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\markboth{}{Buzzard-Stein (\today)}%{Icosahedral Galois representations}
\title{Mod five approaches to modularity of icosahedral
Galois representations}
\author{Kevin Buzzard and William A. Stein}

% all of our conversations have been moved into an edit'' macro
\newcommand{\edit}{\footnote{#1}}
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\begin{document}
\maketitle
\begin{abstract}
Using a theorem of Buzzard and Taylor, we compute new icosahedral
examples of Artin's conjecture on holomorphicity of the $L$-function
associated to certain Galois representations.  We give in detail one
new example of an icosahedral representation of conductor
${\bf 1376}=2^5\cdot 43$ that satisfies Artin's conjecture.  We also
give\edit{\kevin{[[Still let's see what happens...]]}}  seven
additional examples of conductors
${\bf 2416}=2^4\cdot 151$,
${\bf 3184}=2^4\cdot 199$,
${\bf 3556}=2^2\cdot 7\cdot 127$,
${\bf 3756}=2^2\cdot 3\cdot 313$,
${\bf 4108}=2^2\cdot 13\cdot 79$,
${\bf 4288}=2^6\cdot 67$, and
${\bf 5373}=3^3\cdot 199$.
\end{abstract}

\section*{Introduction}\edit{
\kevin{
behavior --$>$ behaviour,
analyze --$>$ analyse. William---what do we do about incompatibilities
in the spelling in our languages?! I once submitted a paper to MRL and
in the typesetting all the UK spelling was changed to US spelling so
in the proofs I just changed it all back again :-) But this is a joint
paper which we'll probably end up submitting to an American journal
so perhaps US spelling is more appropriate...
}
\william{
Let's decide based on the journal we submit to.  Where should we
submit this paper?  In a few days I can drop it in front of Ken
}}
Consider a continuous irreducible Galois representation
$$\rho:\galq\ra\GL_n(\C)$$
with $n > 1$.
Inspired by his reciprocity law,
Artin conjectured in~\cite{artin:conjecture} that
$L(\rho,s)$ has an analytic continuation to the whole complex plane.
Many of the known cases of this conjecture were obtained by
proving the apparently stronger assertion that~$\rho$ is \defn{automorphic},
in the sense that the $L$-function of~$\rho$ is equal to the $L$-function
of a certain automorphic representation (whose $L$-function is known to have
analytic continuation). In the special case where $n=2$ and $\rho$ is in
addition assumed to be odd, the automorphic representation in question
should be the one associated to a classical weight~$1$
modular eigenform, and in fact there is conjectured to be a
bijection between such~$\rho$ and the set of all weight~$1$
cuspidal newforms, which should
preserve $L$-functions. It is this bijection
that we are concerned with in this paper, so assume for the rest
of the paper that $n=2$ and~$\rho$ is odd.

In this special case, the construction
of~\cite{deligne-serre} shows how to construct a continuous irreducible
odd 2-dimensional representation from a weight~$1$ newform, and the problem
is to go the other way. Say that a representation is \defn{modular}
if it arises in this way.

If the image of~$\rho$ is solvable,
then~$\rho$  is known to be modular
\cite{langlands:basechange, tunnell:artin};
if the image is not solvable, then $\im(\rho)$ in $\PGL_2(\C)$
is isomorphic to the
alternating group~$A_5$, and the modularity of~$\rho$
is, in general, unknown. We call such a 2-dimensional representation an
icosahedral representation''.
The published literature contains only eight examples (up to twist)
of odd icosahedral Galois representations that are known to satisfy Artin's
conjecture: one of conductor $800=2^5\cdot 5^2$
(see \cite{buhler:thesis}), and seven of conductors:
$2083,\, 2^2\cdot 487,\, 2^2\cdot 751,\, 2^2 \cdot 887,\, 2^2\cdot 919,\, 2^5\cdot 73,\,\text{ and } 2^5\cdot 193$
(see \cite{freyetal}).

After the first draft of this paper was written, the
preprint~\cite{bdsbt} appeared, which contains a general theorem that
yields infinitely many (up to twist) modular icosahedral representations.
However, we feel that our work, although much less powerful, is still
of some worth, because it gives an effective computational approach to
proving that certain mod~5 representations are modular, without
computing any spaces of weight~1 forms or using effective versions of
the Chebotarev\edit{\kevin{William, I don't have my TeXbook with me so
don't know how to put the accent on the C.}
\william{Kevin, How should we spell this?? I just checked the titles
of recent papers with this word in the MathSciNet database and found:
Chebotarev (Murty, etc.), Chebotar\"{e}v (Lenstra, however the funniness
of the e is dropped in the AMS review),  Tschebotareff (by a Japanese author),
Tchebotarev (Lang), Tschebotarev (van der Ploeg),
Tchebotareff (a Frenchman, but translated to Chebotarev by the AMS),
\v{C}ebotarev (Moshe).
I say we go with Chebotarev,'' because it is seems to be the
most common, and simply looks nice to me.  Also, it is the
closest to Hendrik's without using special characters.  It
seems strange to me to transliterate into English from Russian and use special
characters from, I guess, French.
}}
density theorem.  Finally, the
main theorem of~\cite{bdsbt} does not apply to any of the examples
considered in the present paper.

In this paper we give eight new examples that were computed
by applying the main theorem  of~\cite{buzzard-taylor} to
the mod~$5$ reduction of~$\rho$.
\edit{In seven of the eight examples, we assume an unknown
congruence bound in order to keep the computations
manageable.\kevin{Do we? We'll have to sort this out---see
the lemma later on in section 1.3}
\william{The congruence bound is now known!}}
We verify modularity mod~$5$ on a case-by-case basis. Later we shall
explain our approach more carefully, but let us briefly summarise it here.
By~\cite{buzzard-taylor},
the problem is to show that the mod~5 reduction of~$\rho$ is modular.
\edit{\william{By the way, as I'm sure you've spotted, I have added this
para.}}
We do this by finding a candidate mod~5 modular form at weight~5
and then, using the table of icosahedral extensions of $\Q$ in~\cite{freyetal}
and what we know about the 5-adic representation attached to our candidate
form, we deduce that the mod~5 representation attached to our candidate
form must be the reduction of~$\rho$. In particular, this paper gives
a computational methods for checking the modularity of certain mod~5
representations whose conductors are not too large. We now give
more details.

In each of our examples it is easy to compute a few Hecke operators
and be morally convinced that a mod~$5$ representation should be modular;
it is far more difficult to prove this.
Effective variants of the Chebotarev density theorem require
that we check vastly more traces of Frobenius than is practical.
Instead we use the Local Langlands theorem for $\GL_2$, the
theory of companion forms, the main theorem of \cite{buzzard-taylor},
and Table~2 of~\cite{freyetal}, to provide proofs of modularity
in certain cases.

More precisely, let~$K$ be an icosahedral extension of~$\Q$ that is not
totally real, and consider a minimal lift $\rho:\GQ\ra \GL_2(\C)$
of
$$\GQ\ra \Gal(K/\Q)\ncisom{}A_5\subset \PGL_2(\C);$$
the lift is minimal in the sense  that its conductor is minimal.
Assume that~$5$ does not ramify in~$K$, and that
a Frobenius element at~$5$ in $\Gal(K/\Q)$ does not have order~$1$ or~$5$.
Inspired by the possibility that~$\rho$ is modular,
we search for a mod~$5$ modular form of weight~$5$ whose existence would
be forced by modularity of~$\rho$.  Indeed, we find
a candidate mod~$5$ form~$f$, and then prove that the fixed field
of the kernel of the projective mod~$5$ representation
associated to a certain twist of~$f$  must be~$K$.
This proves that the mod~$5$ reduction of a twist
of~$\rho$ is modular, and the main theorem
of \cite{buzzard-taylor} then implies
that~$\rho$ is modular.
We carried out this program for icosahedral representations
of the following conductors:
${\bf 1376} = 2^5\cdot 43$,
${\bf 2416}=2^4\cdot 151$,
${\bf 3184}=2^4\cdot 199$,
${\bf 3556}=2^2\cdot 7\cdot 127$,
${\bf 3756}=2^2\cdot 3\cdot 313$,
${\bf 4108}=2^2\cdot 13\cdot 79$,
${\bf 4288}=2^6\cdot 67$, and
${\bf 5373}=3^3\cdot 199$.

We choose an icosahedral field~$K$ and representation~$\rho$,
then proceed as follows:
\vspace{.5ex}
\begin{numlist}
\item Search for a form~$f \in S_5(N,\eps;\Fbar_5)$ whose
associated mod~$5$ Galois representation looks like
it is the mod~$5$ reduction of~$\rho$.
\item Twist~$f$ to obtain an eigenform~$g$ with coefficients in~$\F_5$.
\item Prove that~$\rho_g$ is unramified at~$5$ by finding a companion form.
\item Prove that the image of $\proj\rho_g$ is~$A_5$ by ruling out all
other possibilities.
\item Prove that the fixed field~$L$ of $\proj\rho_g$ has
root field of discriminant at most $2083^2$,
so~$L$ is in Table~2 of~\cite{freyetal}; deduce that~$L=K$.
\item Apply the main theorem of~\cite{buzzard-taylor}
to a lift of $\rhobar=\rho_g$
to conclude that~$\rho$ is modular.
\end{numlist}

\section{Modularity of an icosahedral representation of
conductor~$1376=2^5\cdot 43$}\label{sec:1376}
In this section we prove the following theorem.
\begin{theorem}\label{thm:1376}
The icosahedral representations whose corresponding
icosahedral extension
is the splitting field of $x^5 + 2x^4+6x^3+8x^2+10x+8$
are modular.
\end{theorem}

Let~$K$ be the splitting field of $h=x^5 + 2x^4+6x^3+8x^2+10x+8$.
The Galois group of~$K$ is~$A_5$, so we obtain a homomorphism
$G_\Q\ra{}A_5\subset \PGL_2(\C)$;
let $\rho:G_\Q\ra\GL_2(\C)$ be a minimal lift, minimal
in the sense that the Artin conductor of~$\rho$ is minimal.
By Table~$A_5$ of~\cite{buhler:thesis}, the conductor of~$\rho$
is $N=1376=2^5\cdot 43$.  Since
$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5}$,
and ${\rm disc}(h)$ is coprime to~$5$,
any Frobenius element at~$5$ in $\Gal(K/\Q)$ has order~$2$.

We use the notation of Tables 3.1 and 3.2 of~\cite[pg. 46]{buhler:thesis};
from Table 3.2 we see that the type of~$\rho$ at~$2$
is~$17$ and the type at~$43$ is~$2$.
The mod~$N$ Dirichlet character~$\eps=\det(\rho)$
factors as~$\eps=\eps_2\cdot \eps_{43}$ where~$\eps_2$ is
a character mod~$2^5$ and~$\eps_{43}$ is a character mod~$43$.
Corresponding to each type in Buhler's table, there is a character,
and fortunately Buhler's level $800$ example also was of type~$17$ at~$2$
(see the first line of~\cite[Table~3.2]{buhler:thesis}).
By~\cite[pg.~80]{buhler:thesis} $\eps_2$ is the unique
character\edit{
\kevin{William: we should
check that Buhler's example turned out to be a weight~$1$ form
whose character at~$2$ was $\eps_2$.}
\william{\em Yep, by Table 3.2.}
\kevin{I didn't
mean this---but after reading Buhler I found that in fact the answer
to the question I meant to ask was Yep, by the beginning of chapter 6''
(and this is probably the p80 you mention above). So we can kill
this footnote. By the way, how long does
it take you to check the fact'' near the beginning of Chapter 6, that
Buhler talks philosophically about? (not that it matters)}
\william{
On a PII 350, it takes a total of 258 seconds total
for my program to compute a basis of newforms
for each space of level dividing $800$, and from this to then
compute a Gauss-reduced basis of $q$-expansions, up to precision $360$
for the space $S_2(800;\C)$. I didn't check how long it
takes to verify that the $h_{i,d}$ lie in this space, but
this should be completely trivial because my basis is
Gauss-reduced.}}
of conductor~$4$ and order~$2$.
A local computation shows that the image
of~$\eps_{43}$ has order~$3$.

