 0,
\end{equation*}
and
\begin{align*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| &=|\frac{k-1}{12}
  A q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2)| \to \infty \textrm{ as } q
  \to \infty.
\end{align*}

\textbf{Case 3.} $\mu_{r-2} < \nu \le \mu_r$.

If $\nu+r$ is even, the condition translates to $\frac{\nu+r}{2}-1 <
\nu \le \frac{\nu+r}{2}$, so that $\mu_r = \frac{\nu+r}{2} =\nu = r$,
and $\mu_{r-1} = \mu_{r-2} = \mu_r - 1$.  If $\nu+r $ is odd, it
translates to $\frac{\nu+r-1}{2}-1 < \nu \le \frac{\nu+r-1}{2}$, so
that $\mu_r = \mu_{r-1} = \frac{\nu+r-1}{2} =\nu = r-1$, and
$\mu_{r-2} = \mu_r -1$.

If $\nu+r$ is even, we have
\begin{align*}
  \mathcal P(q) &= (q^{\mu_r} + q^{\mu_r - 1}) - 4q^{r-1} +
  2q^{r-2}\\
  &= (q^{\nu} + q^{\nu - 1}) -4 q^{\nu-1} + 2q^{\nu -2}\\
  &= q^{\nu-2 }(q^2 - 3q + 2) = q^{\nu-2}(q-1)(q-2),
\end{align*}
so
\begin{align*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| &=|\frac{k-1}{12}
  A
  q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2) - B( q^{\nu-2 }(q-1)(q-2) )|\\
  &=q^{\nu-2}(q-1)[|\frac{k-1}{12} A q^r(q-1) (1+\frac{1}{q}) -
  B(q-2)|] \to \infty \textrm{ as } q \to \infty.
\end{align*}

If $\nu+r$ is odd, we have
\begin{align*}
  \mathcal P(q) &= 2q^{\mu_r} - 2(q^{\mu_{r-1}} + q^{\mu_{r-2}}) + 2q^{r-2}\\
  &= 2q^\nu - 2(q^\nu + q^{\nu-1}) + 2q^{\nu-1} = 0,
\end{align*}
so
\begin{align*}
  |\frac{k-1}{12} A \mathcal M(q) - B\mathcal P(q)| &=|\frac{k-1}{12}
  A q^{\nu+r-2}(1+\frac {1}{q})(q-1)^2)| \to \infty \textrm{ as } q
  \to \infty.
\end{align*}

\end{proof}

% \newpage

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\small\singlespace \bibliographystyle{amsplain} \bibliography{SE}

% ------------------------------------------- Begin Closing
% Information -------------------------------------------

\address{Department of Mathematics, 6188 Kemeny Hall, Dartmouth
  College, Hanover, NH 03755}\newline \fax{(603) 646-1312}\newline
\email {thomas.r.shemanske@dartmouth.edu}\newline
\URL{http://www.math.dartmouth.edu/\symbol{126}trs/}

\address{Department of Mathematics, 6188 Kemeny Hall, Dartmouth
  College, Hanover, NH 03755}\newline \fax{(603) 646-1312}\newline
\email {stephanie.a.treneer@dartmouth.edu}\newline
\URL{http://www.math.dartmouth.edu/\symbol{126}stephanie/}

\address{Department of Mathematics, University Walk Bristol,BS8 1TW,
  UK}\newline \fax{+44 (0)117 928-7999}\newline \email
{malhw@bristol.ac.uk}\newline
\URL{http://www.maths.bris.ac.uk/people/faculty/malhw/}

% ------------------------------------------- End Closing Information
% -------------------------------------------

\end{document}

%Junk, some false, some true

Next we continue the argument initiated above in which we are
decomposing a given class $[h]$.  We have shown that when
$\ord_q(N/N_0) = 1$, $[h]$ is the direct sum of eigenspaces for $T_q^N$:
$[h;\kappa_q^{(1)}]$ and $[h;\kappa_q^{(2)}]$ having equal dimensions provided
$\lambda_q^2 \ne 4\psi(q)q^{k-1}$.  Given another prime $q' \mid
N/N_0$, we iterate this process.  Presuming $\lambda_{q'}^2 \ne
4\psi(q')q'^{k-1}$ and $\ord_{q'}(N/N_0) = 1$, each subspace
$[h;\kappa_q^\pm]$ decomposes into two eigenspaces for $T_{q'}^N$ (of equal
dimension) $[h,\kappa_q^\pm,\kappa_{q'}^+]$ and
$[h,\kappa_q^\pm,\kappa_{q'}^-]$.