If~$\rho$ is modular,  then there is a weight~$1$
newform $f_?\in S_1(N,\eps;\Qbar)$ that gives rise to~$\rho$.
Suppose for the moment that~$\rho$ is modular, so that~$f_?$ exists.
Choose a prime of~$\overline{\Z}$ lying
over~$5$, and denote by~$\fbar_?$ the reduction
of $f_?$ modulo this prime. The Eisenstein series
$E_4\in M_4(1;\F_5)$  is congruent to~$1$ modulo~$5$, so
$E_4\cdot{}\fbar_?\in S_5(N,\eps;\Fbar_5)$ has the same $q$-expansion
as $\fbar_?$.  Using a computer, we can search for a
form $f\in S_5(N,\eps,\Fbar_5)$ that has the same
$q$-expansion as the conjectural form $E_4\cdot{}\fbar_?$.

Instead of multiplying $\fbar_?$ by~$E_4$, we could have multiplied
it by an  Eisenstein series of weight~$1$, level~$5$, and character $\eps'$.
We used $E_4$ because the dimension of $S_5(N,\eps;\Fbar_5)$
is~$696$ whereas the dimension of the relevant space
$S_2(5\cdot 1376, \eps_{43})$ of weight~$2$ cusp forms is~$1040$.

\subsection{Searching for the newform~$f$}
Using modular symbols (see Section~\ref{sec:modsym}) we
compute (at least up to semi-simplification) the space
$S_5(1376,\eps;\F_{25})$. Note that there is injective map
from the image of~$\eps$ into $\F_{25}^*$.  By computing
the kernels of various Hecke operators on this space, we find~$f$.
In the following computations, we represent nonzero elements of~$\F_{25}$
as powers of a generator~$\alp$ of~$\F_{25}^*$, which satisfies
$$\alp^2 + 4\alp + 2=0.$$
Our character $\eps_{43}$ was represented
as the map sending $(1,3)\in(\Z/2^5\Z)^*\cross(\Z/43\Z)^*$ to
$2\alp+1$. Note that~3 is a primitive root mod~43, and that $2\alp+1$
has order~3.

If the least common multiple of the degrees of the factors of
the polynomial~$h$ modulo an
unramified prime~$p$ is~$2$, then $\Frob_p\in\Gal(K/\Q)$
has order~$2$.  The minimal polynomial of $\rho(\Frob_p)\in\GL_2(\C)$
is then $x^2-1$, so $\rho(\Frob_p)$ has trace~$0$.
The first three primes $p \nmid 5\cdot 1376$ such
that $\rho(\Frob_p)$ has  order~$2$ are $p= 19,31,97$.
We computed the mod~$5$ reduction $\sS_5(1376,\eps;\F_{25})^{+}$
of the $\Z_5[\zeta_3]$-lattice of
modular symbols of level~$1376$ and
character~$\tilde{\eps}$ where
complex conjugation acts as $+1$.
Here~$\tilde{\eps}$ denotes the Teichm\"uller lift of~$\eps$.

Let~$V$ be the intersection of the kernels of $T_{19}$, $T_{31}$, and
$T_{97}$ inside of the space $\sS_5(1376,\eps;\F_{25})^{+}$ of mod~5
modular symbols.
The space~$V$ is $8$-dimensional,
and no doubt all the eigenforms in this space give rise to~$\rho$ or one
of its twists. One of the eigenvalues of~$T_3$ on this space
is~$\alp^{16}$, and the kernel $V_1$ of $T_3-\alp^{16}$ is $2$-dimensional
over $\F_{25}$. The Hecke operator~$T_5$ acted as a diagonalisable matrix on
$V_1$, with eigenvalues $\alp^{10}$ and $\alp^{22}$, so the corresponding
two systems of eigenvalues must correspond to mod~$5$ modular eigenforms,
and furthermore we must have found all mod~$5$ modular eigenforms
of this level, weight and character,
such that $a_{19}=a_{31}=0$ and $a_3=\alp^{16}$.

\begin{remark}
The careful reader might wonder how we know that the
systems of mod~$5$ eigenvalues really do correspond to mod~$5$ modular
forms, and not to perhaps some strange mod~$5$ torsion in the space of
modular symbols. However, we eliminated this possibility by
computing the dimension of the full space of mod~$5$ modular
symbols where complex conjugation acts as~$+1$, and checking that it
equals $696$, the dimension of $S_5(1376,\tilde{\eps},\C)$, which we
computed using the formula in \cite{cohen-oesterle}.
\edit{
\kevin{I still have never seen this
paper, so can't give a precise reference. Did you check it all in magma?
My pari port gives 696 :-) :-) :-)}
\william{I added the biblio reference.
I have carefully re-checked all of these numbers in Magma, using
Cohen-Oesterle.  See the end of this tex file for the Magma code.}
}
\end{remark}

Let~$f$ be the eigenform in~$V_1$ that satisfies
$a_5=\alp^{22}$; the $q$-expansion of~$f$ begins
$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 + \alp^{14}q^9 + 4q^{11}+\cdots.$$
Further eigenvalues are given in Table~\ref{table:1376} below.
The primes~$p$ in the table such that~$a_p=0$ are
exactly those
predicted by considering the splitting behavior of~$h$.
This is strong evidence that~$\rho$ is modular,
and also that our modular symbols algorithm have been correctly
implemented.
\newpage

\begin{center}
\label{table:1376}
{\bf Table~\ref{table:1376}}
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alpha^{16}\\ 5&\alpha^{22}\\ 7&\alpha^{14}\\ 11&4\\ 13&\alpha^{14}\\ 17&\alpha^{14}\\ 19&0\\ 23&\alpha^{16}\\ 29&\alpha^{8}\\ 31&0\\ 37&\alpha^{10}\\ 41&1\\ 43&\alpha^{10}\\ 47&1\\ 53&\alpha^{22}\\ 59&4\\ 61&\alpha^{14}\\ 67&\alpha^{4}\\ 71&\alpha^{20}\\ 73&\alpha^{2}\\ \hline\end{array} \begin{array}{|rl|}\hline 79&\alpha^{20}\\ 83&\alpha^{4}\\ 89&\alpha^{10}\\ 97&0\\ 101&\alpha^{8}\\ 103&\alpha^{14}\\ 107&0\\ 109&\alpha^{10}\\ 113&2\\ 127&0\\ 131&2\\ 137&0\\ 139&\alpha^{22}\\ 149&\alpha^{4}\\ 151&1\\ 157&\alpha^{14}\\ 163&0\\ 167&\alpha^{22}\\ 173&4\\ 179&\alpha^{2}\\ 181&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 191&\alpha^{10}\\ 193&4\\ 197&0\\ 199&3\\ 211&0\\ 223&0\\ 227&\alpha^{10}\\ 229&0\\ 233&\alpha^{14}\\ 239&0\\ 241&\alpha^{2}\\ 251&\alpha^{2}\\ 257&3\\ 263&\alpha^{16}\\ 269&2\\ 271&\alpha^{8}\\ 277&0\\ 281&\alpha^{16}\\ 283&0\\ 293&3\\ 307&\alpha^{4}\\ \hline\end{array} \begin{array}{|rl|}\hline 311&\alpha^{22}\\ 313&0\\ 317&0\\ 331&\alpha^{14}\\ 337&0\\ 347&\alpha^{16}\\ 349&\alpha^{4}\\ 353&0\\ 359&0\\ 367&\alpha^{22}\\ 373&0\\ 379&3\\ 383&3\\ 389&1\\ 397&\alpha^{16}\\ 401&0\\ 409&2\\ 419&3\\ 421&\alpha^{20}\\ 431&4\\ 433&\alpha^{4}\\ \hline\end{array} \begin{array}{|rl|}\hline 439&\alpha^{20}\\ 443&0\\ 449&0\\ 457&0\\ 461&0\\ 463&\alpha^{10}\\ 467&0\\ 479&0\\ 487&\alpha^{8}\\ 491&\alpha^{2}\\ 499&\alpha^{20}\\ 503&\alpha^{2}\\ 509&\alpha^{8}\\ 521&\alpha^{10}\\ 523&\alpha^{14}\\ 541&\alpha^{20}\\ 547&\alpha^{22}\\ 557&3\\ 563&1\\ 569&\alpha^{16}\\ 571&\alpha^{22}\\ \hline\end{array} \begin{array}{|rl|}\hline 577&\alpha^{14}\\ 587&\alpha^{20}\\ 593&0\\ 599&\alpha^{22}\\ 601&0\\ 607&\alpha^{16}\\ 613&2\\ 617&0\\ 619&\alpha^{20}\\ 631&\alpha^{20}\\ 641&4\\ 643&1\\ 647&4\\ 653&1\\ 659&\alpha^{14}\\ 661&2\\ 673&\alpha^{8}\\ 677&4\\ 683&0\\ 691&\alpha^{16}\\ 701&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 709&4\\ 719&\alpha^{4}\\ 727&4\\ 733&0\\ 739&2\\ 743&\alpha^{22}\\ 751&\alpha^{4}\\ 757&\alpha^{2}\\ 761&\alpha^{2}\\ 769&0\\ 773&0\\ 787&\alpha^{20}\\ 797&\alpha^{16}\\ 809&3\\ 811&\alpha^{16}\\ 821&2\\ 823&\alpha^{10}\\ 827&\alpha^{10}\\ 829&\alpha^{22}\\ 839&0\\ 853&\alpha^{14}\\ \hline\end{array} \comment{ \begin{array}{|rl|}\hline 857&0\\ 859&0\\ 863&\alpha^{4}\\ 877&\alpha^{8}\\ 881&1\\ 883&0\\ 887&2\\ 907&0\\ 911&1\\ 919&1\\ 929&0\\ 937&\alpha^{2}\\ 941&\alpha^{4}\\ 947&2\\ 953&\alpha^{8}\\ 967&4\\ 971&\alpha^{2}\\ 977&0\\ 983&0\\ 991&3\\ 997&3\\ \hline\end{array}}$$
\end{center}

\subsection{Twisting into $\GL(2,\F_5)$}
Although there is a representation
$\rho_f:\GQ\ra\GL(2,\F_{25})$ attached to $f$,
it is difficult to say anything about its image without further
work. We use a trick to show that the image of $\rho_f$ is small.
Firstly, for a character~$\chi:\GQ\to\Fbar_5$, let~$\tilde\chi$
denote its Teichm\"uller lift to~$\Qbar_5$. By a result of Carayol,
there is a characteristic 0 eigenform
$\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$ lifting $f$.
The twist $\tilde{g}=\tilde{f} \tensor \tilde{\eps}_{43}$ is, by
\cite[Prop. 3.64]{shimura:intro}, an eigenform in
$S_5(43N, \tilde{\eps}_2; \Qbar_5)$, and its reduction is
a form $g\in S_5(43N,\eps_2,\F_{25})$.
The eigenvalues $a_p(g) = a_p(f) \eps_{43}(p)$, for
$p\nmid 5N$, are given in Table~\ref{table:1376twist}.
\begin{center}\label{table:1376twist}
{\bf Table~\ref{table:1376twist}}
$$\begin{array}{|rl|}\hline 2&*\\%0\\ 3&1\\ 5&*\\%3\\ 7&2\\ 11&4\\ 13&2\\ 17&2\\ 19&0\\ 23&1\\ 29&1\\ 31&0\\ 37&3\\ 41&1\\ 43&*\\%0\\ 47&1\\ 53&2\\ \hline\end{array} \begin{array}{|rl|}\hline 59&4\\ 61&2\\ 67&4\\ 71&4\\ 73&3\\ 79&4\\ 83&4\\ 89&3\\ 97&0\\ 101&1\\ 103&2\\ 107&0\\ 109&3\\ 113&2\\ 127&0\\ 131&2\\ \hline\end{array} \begin{array}{|rl|}\hline 137&0\\ 139&2\\ 149&4\\ 151&1\\ 157&2\\ 163&0\\ 167&2\\ 173&4\\ 179&3\\ 181&2\\ 191&3\\ 193&4\\ 197&0\\ 199&3\\ 211&0\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&3\\ 229&0\\ 233&2\\ 239&0\\ 241&3\\ 251&3\\ 257&3\\ 263&1\\ 269&2\\ 271&1\\ 277&0\\ 281&1\\ 283&0\\ 293&3\\ 307&4\\ 311&2\\ \hline\end{array} \begin{array}{|rl|}\hline 313&0\\ 317&0\\ 331&2\\ 337&0\\ 347&1\\ 349&4\\ 353&0\\ 359&0\\ 367&2\\ 373&0\\ 379&3\\ 383&3\\ 389&1\\ 397&1\\ 401&0\\ 409&2\\ \hline\end{array} \begin{array}{|rl|}\hline 419&3\\ 421&4\\ 431&4\\ 433&4\\ 439&4\\ 443&0\\ 449&0\\ 457&0\\ 461&0\\ 463&3\\ 467&0\\ 479&0\\ 487&1\\ 491&3\\ 499&4\\ 503&3\\ \hline\end{array} \begin{array}{|rl|}\hline 509&1\\ 521&3\\ 523&2\\ 541&4\\ 547&2\\ 557&3\\ 563&1\\ 569&1\\ 571&2\\ 577&2\\ 587&4\\ 593&0\\ 599&2\\ 601&0\\ 607&1\\ 613&2\\ \hline\end{array} \begin{array}{|rl|}\hline 617&0\\ 619&4\\ 631&4\\ 641&4\\ 643&1\\ 647&4\\ 653&1\\ 659&2\\ 661&2\\ 673&1\\ 677&4\\ 683&0\\ 691&1\\ 701&2\\ 709&4\\ 719&4\\ \hline\end{array} \comment{ \begin{array}{|rl|}\hline 727&4\\ 733&0\\ 739&2\\ 743&2\\ 751&4\\ 757&3\\ 761&3\\ 769&0\\ 773&0\\ 787&4\\ 797&1\\ 809&3\\ 811&1\\ 821&2\\ 823&3\\ 827&3\\ \hline\end{array} \begin{array}{|rl|}\hline 829&2\\ 839&0\\ 853&2\\ 857&0\\ 859&0\\ 863&4\\ 877&1\\ 881&1\\ 883&0\\ 887&2\\ 907&0\\ 911&1\\ 919&1\\ 929&0\\ 937&3\\ 941&4\\ \hline\end{array} \begin{array}{|rl|}\hline 947&2\\ 953&1\\ 967&4\\ 971&3\\ 977&0\\ 983&0\\ 991&3\\ 997&3\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ \hline\end{array}}$$
\end{center}

\begin{proposition}\label{prop:1376-g}
Let $g=f\tensor \eps_{43}$.  Then $a_p(g)\in \F_5$
for all~$p\nmid \ell N$.
\end{proposition}
\begin{proof}
Consider an eigenform $\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$
lifting~$f$ as above.
Associated to~$\tilde{f}$ there is an automorphic
representation~$\pi=\tensor_v'\pi_v$ of $\GL(2,\bA)$, where~$\bA$
is the ad\{e}le ring of~$\Q$.
Because $43\mid\mid N$, and~$43$ divides the conductor
of $\eps$, we see that the local component $\pi_{43}$ of $\pi$ at
$43$ must be ramified principal series. By Carayol's theorem,
$\rho_{\tilde{f}}|_{D_{43}} \sim \abcd{\Psi_1}{0}{0}{\Psi_2}$
with, without loss of generality,~$\Psi_2$ unramified.  We have
$(\Psi_1\cdot \Psi_2)|_{I_{43}}=\tilde{\eps}|_{I_{43}}=\tilde{\eps}_{43}$,
therefore, $\rho_{\tilde{f}}|_{I_{43}} \sim \abcd{\tilde{\eps}_{43}}{0}{0}{1}$.

Now twist~$\tilde{f}$ by $\tilde{\eps}_{43}^{-1}$; we find that
$\rho_{\tilde{f}\tensor\tilde{\eps}_{43}^{-1}}|_{I_{43}} \sim \abcd{1}{0}{0}{\tilde{\eps}^{-1}_{43}}$.
In particular, there is an
eigenform~$\tilde{f}'\in S_5(N,\tilde{\eps}_2\tilde{\eps}^{-1}_{43},\Qbar_5)$
whose associated Galois representation is the twist by $\tilde{\eps}^{-1}_{43}$
of that of $\tilde{f}$ (recall that $N=1376$ and so~$43$ divides~$N$
exactly once). Let~$f'$ denote the mod~$5$ reduction of~$\tilde{f}'$. Then
one checks easily that $f'\in S_5(N,\eps_2\eps^{-1}_{43},\F_{25})=S_5(N,\eps^5,\F_{25})$.

For all primes $p\nmid5N$ we have $a_p(f')=\eps_{43}(p)^{-1}a_p(f)$.
In particular, we have $a_p(f')=0$ for
$p=19,31$.
Also, $\eps_{43}(3)=\alp^8$ and $\eps_{43}(5) =\alp^8$, so
$$a_3(f')=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
$$a_5(f')=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5.$$
Now if $\sigma$ is the non-trivial automorphism of $\F_{25}$,
then $\sigma(f')$ and $f$ both lie in
$S_5(1376,\eps;\F_{25})$ and have same~$a_p$ for
$p=3,5,19,31$, so they are equal because we found~$f$
by computing the unique eigenform with given~$a_p$ for $p=3,5,19,31$.
So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
Thus for all $p\nmid 5N$, we see that
$a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10} = a_p(f) \eps_{43} = a_p(g)$.
\end{proof}

\subsection{Proof that~$\rho_g$ is unramified at~$5$}
We begin with a generalisation of~\cite{sturm:cong}.\edit{\kevin{Are you happy
to put this lemma here? I expect you're just relieved to see the
lemma at all :-)}
\william{It fits naturally here.  And {\em yes}, I am very relieved
to see the lemma at all!}}
Let $M>4$ be an integer, and let $h=\sum_{n\geq1}c_nq^n$ be a
normalised cuspidal eigenform
of some weight~$k\geq1$, level~$M$ and character~$\chi$, defined over some
field of characteristic not dividing~$M$.
Let~$I$ be the set of primes~$p$
dividing~$M$ such that~$h$ is $p$-new, and
such that either~$p$ does not divide $M/\cond(\chi)$, or~$p$
divides~$M$ exactly once. Let~$C$ denote
the orbit of the cusp~$\infty$ in $X_1(M)$
under the action of the group generated by $w_p$ for $p\in I$, and
the diamond operators $\langle d\rangle_M$. By Corollary~4.6.18
of \cite{miyake}\edit{\kevin{don't know the publisher/year and don't have
it with me, sorry. Working offline and at home is hard!}\william{Fixed.}},
we see that the first~$t$ terms of the $q$-expansion
of~$h$ at any cusp in~$C$ are determined by~$M$,~$k$, $\chi$, $c_p$
for~$p$ in~$I$, and $c_n$ for $1\leq n\leq t$. The size of~$C$ is
$\phi(M).2^{|I|-1}.$ The usefulness of this result is that
if $h_1$ and $h_2$ are two normalised eigenforms of the same level,
weight and character as above, both new at all primes in~$I$,
and the coefficients of $q^n$ in the
$q$-expansions of $h_1$ and $h_2$ agree for $n\in I$ and $n\leq t$,
then $h_1-h_2$ has a zero of order at least $t+1$ at all cusps in $C$,
and in particular if $\phi(M).2^{|I|-1}(t+1)>k/12[\SL_2(\Z):\Gamma_1(M)]$
then $h_1=h_2$. Using the fact that $[\Gamma_0(M):\Gamma_1(M)]=\phi(M)/2$,
we deduce
\begin{lemma}\label{lem:bound}
Let $h_1$ and $h_2$ be two normalised eigenforms as above.
If the coefficients of $q^n$ in the $q$-expansions of $h_1$ and $h_2$ agree
for all primes in $I$ and for all
$n\leq\frac{k}{12}[\SL_2(\Z):\Gamma_0(M)]/2^{|I|}$ then $h_1=h_2$.
\end{lemma}\edit{\kevin{\bf William: I have been a
bit vague here---but tell me whether this
result (a) looks right and (b) looks useful, and then we can go from
there. I could strengthen it a little in the case when $p$ is in $I$
and $p^2$ divides $M$ but I don't know whether this is necessary. Note that
it gives exactly what Sturm says in the squarefree case. Note also that
it's easy to check that e.g. our first form is new at 43 because the
char is non-trivial! I could give details of Miyake's argument (which
is in fact over $\C$ but which I believe I can generalise to an arbitrary
field if necessary) if you want.}
\william{
I think that both (a) and (b) are true.
I would like to see details of Miyake's argument over an arbitrary field.
Even if we don't put them in the paper, I need to understand them
because surely somebody will grill me about this...  Anyway, I can
also use this result to improve my MAGMA software.