Since Theorem~\ref{thm:classify-for-T_q} shows there can be no basis of
$[h]$ consisting of simultaneous eigenforms for all Hecke operators
unless $N/N_0$ is square-free, we write $N/N_0 = q_1\cdots q_r,$ with
distinct primes $q_i$.  For each $q_i$ we assume that $\lambda_{q_i}^2
\ne 4 \psi(q_i) q_i^{k-1}$. Then $[h]$ decomposes into $2^r$ eigenspaces
each of equal dimension denoted $[h;\kappa_{q_1}^\pm, \dots,
\kappa_{q_r}^\pm]$.  The elements of these subspaces are simultaneous
eigenforms for all the Hecke operators $T_p^N$ (for all primes $p$),
and so any nonzero element must have nonzero first Fourier
coefficient.  The Fourier coefficients are then characterized by the
eigenvalues, so these spaces are one-dimensional which shows that
$T_q^N$ can be diagonalized on $[h]$.


\begin{prop}\label{prop:dimension-newforms}
  Let $q$ be a prime, $r$ an integer $r\ge 2$, and $\psi$ a character
  with conductor $q^\nu$, $\nu \ge 0$. Then except for finitely many
  values of $k \ge 2$ with $\psi(-1) = (-1)^k$ and finitely many
  values of $q$, we have $\dim S_k^+(q^{\nu+r},\psi) \ge 1$.
\end{prop}

\begin{proof}
  We consider the Hecke algebra generated by all operators $T_p$ with
  $p \ne q$, and recall our shorthand of writing $mS$ for
  $\bigoplus_{i=1}^m S$ in any isomorphism of modules for the Hecke
  algebra.  Thus

\[
S_k(q^{\nu+r},\psi) = \bigoplus_{i = 0}^r\ \bigoplus_{d \mid q^{r-i}}
S_k^+(q^{\nu+i}, \psi) \slash B_d \cong \bigoplus_{i = 0}^r (r - i +
1) S_k^+(q^{\nu+i},\psi).
\]

Using the above, an easy exercise shows that
\begin{equation}
  \label{eqn:dimen-new}
  \dim S_k^+(q^{\nu+r},\psi) = \dim S_k(q^{\nu+r},\psi) - 2\dim
  S_k(q^{\nu+r -1},\psi) + \dim S_k(q^{\nu+r-2},\psi).
\end{equation}


We now use Corollary 8 of \cite{Ross} for a formula for the dimension
of the space of cusp forms with arbitrary character.  In our context,
it yields:

\[
\dim S_k(q^{\nu+r},\psi) = -s_0^{(r)} - s_1^{(r)} + d + m_r - p_r,
\]
where $|s_0^{(r)}| \le 1/2$, $|s_1^{(r)}| \le 2/3$, $m_r =
\frac{k-1}{12}q^{\nu+r}(1+\frac1q)$, $d =
\begin{cases}
  1&\textrm{ if } k=2 \textrm{ and } \psi \textrm{ is trivial, }\\
  0&\textrm{ otherwise,}
\end{cases}
$ and setting $\mu_r = \lfloor{\frac{\nu+r}{2}}\rfloor$,
\begin{equation}\label{eqn:pr}
  p_r = 
  \begin{cases}
    q^r& \textrm{ if } \nu \ge \mu_r + 1,\\
    \frac12(q^{\mu_r} + q^{\mu_r-1})& \textrm{ if } \nu \le \mu_r
    \textrm{ and }
    \nu+r \textrm{ is even,}\\
    q^{\mu_r}& \textrm{ if } \nu \le \mu_r \textrm{ and }
    \nu+r \textrm{ is odd}.\\
  \end{cases}
\end{equation}


It is then clear that
\[
(\frac{k-1}{12})(1+\frac1q) q^{\nu+r} + d - 2 - p_r \le\ \dim
S_k(q^{\nu+r},\psi) \le\ (\frac{k-1}{12})(1+\frac1q) q^{\nu+r} +d + 2
- p_r.
\]

From Equation~\ref{eqn:dimen-new}, we deduce that

\begin{align}
  \dim S_k^+(q^{\nu+r},\psi) &\ge (\frac{k-1}{12})(1+\frac
  1q)q^{\nu+r} -2 - p_r -2\left( (\frac{k-1}{12})(1+\frac
    1q)q^{\nu+r-1}
    +2 - p_{r-1} \right)\notag\\
  &\hspace{2.2in}+ (\frac{k-1}{12})(1+\frac 1q)q^{\nu+r-2} -2 -
  p_{r-2}\notag\\
  &= (\frac{k-1}{12})(1+\frac 1q) q^{\nu+r-2}(q-1)^2 -8 - (p_r
  -2p_{r-1} + p_{r-2})
\end{align}

We need to compute $(p_r -2p_{r-1} + p_{r-2})$, and do so via three
cases; recall $\mu_r = \lfloor{\frac{\nu+r}{2}}\rfloor$.