Am I mistaken in believing that multiplicity one'' implies that
the forms we consider are all new?
}}

We now go back to the explicit situation we are concerned with.
Although~$g$ is an eigenform of level $59168=2^5\cdot 43^2$,
we can still consider the corresponding representation
$\rho_g :\GQ\ra \GL(2,\F_5)$, and then directly analyze
its ramification.
\begin{proposition}
The representation~$\rho_g$ is unramified at~$5$.
\end{proposition}
\begin{proof}
Continuing the modular symbols computations as above,
we find that~$V_1$ is spanned by the two eigenforms
\begin{align*}
f\,\,&=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
+ \alp^{14}q^9 + 4q^{11}+\cdots\\
f_1&=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7
+ \alp^{14}q^9 + 4q^{11}+\cdots.
\end{align*}
For $p\neq 5$ and $p\leq 997$, we have $a_p(f_1)=a_p(f)$.
To check that $a_p(f) = a_p(f_1)$ for all $p\neq 5$,
it suffices to show that the difference~$f-f_1$ has
$q$-expansion involving only powers of~$q^5$;
for this we use the $\theta$-operator
$q\frac{d}{dq}:S_5(1376,\eps,\F_{25})\ra S_{11}(1376,\eps;\F_{25})$.
Since~$\theta$ sends normalized eigenforms to normalized eigenforms,
it suffices to check that the subspace of
$S_{11}(1376,\eps;\F_{25})$ generated by~$\theta(f)$
and~$\theta(f_1)$ has dimension~$1$.
Lemma~\ref{lem:bound} implies that it suffices to verify that the
coefficients $a_p(\theta(f))$ and $a_p(\theta(f_1))$ are equal for all
$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(1376)]\cdot \frac{1}{4} = 484.$$
The eigenform~$f$ must be new because we computed it by finding
the intersections of the kernels of Hecke operators $T_p$ with
$p\nmid 1376$; if~$f$ were an oldform then the intersection of the
kernels of these Hecke operators
would necessarily have dimension greater than~$1$.
Because it takes less than a second
to compute each $a_p(\theta(f))$, we were easily able to verify that the
space generated by $\theta(f)$ and $\theta(f_1)$ has dimension~$1$.

\begin{remark}
It is possible to avoid appealing to Lemma~\ref{lem:bound} by using
one of the following two alternative methods:
\begin{enumerate}
\item Define~$\theta$ directly on modular symbols and compute it.
\item Compute the intersection
$$\bigcap_{p\geq 2} \ker(T_p - pa_p(f)) \subset S_{11}(1376,\eps;\F_{25}).$$
Since~$\theta(f)$ and~$\theta(f_1)$ both lie
in the intersection, the moment the dimension
of a partial intersection is~$1$, it follows
that $\theta(f-f_1)=0$.
\end{enumerate}
We successfully carried out both alternatives.
For the first, we showed that~$\theta$ on modular symbols is
induced by multiplication by
$X^5Y - Y^5X$.
For the second, we find that after intersecting
kernels for $p\leq 11$, the dimension is already~$1$.
The first of these two methods took much less
time than the second.
\end{remark}
\edit{\kevin{Have I done this? That is, is my result strong enough?
The point is that for arbitrary forms on $\Gamma_1(N)$ you have to
check a long way but for 2 forms which are eigenforms for the diamond
ops with the same character you can check
they're the same just by using the $\Gamma_0(N)$ bounds, because if the
q-exps agree at one cusp then they agree at lots of cusps
($\phi(N)/2$ cusps, to be precise, the orbit of infinity under the
diamond operators) Is this all you needed? It's kind
of like why you don't need to deal with a huge huge space when working
with modular symbols for $\Gamma_1(N)$ with a character. Anyway, if my
result is strong enough then perhaps this section needs mild rewriting.}
}

Next we use that $\theta(f-f_1)=0$ to show that $\rho_g$ is unramified,
thus finishing the proof of the proposition.
Since~$f$ is ordinary, Deligne's theorem (see~\cite[\S12]{gross:tameness})
implies that
$$\rho_f|_{D_5}\sim \mtwo{\alp}{*}{0}{\beta}\qquad\text{over \Fbar_5}$$
with both~$\alp, \beta$ unramified,
$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$, and
$\beta(\Frob_5)=\alp^{22}$.
Since $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
with
$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
$\beta'(\Frob_5)=\alp^{10}$;
in particular, $\alp'=\beta$.
Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so
$\rho_f|_{D_5}\sim\alp\oplus\beta$ and hence there is a choice
of basis so that $*=0$.

\end{proof}

\subsection{The image of $\proj \rho_g$}
\begin{proposition}
The image of $\proj \rho_g$ is $A_5$.
\end{proposition}
\begin{proof}
The image~$H$ of $\proj \rho_g$ in $\PGL_2(\F_5)$ is easily checked to
lie in $\PSL_2(\F_5)\cong A_5$ because of what we know about the determinant
of $\rho_g$. Hence $H$ is a subgroup of $A_5$ that
contains an element of order~$2$ (complex conjugation)
and an element of order~$3$ (for example, $\rho_g(\Frob_7)$ has
characteristic polynomial $x^2-2x-1$).
This proves that~$H$ is isomorphic to either~$S_3$,~$A_4$, or~$A_5$.
Let $L$ be the number field cut out by~$H$.
If~$L$ were an $S_3$ extension, then there would be a
quadratic extension contained in it which is unramified
outside $2\cdot 5\cdot 43$; it is furthermore unramified at~$5$ by
the previous section and unramified at $43$ because $I_{43}$
has order~$3$.  Thus it is one of the three quadratic
fields unramified outside~$2$. In particular, the trace of $\Frob_p$
would be zero for all primes in a certain congruence class
modulo~8.
However, there are primes~$p$ congruent to $3$, $5$, and $7$
mod $8$ such that $a_p(g)\neq 0$, e.g., $3$, $7$, and $13$.

If $H$ were isomorphic to $A_4$,
then let~$M$ denote the cyclic extension of degree~3 over~$\Q$ contained
in~$L$. Now~$M$ is unramfied at~2 and~5, and hence is the subfield
of $\Q(\zeta_{43})$ of degree~3.
Choose $p\nmid 1376\cdot 5$ that is inert in~$M$.  The
order of $\rho_g(\Frob_p)$ in $\GL_2(\F_5)$ must be divisible
by~$3$.  However, a quick check using
Table~\ref{table:1376twist} shows that this is
usually not case, even for $p=3$.
\end{proof}

\subsection{Bounding the ramification at~$2$ and~$43$}
Let~$L$ be the fixed field of $\ker(\proj(\rho_g))$. We have just
shown that $\Gal(L/\Q)$ is isomorphic to $A_5$.
By a root field for~$L$, we mean
a non-Galois extension of $\Q$ of degree~5 whose Galois closure is~$L$.
\begin{proposition}
The discriminant of a root
field for~$L$ divides $(43\cdot 8)^2=344^2$, and
in particular,~$L$ must be mentioned in Table~1
of \cite[pg 122]{freyetal}.\end{proposition}
\begin{proof}
The analysis of the local behavior of~$\rho_f$ at~$43$ given in
Proposition~\ref{prop:1376-g}
shows that the inertia group at~$43$ in $\Gal(L/\Q)$ has order~$3$. Using
Table~3.1 of~\cite{buhler:thesis}, we see that if
$\Gal(L/\Q)\isom A_5$
then it must be type~$2$'' at 43, and hence the discriminant of a root
field of~$L$, that is, of a non-Galois extension of~$\Q$ of degree~$5$
whose Galois closure is~$L$, must be $43^2$ at~$43$.

At~$2$ the behavior of~$\rho$ is more subtle and we shall not analyze
it fully. But we can say that, because~$\rho$ has arisen from
a form of level $1376=2^5.43$, we must be either of type~$5$
or one of types~$14$--$17$. In particular, the discriminant at~$2$ of a root
field for~$L$ will be at most~$2^6$.

Finally,~$L$ is unramified at all other primes, because~$\rho$ is.
Hence the discriminant of a root field for~$L$, assuming that
$\Gal(L/\Q)\cong A_5$, divides $(43.8)^2=344^2$.
\end{proof}

We know that~$L$ is an icosahedral extension of~$\Q$ with
discriminant dividing $43^2\cdot 2^6$.  Table~1 of \cite[pg 122]{freyetal}
contains all icosahedral extensions, such that the discriminant
of a root field is bounded by $2083^2$.  The table
must contain~$L$; there is only one icosahedral extension with
discriminant dividing $43^2\cdot 2^6$, so $L=K$.

\subsection{Obtaining a classical weight one form}
We have shown that a twist of the icosahedral
representation $\rho:\GQ\ra\GL(2,\C)$,
obtained by lifting $\GQ\ra \Gal(K/\Q)\ncisom A_5$,
has a mod~$5$ reduction $\rho_g:\GQ\ra \GL_2(\F_5)$ that
is modular.  Since~$\rho$ ramifies at only finitely many primes,
and~$\rho$ can be chosen to be unramified at~$5$ with distinct eigenvalues,
\cite{buzzard-taylor} implies that~$\rho$ arises from
a classical weight~$1$ newform.\edit{\kevin{You write say something
about choice of iso $\overline{\Q}_5=\C$'' but I think that we can just
say nothing.}\william{I'm happy with saying nothing, as the reader
will know what needs to be said upon looking at [BT].}}

\section{More examples}
\subsection{Conductor~$2416=2^4\cdot 151$}
\label{sec:2416}
Consider the icosahedral extension~$K$ defined by $h=x^5-2x^3+2x^2+5x+6$.
A Frobenius element at~$5$ has order~$2$.
The first three primes~$p$ such that $a_p=0$ are
$53$, $97$, and $127$.
The type at~$2$ is~16, and the type at~$151$ is~2.
The order of $\eps_{151}$ is~$3$, and a local computation would tell us
what $\eps_2$ was but we preferred to guess---if we guessed wrong then
we would almost certainly find no forms and we could just guess again.
We guessed that $\eps_2$ was the character of conductor~$4$, and this
turned out to be correct.

Using \cite{cohen-oesterle}\edit{\kevin{Computing in general is currently
a real pain for me. I am currently spending as much time as possible
at home (Joel is too young to be put in a creche at the minute)
and hence am (a) using Windows 98 and (b) usually offline (it's
expensive in France to use the net) so access to magma is a pain! Added
to this is the intense irritation caused by the fact that my office in
London is being used by someone who has rebooted the computer into
Windows so I can't telnet into it, and you can see why doing any computations
at all is a real pain! Fortunately I have gp and I have ported all the
DimensionCuspForms stuff (it's a pain because evaluate'' typically gives
complex numbers to 30dps!!) and indeed it's 1210. What's the precise
reference to [CO] by the way?}} we find $\dim S_5(2416,\eps)=1210$; this
is the same as the dimension obtained using modular symbols, so
there is no spurious $5$-torsion.
We compute the space of modular symbols corresponding to
$S_5(2416,\eps;\F_{25})$, and then the $8$-dimensional
intersection~$V$ of the kernels of~$T_{53}$, $T_{97}$, and~$T_{127}$.
The characteristic polynomial of~$T_3$ is
$(x+2)^4(x+3)^4$.  The kernel~$V_1$ of $T_3+2$ has dimension~$4$.
The characteristic polynomial of~$T_7$ on~$V_1$
is $(x+\alp^4)^4$; the kernel~$V_2\subset V_1$ of $T_7+\alp^4$ has
dimension~$2$.
The characteristic polynomial of~$T_5$ on~$V_2$ is
$(x+\alp^{10})(x+\alp^{22})$.
Let~$f$ be the newform such that $a_5+\alp^{10}=0$.
We have
$$f = q +3q^3 + \alp^{22}q^5 + \alp^{16}q^7 + \alp^{4} q^{11} + \cdots.$$

There is one other newform $f'$ in~$V_2$, and it is a companion of~$f$.
We verified this using Lemma~\ref{lem:bound} by
checking that $a_p(f)=a_p(f')$ for those primes~$p\neq 5$ such that
$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(2416)]\cdot \frac{1}{4} = 836.$$

Let $g=f\tensor\eps_{151}$.
We have $a_p(f\tensor\eps_{151}^{-1})=0$ for $p=53,97,127$.
Also,
$\eps_{151}(3)=1$,
$\eps_{151}(5)=\alp^8$,
$\eps_{151}(7) =\alp^{20}$, so
$a_3(f\tensor\eps_{151}^{-1})=3/1=3=3^5$,
$a_5(f\tensor\eps_{151}^{-1})= \alp^{22}/\alp^8 = \alp^{14}=(\alp^{22})^5$,
$a_7(f\tensor\eps_{151}^{-1})= \alp^{16}/\alp^{8} = \alp^{8} = (\alp^{16})^5$.
Since $f\tensor\eps_{151}^{-1}$ and~$\sigma(f)$ both lie
in $S_5(2416,\eps_2\eps_{151}^2;\F_{25})$ and have same~$a_p$ for
$p=3,5,7,53,97,127$, they are equal.
Exactly as in Proposition~\ref{prop:1376-g} we see
that $g\in S_5(151N,\eps_2;\F_5)$.

We have not yet bounded ramification or checked that the image
of~$\rho_g$ is~$A_5$.\edit{\kevin{This shouldn't take too long but I didn't
want to waste time on it until you confirmed that my bound was good enough.
By the way, do you want to move the lemma to a different section or
anything?}\william{{\bf If I understand it correctly,
you're bound is definitely good enough.}}}
This computation can be done with the
help of the following table:
{\arraycolsep=.36em
$$\begin{array}{|rl|}\hline 2&0\\ 3&3\\ 5&\alp^{22}\\ 7&\alp^{16}\\ 11&\alp^{4}\\ 13&\alp^{2}\\ 17&\alp^{22}\\ 19&3\\ \hline\end{array} \begin{array}{|rl|}\hline 23&\alp^{22}\\ 29&3\\ 31&\alp^{16}\\ 37&\alp^{22}\\ 41&2\\ 43&\alp^{8}\\ 47&\alp^{8}\\ 53&0\\ \hline\end{array} \begin{array}{|rl|}\hline 59&1\\ 61&\alp^{8}\\ 67&3\\ 71&\alp^{8}\\ 73&2\\ 79&2\\ 83&2\\ 89&\alp^{20}\\ \hline\end{array} \begin{array}{|rl|}\hline 97&0\\ 101&2\\ 103&\alp^{16}\\ 107&1\\ 109&\alp^{20}\\ 113&\alp^{22}\\ 127&0\\ 131&4\\ \hline\end{array} \begin{array}{|rl|}\hline 137&\alp^{20}\\ 139&0\\ 149&\alp^{22}\\ 151&2\\ 157&\alp^{22}\\ 163&\alp^{8}\\ 167&\alp^{10}\\ 173&0\\ \hline\end{array} \begin{array}{|rl|}\hline 179&2\\ 181&\alp^{2}\\ 191&\alp^{10}\\ 193&0\\ 197&\alp^{14}\\ 199&\alp^{16}\\ 211&0\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&\alp^{20}\\ 229&3\\ 233&\alp^{22}\\ 239&\alp^{10}\\ 241&\alp^{2}\\ 251&\alp^{4}\\ 257&0\\ 263&\alp^{2}\\ \hline\end{array} \begin{array}{|rl|}\hline 269&0\\ 271&\alp^{10}\\ 277&\alp^{20}\\ 281&\alp^{16}\\ 283&0\\ 293&2\\ 307&\alp^{4}\\ 311&2\\ \hline\end{array}$$}

\subsection{Conductor~$3184=2^4\cdot 199$}
Consider the icosahedral extension~$K$ defined by
$h=x^5+5x^4+8x^3-20x^2-21x-5$.
A Frobenius element at~$5$ has order~$2$.
The first~$3$ primes~$p$ such that $a_p=0$ are
$31$, $89$, and $97$.
The type at~$2$ is~$16$ and the type at~$199$ is~$2$.
As in Section~\ref{sec:2416}, $\eps_2$ is of
conductor~$4$ and order~$2$, and the order of~$\eps_{199}$ is~$3$.
\edit{\kevin{We don't need to
guess any more---I think that the calculation in the last section
has convinced us that type 16 is associated to char non-trivial of
conductor~4. Probably I should think of some way of explaining this
that makes us look less like idiots.}\william{I fixed this.}}

Using \cite{cohen-oesterle}
we find $\dim S_5(3184,\eps)=1594$,
which agrees with the dimension computed using modular symbols.
We compute $S_5(3184,\eps;\F_{25})$ and then the
$8$-dimensional intersection~$V$ of the kernels of
$T_{31}$, $T_{89}$, and $T_{97}$.
The characteristic polynomial of~$T_3$ on~$V$ is
$(x+\alp^4)^4(x+\alp^{16})^4$.  The dimension of
$V_1=\ker(T_3+\alp^4)$ is~$2$. It appears that all
Hecke operators~$T_p$ act diagonally on $V_1$, except~$T_5$
which has the distinct eigenvalues~$2$ and~$3$.
We thus isolate a one-dimensional eigenspace, spanned by
a form~$f$ whose Hecke eigenvalues are:
{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alpha^{16}\\ 5&3\\ 7&\alpha^{22}\\ 11&3\\ 13&\alpha^{22}\\ 17&3\\ 19&\alpha^{16}\\ \hline\end{array} \begin{array}{|rl|}\hline 23&\alpha^{4}\\ 29&\alpha^{16}\\ 31&0\\ 37&\alpha^{8}\\ 41&\alpha^{2}\\ 43&\alpha^{10}\\ 47&\alpha^{4}\\ 53&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 59&3\\ 61&1\\ 67&4\\ 71&\alpha^{22}\\ 73&\alpha^{10}\\ 79&\alpha^{16}\\ 83&3\\ 89&0\\ \hline\end{array} \begin{array}{|rl|}\hline 97&0\\ 101&2\\ 103&2\\ 107&3\\ 109&0\\ 113&\alpha^{2}\\ 127&0\\ 131&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 137&1\\ 139&1\\ 149&\alpha^{22}\\ 151&\alpha^{2}\\ 157&0\\ 163&\alpha^{16}\\ 167&0\\ 173&0\\ \hline\end{array} \begin{array}{|rl|}\hline 179&\alpha^{8}\\ 181&2\\ 191&3\\ 193&\alpha^{14}\\ 197&0\\ 199&2\\ 211&4\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&1\\ 229&\alpha^{2}\\ 233&0\\ 239&3\\ 241&2\\ 251&0\\ 257&0\\ 263&1\\ \hline\end{array} \begin{array}{|rl|}\hline 269&\alpha^{2}\\ 271&\alpha^{2}\\ 277&0\\ 281&1\\ 283&\alpha^{10}\\ 293&\alpha^{8}\\ 307&\alpha^{20}\\ 311&\alpha^{20}\\ \hline\end{array}$$}

The bound from Lemma~\ref{lem:bound} is
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3184)]\cdot \frac{1}{4} = 1100.$$

\subsection{Conductor~$3556=2^2\cdot 7\cdot 127$}
Consider the icosahedral extension~$K$ defined by
$h=x^5+3x^4+9x^3-6x^2-4x-40$.
A Frobenius element at~$5$ has order~$3$.
The first~$3$ unramified~$p$ such that $a_p=0$ are~$19$,~$29$, and~$89$.
The type at~$2$ is~5; the type at~$7$ is~3; the type at~$127$ is~2.
The order of~$\eps_{127}$ is~$3$.
A local analysis shows that the order
of~$\eps_{2}$ is~$1$ and of~$\eps_7$ is~$2$.\edit{\kevin{William: I took
out some bad guesses here! If the type is $\leq15$ and not 6 then
I think I can work out the character, and indeed I can verify that your
first guess is wrong!}}
The formula of~\cite{cohen-oesterle}
gives $\dim S_5(3556,\eps)=2042$, which agrees
with the dimension of the corresponding space of modular symbols.

The intersection~$V$ of the kernels of~$T_{19}$,~$T_{29}$, and~$T_{89}$
has dimension~$8$.
The characteristic  polynomial of~$T_{3}$ on~$V$
is $(x+\alp^4)^4(x+\alp^{16})^4$,
and the kernel~$V_1$ of $T_3+\alp^4$ is of dimension~$2$.
It appears that all Hecke operators~$T_p$ act
diagonally on~$V_1$ except~$T_5$, whose eigenvalues
\edit{\kevin{How come it says $\alpha^{14}$ in
the table then? Are we out by a sign here?}\william{I gave the
signs incorrectly in this particular sentence.  The table is fine,
and the sentence is now fixed.}}
are~$\alp^{14}$ and~$\alp^{22}$. The eigenvalues of the
newform with $a_5 = \alp^{14}$ are given in the following table.
{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alpha^{16}\\ 5&\alpha^{14}\\ 7&\alpha^{10}\\ 11&\alpha^{2}\\ 13&\alpha^{22}\\ 17&\alpha^{14}\\ 19&0\\ \hline\end{array} \begin{array}{|rl|}\hline 23&\alpha^{10}\\ 29&0\\ 31&\alpha^{16}\\ 37&\alpha^{20}\\ 41&\alpha^{14}\\ 43&\alpha^{2}\\ 47&1\\ 53&\alpha^{2}\\ \hline\end{array} \begin{array}{|rl|}\hline 59&\alpha^{8}\\ 61&4\\ 67&\alpha^{10}\\ 71&\alpha^{4}\\ 73&4\\ 79&\alpha^{16}\\ 83&\alpha^{8}\\ 89&0\\ \hline\end{array} \begin{array}{|rl|}\hline 97&\alpha^{16}\\ 101&\alpha^{4}\\ 103&\alpha^{4}\\ 107&0\\ 109&\alpha^{14}\\ 113&\alpha^{4}\\ 127&\alpha^{20}\\ 131&1\\ \hline\end{array} \begin{array}{|rl|}\hline 137&2\\ 139&\alpha^{22}\\ 149&\alpha^{2}\\ 151&0\\ 157&0\\ 163&\alpha^{14}\\ 167&0\\ 173&\alpha^{10}\\ \hline\end{array} \begin{array}{|rl|}\hline 179&\alpha^{4}\\ 181&1\\ 191&1\\ 193&0\\ 197&\alpha^{22}\\ 199&\alpha^{2}\\ 211&\alpha^{8}\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&3\\ 229&0\\ 233&0\\ 239&0\\ 241&\alpha^{10}\\ 251&0\\ 257&\alpha^{16}\\ 263&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 269&\alpha^{4}\\ 271&\alpha^{20}\\ 277&\alpha^{16}\\ 281&0\\ 283&\alpha^{20}\\ 293&\alpha^{14}\\ 307&\alpha^{8}\\ 311&0\\ \hline\end{array}$$
}
By checking that the Hecke operators $T_p$ act diagonally
on $V_1$, for $p\neq 5$ with
$$p\leq\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3556)]\cdot \frac{1}{8} = 704,$$
we see that the other newform in~$V_1$ is a companion of~$f$.