\textbf{Case 1.}  $\nu \ge \mu_r + 1$.  Then $\nu \ge \mu_r + 1 \ge
\mu_{r-1} + 1 \ge \mu_{r-2} + 1$.  Note that this case cannot occur
unless $\nu \ge 3$.

By Equation~\ref{eqn:pr}, $p_t = q^t$ for $t = r, r-1, r-2$, so $(p_r
-2p_{r-1} + p_{r-2}) = q^{r-2}(q-1)^2$.  From
Equation~\ref{eqn:dimen-new}, we have
\[
\dim S_k^+(q^{\nu+r},\psi) \ge q^{r-2}(q-1)^2 [
(\frac{k-1}{12})(1+\frac 1q) q^{\nu} - 1] -8 \ge 1
\]
for large $q$ or $k$.

\textbf{Case 2.}  $\nu \le \mu_{r-2}$.  Then $\nu \le \mu_{r-2} \le
\mu_{r-1} \le \mu_{r}$.

\noindent When $\nu + r$ is even,
\[
p_r = \frac12\left(q^\frac{\nu+r}{2} + q^{\frac{\nu+r}{2}-1} \right),
\qquad p_{r-1} = q^{\frac{\nu+r}{2} - 1},\qquad p_{r-2} =
\frac12\left(q^{\frac{\nu+r}{2}-1} + q^{\frac{\nu+r}{2}-2} \right),
\]
so $(p_r -2p_{r-1} + p_{r-2}) = \frac12 q^{\frac{\nu+r}{2}-2}(q-1)^2$,
and by Equation~\ref{eqn:dimen-new}, we have
\[
\dim S_k^+(q^{\nu+r},\psi) \ge q^{\frac{\nu+r}{2}-2}(q-1)^2 [
(\frac{k-1}{12})(1+\frac 1q) q^{\frac{\nu+r}{2}} - \frac12] -8 \ge 1
\]
for large $q$ or $k$ as $\nu + r \ge 2$.


\noindent When $\nu + r$ is odd,
\[
p_r=q^{\frac{\nu+r-1}{2}}, \qquad p_{r-1} =
\frac12\left(q^{\frac{\nu+r-1}{2}} + q^{\frac{\nu+r-1}{2}-1}\right),
\qquad p_{r-2} = q^{\frac{\nu+r-1}{2}-1},
\]
so $(p_r -2p_{r-1} + p_{r-2}) = 0$, and by
Equation~\ref{eqn:dimen-new}, we have

\[
\dim S_k^+(q^{\nu+r},\psi) \ge [ (\frac{k-1}{12})(1+\frac 1q)
q^{\nu+r-2}(q-1)^2] -8 \ge 1
\]
for large $q$ or $k$.

\textbf{Case 3.} $\mu_{r-2} < \nu \le \mu_r$.

If $\nu+r$ is even, this translates to $\frac{\nu+r}{2}-1 < \nu \le
\frac{\nu+r}{2}$, that is $\nu = \frac{\nu+r}{2} = r$.  If $\nu+r $ is
odd, this translates to $\frac{\nu+r-1}{2}-1 < \nu \le
\frac{\nu+r-1}{2}$, that is $\nu = \frac{\nu+r-1}{2} = r-1$.

If $\nu+r$ is even, we have

\[
p_r = \frac12(q^\nu + q^{\nu-1}), \qquad p_{r-1} = q^{r-1} =
q^{\nu-1}, \qquad p_{r-2} = q^{r-2} = q^{\nu-2},
\]
so $(p_r -2p_{r-1} + p_{r-2}) = \frac12 q^{\nu-2}(q^2 - 3q+2)$, and by
Equation~\ref{eqn:dimen-new}, we have
\[
\dim S_k^+(q^{\nu+r},\psi) \ge q^{\nu-2}(q-1) [
(\frac{k-1}{12})(1+\frac 1q) q^r (q-1) - \frac12(q-2)] -8 \ge 1
\]
for large $q$ or $k$; recall $\nu = r \ge 2$.

Finally, if $\nu+r$ is odd, we have
\[
p_r = q^{\mu_r} = q^\nu, \qquad p_{r-1} = \frac12 (q^\nu + q^{\nu-1}),
\qquad p_{r-2} = q^{r-2} = q^{\nu-1},
\]
so $(p_r -2p_{r-1} + p_{r-2}) = 0$, and by
Equation~\ref{eqn:dimen-new}, we have
\[
\dim S_k^+(q^{\nu+r},\psi) \ge q^{\nu-2} [ (\frac{k-1}{12})(1+\frac
1q) q^{\nu+r-2} (q-1)^2] -8 \ge 1
\]
for large $q$ or $k$ since $\nu = r-1$ implies $q^{2\nu + r -4}(q-1)^2
\sim q^{3r-4}$ (as $q \to \infty$).
  
\end{proof}

A generalization