Let $g=f\tensor\eps_{127}$.
We have $a_p(f\tensor\eps_{127}^{-1})=0$ for $p=19, 29, 89.$
We have $\eps_{127}(3)=\alp^8$ and $\eps_{127}(5)=1$, so
$a_3(f\tensor\eps_{127}^{-1})= \alp^{16}/\alp^8 = \alp^{8}$,
and $a_{5}(f\tensor\eps_{127}^{-1})= \alp^{14}/1=\alp^{14}$.
At first glance this form looks like the Galois conjugate
of~$f$; and if it were the Galois conjugate then the form
twisted by the character at~$127$ will, as in
Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
for $p \nmid \ell N$ and hence~$\rho_g$
will go into $\GL(2,\F_5)$.  Unfortunately the
newform associated to the twisted form is not
quite the conjugate of our original form; the coefficient
of~$a_5$ is wrong!  The conjugate form has
$a_5=(\alp^{14})^5 = \alp^{22}$.
The other newform $f_1\in V_1$, which is a companion
of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
This proves that~$g=\sigma(f_1)$.
An argument as in
Proposition~\ref{prop:1376-g} shows that $a_p(g)\in\F_5$
for all $p\nmid \ell N$.

\edit{\william{[[Kevin has an argument that
uses that~$\Frob_5$ has order~$3$
in the~$A_5$ field to show that~$\rho_f$ is unramified at~$5$.
Unfortunately his argument uses that the $a_p(g)$ are in~$\F_5$,
for $p\nmid \ell N$!
It might be possible to appeal to the Gross and Coleman-Voloch theorems
to prove that~$f_1$ is indeed a companion form.  This would
avoid any further computation.]]}
\kevin{I'm not sure I believe you.
You see: if we want to go down this path, we have to check that the
repn is unram at 5. But we don't know a darn thing about the repn yet.
Am I right in thinking that we are stuck in this case?}
\william{Given the lemma 1.3, I think we can just use the standard
technique and not go down this path!?}}
\comment{
\edit{\william{This shouldn't be needed.}}
Now $\Frob_5$ has order~$3$ in the~$A_5$
extension in Buhler so I think what should happen in your computations is
that the form of level $3556\cdot 127$
given by twisting the form of level~$3556$
will have all~$a_p$ in $\Z/5\Z$ apart from $a_5$ which will be
in~$\F_{25}$ and nonzero. The image of a
decomposition group at~$5$ can then be shown
to be upper-triangularisable (Deligne) and the chars on the diagonal
are unramified of order~$3$. So the image of a decomp group at~$5$
will have order $3\cdot\text{(power of$5$)}$. But there
are no subgroups of the subgroup of $\GL(2,\Z/5\Z)$ consisting of elements
of det $\pm 1$'' of order~$15$ because their image in
$\PSL(2,\Z/5\Z)$ would also
have order~$15$ and~$A_5$ has no such subgroups. So no companion form
argument is needed---the repn will be unramified at~$5$.
}

\subsection{Conductor~$3756=2^2\cdot 3\cdot 313$}
Consider the icosahedral extension~$K$ defined by
$h=x^5 - 3x^3 + 10x^2 + 30x - 18$.
A Frobenius element at~$5$ has order~$3$.
The first~$3$ unramified~$p$ such that $a_p=0$ are $17, 61, 67$.
The type at~$2$ is~$5$; the type at~$3$ is~$3$; the type
at~$127$ is~$2$.
The order of~$\eps_2$ is~$1$,
the order of~$\eps_3$ is~$2$, and
the order of~$\eps_{127}$ is~$3$;

Using \cite{cohen-oesterle} we find $\dim S_5(3756,\eps)=2506$;
this is the same as the dimension computed using modular symbols,
so there is no spurious $5$-torsion.
We find that~$V$ has dimension~$8$,
and the characteristic polynomial of~$T_{11}$ is
$(x+\alp^{4})^4(x+\alp^{16})^4$
(we do not use $T_3$ since $3\mid N$).
The kernel~$V_1$ of $T_{11}+\alp^{4}$ has dimension~$2$.
The characteristic polynomial of~$T_5$ on~$V_1$
is $(x+\alp^2)(x+\alp^{10})$.
The first few~$a_p$ of the newform~$f$
in~$V_2$ such that $a_5+\alp^2=0$ are given in
the following table.
{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alp^{14}\\ 5&\alp^{14}\\ 7&3\\ 11&\alp^{16}\\ 13&\alp^{10}\\ 17&0\\ 19&3\\ \hline\end{array} \begin{array}{|rl|}\hline 23&3\\ 29&\alp^{2}\\ 31&\alp^{22}\\ 37&\alp^{22}\\ 41&\alp^{20}\\ 43&\alp^{16}\\ 47&\alp^{4}\\ 53&4\\ \hline\end{array} \begin{array}{|rl|}\hline 59&\alp^{8}\\ 61&0\\ 67&0\\ 71&0\\ 73&0\\ 79&3\\ 83&\alp^{10}\\ 89&\alp^{20}\\ \hline\end{array} \begin{array}{|rl|}\hline 97&4\\ 101&0\\ 103&2\\ 107&\alp^{4}\\ 109&\alp^{20}\\ 113&2\\ 127&\alp^{10}\\ 131&\alp^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 137&0\\ 139&0\\ 149&0\\ 151&2\\ 157&\alp^{20}\\ 163&1\\ 167&0\\ 173&\alp^{20}\\ \hline\end{array} \begin{array}{|rl|}\hline 179&0\\ 181&\alp^{14}\\ 191&0\\ 193&\alp^{20}\\ 197&\alp^{22}\\ 199&0\\ 211&\alp^{10}\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&\alp^{8}\\ 229&\alp^{16}\\ 233&\alp^{4}\\ 239&\alp^{20}\\ 241&\alp^{10}\\ 251&\alp^{20}\\ 257&1\\ 263&\alp^{10}\\ \hline\end{array} \begin{array}{|rl|}\hline 269&2\\ 271&1\\ 277&3\\ 281&\alp^{20}\\ 283&2\\ 293&\alp^{20}\\ 307&1\\ 311&\alp^{16}\\ \hline\end{array}$$}

It appears that the other newform in~$V_1$ is a companion of~$f$,
but we have not proved this.

Let $g=f\tensor\eps_{313}$.
We have $a_p(f\tensor\eps_{313}^{-1})=0$ for $p=17, 61, 67.$
We have $\eps_{313}(5)=1$ and $\eps_{313}(11)=\alp^8$, so
$a_5(f\tensor\eps_{313}^{-1})= \alp^{14}/1 = \alp^{14}$,
and $a_{11}(f\tensor\eps_{313}^{-1})= \alp^{16}/\alp^{8}=\alp^8$.
At first glance this form looks like the Galois conjugate
of~$f$; and if it were the Galois conjugate then the form
twisted by the character at~$313$ will, as in
Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
for $p \nmid \ell N$ and hence~$\rho_g$
will go into $\GL(2,\F_5)$.  Unfortunately the
newform associated to the twisted form is not
quite the conjugate of our original form; the coefficient
of~$a_5$ is wrong!  The conjugate form has
$a_5=(\alp^{14})^5 = \alp^{22}$.
The other newform $f_1\in V_1$, which appears to be a companion
of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
This proves that $g=\sigma(f_1)$.
The bound from Lemma~\ref{lem:bound} is
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3756)]\cdot \frac{1}{8} = 863.5,$$
and we are easily able to verify that $f_1$ is a companion.
An argument as in
Proposition~\ref{prop:1376-g} then shows that $a_p(g)\in\F_5$
for all $p\nmid \ell N$.

For this example it remains to check
the discriminant bound, and surjectivity onto~$A_5$.

\subsection{Conductor~$4108=2^2\cdot 13\cdot 79$}
Consider the icosahedral extension~$K$ defined by
$h=x^5+4x^4+3x^3+9x^2+4x+5$.
A Frobenius element at~$5$ has order~$3$.
The first~$4$ unramified~$p$ such that
$a_p=0$ are $17$, $23$, $31$, and $89$.
The type at~$2$ is~5, the type at~$13$ is~2,
and the type at~$79$ is~3.
The order of~$\eps_2$ is~$1$,
the order of~$\eps_{13}$ is~$3$,
and the order of~$\eps_{79}$ is~$2$.

Using~\cite{cohen-oesterle} we find $\dim S_5(4108,\eps)= 2234$;

The intersection~$V$ of the kernels of $T_{17}$, $T_{23}$,
$T_{31}$, and $T_{89}$ has dimension~$8$.
The characteristic polynomial of~$T_3$ on~$V$ is
$(x+\alp^4)^4(x+\alp^{16})^4$.
The dimension of $V_1=\ker(T_3+\alp^4)$ is~$2$.
The characteristic polynomial of~$T_5$ on~$V_1$ is
$(x+\alp^4)(x+\alp^{20})$ and all of the other~$T_p$ appear
to act as scalars.

{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alpha^{16}\\ 5&\alpha^{11}\\ 7&\alpha^{20}\\ 11&\alpha^{10}\\ 13&4\\ 17&0\\ 19&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 23&0\\ 29&\alpha^{22}\\ 31&0\\ 37&\alpha^{22}\\ 41&\alpha^{10}\\ 43&\alpha^{2}\\ 47&3\\ 53&4\\ \hline\end{array} \begin{array}{|rl|}\hline 59&\alpha^{14}\\ 61&\alpha^{2}\\ 67&\alpha^{10}\\ 71&\alpha^{20}\\ 73&2\\ 79&3\\ 83&3\\ 89&0\\ \hline\end{array} \begin{array}{|rl|}\hline 97&\alpha^{2}\\ 101&\alpha^{22}\\ 103&2\\ 107&\alpha^{16}\\ 109&1\\ 113&\alpha^{8}\\ 127&0\\ 131&4\\ \hline\end{array} \begin{array}{|rl|}\hline 137&\alpha^{8}\\ 139&0\\ 149&\alpha^{8}\\ 151&0\\ 157&4\\ 163&\alpha^{20}\\ 167&0\\ 173&\alpha^{14}\\ \hline\end{array} \begin{array}{|rl|}\hline 179&\alpha^{4}\\ 181&2\\ 191&\alpha^{20}\\ 193&0\\ 197&0\\ 199&\alpha^{20}\\ 211&\alpha^{4}\\ 223&\alpha^{4}\\ \hline\end{array} \begin{array}{|rl|}\hline 227&0\\ 229&0\\ 233&3\\ 239&2\\ 241&\alpha^{8}\\ 251&0\\ 257&\alpha^{10}\\ 263&\alpha^{10}\\ \hline\end{array} \begin{array}{|rl|}\hline 269&0\\ 271&\alpha^{10}\\ 277&0\\ 281&2\\ 283&\alpha^{4}\\ 293&\alpha^{14}\\ 307&4\\ 311&0\\ \hline\end{array}$$}

The bound of Lemma~\ref{lem:bound} is
\edit{\kevin{Before we go any further, which forms can
we deal with rigorously now?}\william{All of them?}}
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4108)]\cdot \frac{1}{8} = 770.$$

\subsection{Conductor~$4288=2^6\cdot 67$}
Consider the icosahedral extension~$K$ defined by
$h=x^5+4x^4+5x^3+8x^2+3x+2$.
A Frobenius element at~$5$ has order~$3$.
The first~$3$ unramified~$p$ such that $a_p=0$ are $19$,
$23$, $47$.
The type at~$2$ is~18, and the type at~$67$ is~2.
The order of~$\eps_{67}$ is~$3$, so in order
for~$\eps$ to be odd, the order of~$\eps_2$
must be~$2$.  First we try letting $\eps_2$ be the character
of conductor~$4$.\edit{\kevin{I don't know anything
about type 18'' and your statement order 2'' is not enough
to tell me the character, so you should perhaps say what it is.
Oh! I've just noticed that this is wrong anyway!}}

Using \cite{cohen-oesterle} we find that\edit{\kevin{These
sections are getting really tedious, aren't they. I think we should
perhaps consider instead summarising things more succinctly,
e.g. saying we tried our approach for 4 other forms and it worked/didn't
work.}
\william{
You may be right.  However, there is a lot of information in each
situation.  One wants to know how many Hecke operators are needed
to cut out the space, what the character is, and so on.  All of
this information would have to be summarized.}}
$\dim S_5(4288,\eps)=2164$.
the intersection~$V$ of the kernels of $T_{19}$, $T_{23}$, and
$T_{47}$ has dimension~$0$ --- we must have chosen~$\eps_2$
incorrectly.  Next we try with $\eps_2$ one of the characters
of conductor~$8$.\edit{\william{The
character $\eps$ is [1,2,2].}}  \edit{\kevin{William: this character
also has order 2!}}  The dimension is also~$2164$.  This time we are
able to compute the intersection~$V$, and it has dimension~$16$.
%There is a quadratic twist of
%conductor~$8$ that preserves the level, so the dimension should
%be~$16$.
\edit{\kevin{William---this is a very optimistic statement!
All these spaces have Hecke acting non-semi-simply so I wouldn't dare
to guess the dimension of the space, just the number of
eigenforms\ldots}\william{I removed the remark about the dimension
being~$16$.}}

The characteristic polynomial of~$T_3$ on~$V$ is
$(x+2)^8(x+3)^8$.  The dimension of the kernel~$V_1$
of $T_3|_V+2$ is~$4$.  The characteristic polynomial
of~$T_7$ on~$V_1$ is $(x+\alp^8)^2(x+\alp^{20})^2$.
The dimension of the kernel~$V_2$
of $T_3|_{V_1}+\alp^8$ is~$2$.
The characteristic polynomial of~$T_5$ on~$V_2$ is
$(x+\alp^2)(x+\alp^{10})$.
By checking up to the bound
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4288)]\cdot \frac{1}{4} = 1496$$
of Lemma~\ref{lem:bound}
we see that all $T_p$, with $p\neq 5$,
act as scalars on~$V_2$.

{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&0\\ 3&3\\ 5&\alpha^{14}\\ 7&\alpha^{20}\\ 11&\alpha^{20}\\ 13&\alpha^{16}\\ 17&\alpha^{16}\\ 19&0\\ \hline\end{array} \begin{array}{|rl|}\hline 23&0\\ 29&\alpha^{14}\\ 31&\alpha^{20}\\ 37&\alpha^{4}\\ 41&\alpha^{8}\\ 43&2\\ 47&0\\ 53&4\\ \hline\end{array} \begin{array}{|rl|}\hline 59&1\\ 61&\alpha^{16}\\ 67&\alpha^{4}\\ 71&\alpha^{20}\\ 73&0\\ 79&\alpha^{20}\\ 83&0\\ 89&2\\ \hline\end{array} \begin{array}{|rl|}\hline 97&\alpha^{10}\\ 101&\alpha^{14}\\ 103&\alpha^{20}\\ 107&2\\ 109&0\\ 113&\alpha^{2}\\ 127&\alpha^{2}\\ 131&1\\ \hline\end{array} \begin{array}{|rl|}\hline 137&1\\ 139&2\\ 149&4\\ 151&\alpha^{4}\\ 157&\alpha^{22}\\ 163&0\\ 167&\alpha^{14}\\ 173&0\\ \hline\end{array} \begin{array}{|rl|}\hline 179&1\\ 181&\alpha^{14}\\ 191&\alpha^{4}\\ 193&3\\ 197&0\\ 199&0\\ 211&\alpha^{16}\\ 223&2\\ \hline\end{array} \begin{array}{|rl|}\hline 227&\alpha^{14}\\ 229&\alpha^{4}\\ 233&\alpha^{2}\\ 239&\alpha^{20}\\ 241&3\\ 251&\alpha^{4}\\ 257&\alpha^{2}\\ 263&4\\ \hline\end{array} \begin{array}{|rl|}\hline 269&0\\ 271&0\\ 277&2\\ 281&\alpha^{10}\\ 283&3\\ 293&1\\ 307&\alpha^{22}\\ 311&3\\ \hline\end{array}$$}

\subsection{Conductor~$5373=3^3\cdot 199$}
There are two $A_5$ entries in Frey's
Table~2 of conductor 5373.  Though neither 2 nor 5 ramify,
the hypothesis of~\cite{bdsbt} do not apply
to the example below because $\Frob_2$ has order~$5$.
This is the only example of odd level treated in this paper.

Consider the icosahedral extension~$K$ defined by
$h=x^5+2x^4+x^3+7x^2+23x-11$.
A Frobenius element at~$5$ has order~$2$.
The first few $p\nmid \ell N$ such that
$a_p=0$ are $7$, $23$,  $37$, $79$, $89$.
The type at~$3$ is~11, and the type at~$199$ is~2.
The order of~$\eps_3$ is~$2$, and
the order of~$\eps_{199}$ is~$3$.

Using~\cite{cohen-oesterle} we find $\dim S_5(5373,\eps)= 2394$;
this agrees with the dimension of the space of modular symbols.

The intersection~$V$ of the kernels of
$T_7$, $T_{23}$, $T_{37}$, $T_{79}$, and $T_{89}$
is of dimension~$8$.
The characteristic polynomial of~$T_2$ on~$V$ is
$(x+\alp^4)^4(x+\alp^{16})^4$.
The kernel~$V_1$ of $T_2|_V+\alp^4$
is of dimension~$2$. It appears that all~$T_p$ act
as scalars on~$V_1$ except~$T_5$ which has
characteristic polynomial $(x+1)(x+4)$.
Let~$f$ be the form with $a_5=-1$; some eigenvalues of~$f$ are
given in the following table.
{\arraycolsep=.4em
$$\begin{array}{|rl|}\hline 2&\alpha^{16}\\ 3&0\\ 5&4\\ 7&0\\ 11&2\\ 13&\alpha^{22}\\ 17&1\\ 19&\alpha^{16}\\ \hline\end{array} \begin{array}{|rl|}\hline 23&0\\ 29&\alpha^{4}\\ 31&\alpha^{14}\\ 37&0\\ 41&\alpha^{14}\\ 43&\alpha^{10}\\ 47&\alpha^{4}\\ 53&\alpha^{8}\\ \hline\end{array} \begin{array}{|rl|}\hline 59&3\\ 61&3\\ 67&4\\ 71&\alpha^{22}\\ 73&\alpha^{10}\\ 79&0\\ 83&3\\ 89&0\\ \hline\end{array} \begin{array}{|rl|}\hline 97&\alpha^{14}\\ 101&3\\ 103&0\\ 107&2\\ 109&3\\ 113&\alpha^{20}\\ 127&\alpha^{14}\\ 131&\alpha^{20}\\ \hline\end{array} \begin{array}{|rl|}\hline 137&2\\ 139&2\\ 149&\alpha^{4}\\ 151&0\\ 157&4\\ 163&\alpha^{16}\\ 167&\alpha^{8}\\ 173&\alpha^{8}\\ \hline\end{array} \begin{array}{|rl|}\hline 179&\alpha^{8}\\ 181&1\\ 191&3\\ 193&\alpha^{20}\\ 197&0\\ 199&1\\ 211&2\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&1\\ 229&\alpha^{20}\\ 233&\alpha^{4}\\ 239&2\\ 241&1\\ 251&4\\ 257&\alpha^{20}\\ 263&4\\ \hline\end{array} \begin{array}{|rl|}\hline 269&\alpha^{2}\\ 271&\alpha^{20}\\ 277&3\\ 281&1\\ 283&\alpha^{22}\\ 293&\alpha^{14}\\ 307&0\\ 311&\alpha^{2}\\ \hline\end{array}$$
}

The bound of Lemma~\ref{lem:bound} is
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(5373)]\cdot \frac{1}{4} = 1650.$$

\comment{
\subsection{Conductor $N=1825=5^2\cdot 73$ and $\ell=7$}
In this example, the method does not apply because $5\mid N$.
Nonetheless, we proceed using~$\ell=7$ instead.\edit{\kevin{We can't
\emph{hope} to finish though, because we'll never show the thing is $A_5$
and not something like $GL_2(\F_7)$. So probably we should kill this
example completely. Somehow our paper gives a method for dealing
with $\GL_2(\F_5)$ (and it would also perhaps work with $\GL_2(\F_4)$)
but doesn't have a chance of working wiht $GL_2(\F_7)$, I shouldn't think.}
\william{It is completely dead.}}

Consider the icosahedral extension~$K$ defined by
$h=x^5+4x^4-x^3-21x^2-x-7$.
A Frobenius element at~$7$ has order~$3$, so
distinct eigenvalues mod~$7$ (I think).
The first three unramified~$p$ such that
$a_p=0$ are $17$, $59$, $67$, $89$.
The type at~$5$ is~6, and the type at~$73$ is~2.
The order of~$\eps_5$ is~$4$ (I think), and
the order of~$\eps_{73}$ is~$3$.

Using~\cite{cohen-oesterle} we find $\dim S_8(1825,\eps)= 1104$, and
this agrees with modular symbols.
The intersection~$V$ of the kernels of $T_{17}$, $T_{59}$,
and~$T_{67}$ has dimension~$8$.
The operator~$T_{89}$ also vanishes on~$V$.
The character takes values in $\F_{7^2}$ which we represent
using a generator~$\alp$ of the multiplicative group such that
$\alp^2-\alp+3=0$.
The characteristic polynomial of~$T_2$ on~$V$ is
$(x+\alp)^2(x+\alp^{19})^2(x+\alp^{25})^2(x+\alp^{43})^2$.
Let $V_1$ be the kernel of $T_2+\alp$; then $V_1$ is spanned
by two newforms~$f$ and~$f_1$:
\begin{align*}
f   &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{46}q^7+\cdots\\
f_1 &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{14}q^7+\cdots
\end{align*}
It appears that~$f$ and~$f_1$ are companions; and undoubtedly this can
be checked.

It doesn't seem possible to twist and obtain a representation
into $\GL(2,\F_7)$. The $\eps_{73}$ part of the character
is $\F_7$-rational.  The $\eps_5$ part has order~$4$, and is
hence not $\F_7$-rational.  Twisting by~$\eps_5$ (or its inverse)
only replaces $\eps_5$ by $\eps_5^3$, which is still not $\F_7$-rational.

\begin{dashlist}
\item Even if we could twist into $\GL(2,\F_7)$,
the order~$60$ of $A_5$ doesn't divide the order of $\GL(2,\F_7)$.
\item  In this $N=1825$ example, we found a modular representation
$\rho_g : \GQ \ra \GL(2,\F_{7^2})$
that (morally) is the modulo a prime-over-$7$ reduction of
the Artin representation  $\rho : \GQ \ra \GL(2,\cO)$.
Here~$\cO$ is the ring of integers in a finite extension of~$\Q_7$.
If someone ever proves modularity of~$\rho$ they will
have proved that~$\rho_g$ admits a lift to a characteristic zero
representation whose image is an extension of~$A_5$;
perhaps we can prove this.  We could then try and analyze the
ramification and conclude that the lift of~$\rho_g$ is (a twist) of~$\rho$.
Then~\cite{buzzard-taylor} implies modularity of~$\rho$.
\end{dashlist}
}

\section{Computations}
\subsection{Higher weight modular symbols}
\label{sec:modsym}
The second author developed software that computes the space of
weight~$k$ modular symbols $\sS_k(N,\eps)$, for $k\geq 2$ and
arbitrary~$\eps$.
See~\cite{merel:1585} for the standard facts about higher weight
modular symbols, and~\cite{stein:phd} for a description of
how to efficiently compute with them.

Let $K=\Q(\eps)$ be the field generated by the values of~$\eps$.
The cuspidal modular symbols $\sS_k(N,\eps)$ are a
finite dimensional vector space over~$K$, which is generated by all
linear combinations of  higher weight modular symbols
$$X^i Y^{k-2-i}\{\alp,\beta\}$$
that lie in the kernel of an appropriate boundary map.  There is an
involution~$*$ that acts on $\sS_k(N,\eps)$, and
$\sS_k(N,\eps)^+\tensor_K\C$ is isomorphic, as a module over the Hecke
algebra, to the space $S_k(N,\eps;\C)$ of cusp forms.

Fix $k=5$.  In each case considered in this paper,
there is a prime ideal~$\lambda$
of the ring of integers $\mathcal{O}$ of~$K$
such that $\mathcal{O}/\lambda\isom \F_{25}$.
Let~$\cL$ be the $\mathcal{O}$-module generated by all modular
symbols of the form $X^iY^{3-i}\{\alp,\beta\}$,
and let
$$\sS_5(N,\eps;\F_{25})=(\cL\tensor_{\mathcal{O}}\F_{25})\cap \sS_5(N,\eps).$$
This is the space that we computed.
The Hecke algebra acts on $\sS_5(N,\eps;\F_{25})$, so when
we find an eigenform we find a maximal ideal of the Hecke algebra.

As an extra check on our computation of
$\sS_5(N,\eps;\F_{25})$, we computed the dimension
of $S_5(N,\eps;\C)$ using both the formula of~\cite{cohen-oesterle}
and the Hijikata trace formula (see~\cite{hijikata:trace})
applied to the identity Hecke operator.

\comment{%it's all in my thesis and it's not that relevant.
The Manin symbols are
$[i, (c,d)]$ where $0\leq i\leq k-2=3$ and
$(c,d)$ vary over points in the projective plane.
The Manin symbol $[i,(c',d')]$ corresponds to the
modular symbol $(g.X^iY^{3-i})\{g(0),g(\infty)\}$
where $g=\abcd{a}{b}{c}{d}\in\SL_2(\Z)$ is a matrix whose lower
two entries are congruent to $(c',d')$ modulo $N$,
and $g.X^iY^{3-i} := (dX-bY)^i(-cX+aY)^{3-i}$.
Let $\sigma=\abcd{0}{-1}{1}{0}$, $\tau={0}{-1}{1}{-1}$
and for $\gamma\in\SL_2(\Z)$, let
$[i,(c,d)]\gamma = [\gamma.X^iY^{3-i}, (c,d)\gamma]$.
Since there are only finitely many
Manin symbols, we  can
compute $\sS_5(N,\eps)$ as the quotient of the $\F$-vector
space generated by Manin symbols modulo
the following relations:
\begin{align*}
{[i,(c,d)] + [i,(c,d)]\sigma} &= 0\\
{[i,(c,d)] + [i,(c,d)]\tau + [i,(c,d)]\tau^2} &= 0\\
{[i,(n c,n d)]}&=\eps(n)[i,(c,d)]
\end{align*}
The quotient was computed by using a fast hashing'' function
to quotient out by the $2$-term relations.  The quotient
by the $3$-term relations was then computed using sparse
Gauss elimination.  One important subtlety is that, e.g., $\sigma$
and~$\tau$ do not commute so, after modding out by
the~$\sigma$ relations, it is important to mod out by~$3$
term relations coming both from~$\tau$ and~$\sigma\tau$.

The main result of~\cite{merel:1585} gives
a way to compute the action of $T_p$ directly
on the Manin symbols.
Suppose $f\in\sS_5(N,\eps;\F_{25})$ is an eigenvector; to
naively compute the action of~$T_p$ on~$f$ requires computing
the action of~$T_p$ on each Manin symbol involved in~$f$,
and then summing the result. This requires roughly
$\dim\sS$ times as long as computing~$T_p$ on a single
Manin symbol.
In order to quickly compute a large number of
Hecke eigenvalues we use the following projection trick.
Let $\vphi\in\Hom(\sS_5(N,\eps;\F_{25}),\F_{25})$ be a (left) eigenvector for all
Hecke operators~$T_p$ having the same eigenvalues as~$f$.
Choose a Manin symbol $x=[i,(c,d)]$ such
that $\vphi(x)\neq 0$.  Since~$x$ is of a very simple form,
it is easy to compute~$T_p(x)$ quickly.  We have
$\vphi(T_p(x)) = (T_p(\vphi))(x) = a_p \vphi(x)$,
so since $\vphi(x)\neq 0$ we divide and find
$a_p = \vphi(T_p(x))/\vphi(x)$.
In fact, we use a generalization of this trick to
quickly compute the action of~$T_p$ on any Hecke stable subspace
$V\subset \Hom(\sS(N,\eps;\F_{25}),\F_{25})$.
}

\subsection{Complexity}
We implemented the modular symbols algorithms mentioned above
in \magma{} (see \cite{magma}) because of its robust support
for linear algebra over small finite fields.

The following table gives a flavor for the complexity of the
machine computations appearing in this paper.
The table indicates how much
CPU time on a 450Mhz SUN Ultra-4 was required to compute all data
for the given level,
including the matrices $T_p$ on the $2$-dimensional spaces,
for $p<2000$.   For example, the total time for level $N=1376$
was~$6$ minutes and~$58$ seconds.
\begin{center}
\begin{tabular}{|cr|}\hline
\vspace{-2ex}&\\
N & time (minutes)\\
\vspace{-2ex}&\\
1376&  6:58\hspace{2.5em}\mbox{}\\
2416&  10:42\hspace{2.5em}\mbox{}\\
3184&  14:16\hspace{2.5em}\mbox{}\\
3556&  19:55\hspace{2.5em}\mbox{}\\
3756&  27:47\hspace{2.5em}\mbox{}\\
4108&  23:11\hspace{2.5em}\mbox{}\\
4288&  15:18\hspace{2.5em}\mbox{}\\
5376&  24:49\hspace{2.5em}\mbox{}\\\hline
\end{tabular}
\end{center}

\subsection{Acknowledgment}
Some of the computing equipment was purchased
using a Vice Chancellor's Research Grant from the University of
California, Berkeley.   We also made use of
the computing facilities at \magma.  Allan Steel, of the~\magma{} group,
rewrote several of the time-critical parts of our program
in C. \edit{\kevin{Do we need to
mention the MAGMA supercomputer thing?}  \william{I just emailed
Cannon asking for the proper attribuation.}}

\edit{
\kevin{OK, I looked through it. I kind of rushed the ending because it kind of
seemed unfinished anyway. I wrote section 2.4 but didn't have time
to write 2.5---indeed, I think you know more about it than I do now,
given that you talked to the master about all this! It looks like
we're on the way now. I chopped and changed some things in the first
few sections i.e. up to "3 More Examples". After that, I only slightly
changed things, removing guesswork when we didn't need to guess
(about characters). See what you think; I'll just wait until you
throw it back to me, but if you essentially instantly throw it
back then I'll have another go when I get some more time, and in fact
this week is looking surprisingly non-busy.
Kevin}
}

\edit{
\william{
I don't think we can reasonably compute any more Artin conjecture
examples using our methods.  There is a large gap between 5376 and the
next example that one might try.  The next two examples after 5376 are
$N=6176=2^5\cdot 193$ and $N=7236=2^2\cdot 3^3\cdot 67$.
There's no point in doing 6176
because Frey already did it.  I just tried the 7236 example on the
computer in Sydney and it ran out of memory at about 2.8GB.  In this
example, the dimension of the weight 5 space is almost 5000.  The next
example in which 5 is unramified and $\Frob_5$ has order $\neq 5$ occurs
beyond level 10,000; I didn't look far enough to find it.
}}

\comment{\bibliographystyle{amsplain}
\bibliography{biblio}}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{10}

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Univ. Hamburg \textbf{3} (1923)\edit{\kevin{Yup.}\william{I just emailed Hendrik.  It says 1924 on my T-shirt, but I think Hendrik complained about that...}}, 89--108.

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Springer-Verlag, Berlin, 1978, Lecture Notes in Mathematics, Vol. 654.

\bibitem{bdsbt}
K.~Buzzard, M.~Dickinson, N.~Shepherd-Barron, and R.~Taylor, \emph{On
icosahedral {A}rtin representations}, in preparation.

\bibitem{buzzard-taylor}
K.~Buzzard and R.~Taylor, \emph{Companion forms and weight one forms}, Ann. of
Math. (2) \textbf{149} (1999), no.~3, 905--919.

\bibitem{cohen-oesterle}
H.~Cohen and J.~Oesterl{\'e}, \emph{Dimensions des espaces de formes
modulaires},  (1977), 69--78. Lecture Notes in Math., Vol. 627.

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\\\protect{\sf http://www.maths.usyd.edu.au:8000/u/magma/}.

\bibitem{deligne-serre}
P.~Deligne and J-P. Serre, \emph{Formes modulaires de poids $1$}, Ann. Sci.
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\bibitem{freyetal}
G.~Frey (ed.), \emph{On {A}rtin's conjecture for odd \protect{$2$}-dimensional
representations}, Springer-Verlag, Berlin, 1994.

\bibitem{gross:tameness}
B.\thinspace{}H. Gross, \emph{A tameness criterion for \protect{G}alois
representations associated to modular forms (mod \protect{$p$})}, Duke Math.
J. \textbf{61} (1990), no.~2, 445--517.

\bibitem{hijikata:trace}
H.~Hijikata, \emph{Explicit formula of the traces of \protect{H}ecke operators
for \protect{$\Gamma_0(N)$}}, J. Math. Soc. Japan \textbf{26} (1974), no.~1,
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L.~Merel, \emph{Universal \protect{F}ourier expansions of modular forms}, On
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\end{thebibliography}

\end{document}

***

[8458981, 509]

// Cohen-Oesterle Dimension computations in MAGMA:

> G<a2,b2,c> := DirichletGroup(1376,CyclotomicField(EulerPhi(1376)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);

> G<a2,b2,c> := DirichletGroup(2416,CyclotomicField(EulerPhi(2416)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1210

> G<a2,b2,c> := DirichletGroup(3184,CyclotomicField(EulerPhi(3184)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1594

> G<a,b,c> := DirichletGroup(3556,CyclotomicField(EulerPhi(3556)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
2042;

> G<a,b,c> := DirichletGroup(3756,CyclotomicField(EulerPhi(3756)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);

> G<a,b,c> := DirichletGroup(4108,CyclotomicField(EulerPhi(4108)));
> eps:=(b^(Order(b) div 3))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);

> G<a,b,c> := DirichletGroup(4288,CyclotomicField(EulerPhi(4288)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);

> G<a,b> := DirichletGroup(5373,CyclotomicField(EulerPhi(5373)));
> eps:=(a^(Order(a) div 2))*(b^(Order(b) div 3));
> DimensionCuspForms(eps,5);

`