CoCalc Public Fileswww / artin2.tex
Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
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3\documentclass{article}
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6\markboth{}{Buzzard-Stein (\today)}%{Icosahedral Galois representations}
7\title{Mod five approaches to modularity of icosahedral
8       Galois representations}
9\author{Kevin Buzzard and William A. Stein}
10
11% all of our conversations have been moved into an edit'' macro
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111\begin{document}
112\maketitle
113\begin{abstract}
114Using a theorem of Buzzard and Taylor, we compute new icosahedral
115examples of Artin's conjecture on holomorphicity of the $L$-function
116associated to certain Galois representations.  We give in detail one
117new example of an icosahedral representation of conductor
118${\bf 1376}=2^5\cdot 43$ that satisfies Artin's conjecture.  We also
119give\edit{\kevin{[[Still let's see what happens...]]}}  seven
121${\bf 2416}=2^4\cdot 151$,
122${\bf 3184}=2^4\cdot 199$,
123${\bf 3556}=2^2\cdot 7\cdot 127$,
124${\bf 3756}=2^2\cdot 3\cdot 313$,
125${\bf 4108}=2^2\cdot 13\cdot 79$,
126${\bf 4288}=2^6\cdot 67$, and
127${\bf 5373}=3^3\cdot 199$.
128\end{abstract}
129
130\section*{Introduction}\edit{
131\kevin{
132behavior --$>$ behaviour,
133  analyze --$>$ analyse. William---what do we do about incompatibilities
134 in the spelling in our languages?! I once submitted a paper to MRL and
135 in the typesetting all the UK spelling was changed to US spelling so
136 in the proofs I just changed it all back again :-) But this is a joint
137 paper which we'll probably end up submitting to an American journal
138 so perhaps US spelling is more appropriate...
139}
140\william{
141Let's decide based on the journal we submit to.  Where should we
142submit this paper?  In a few days I can drop it in front of Ken
144}}
145Consider a continuous irreducible Galois representation
146$$\rho:\galq\ra\GL_n(\C)$$
147with $n > 1$.
148Inspired by his reciprocity law,
149Artin conjectured in~\cite{artin:conjecture} that
150$L(\rho,s)$ has an analytic continuation to the whole complex plane.
151Many of the known cases of this conjecture were obtained by
152proving the apparently stronger assertion that~$\rho$ is \defn{automorphic},
153in the sense that the $L$-function of~$\rho$ is equal to the $L$-function
154of a certain automorphic representation (whose $L$-function is known to have
155analytic continuation). In the special case where $n=2$ and $\rho$ is in
156addition assumed to be odd, the automorphic representation in question
157should be the one associated to a classical weight~$1$
158modular eigenform, and in fact there is conjectured to be a
159bijection between such~$\rho$ and the set of all weight~$1$
160cuspidal newforms, which should
161preserve $L$-functions. It is this bijection
162that we are concerned with in this paper, so assume for the rest
163of the paper that $n=2$ and~$\rho$ is odd.
164
165In this special case, the construction
166of~\cite{deligne-serre} shows how to construct a continuous irreducible
167odd 2-dimensional representation from a weight~$1$ newform, and the problem
168is to go the other way. Say that a representation is \defn{modular}
169if it arises in this way.
170
171If the image of~$\rho$ is solvable,
172then~$\rho$  is known to be modular
173\cite{langlands:basechange, tunnell:artin};
174if the image is not solvable, then $\im(\rho)$ in $\PGL_2(\C)$
175is isomorphic to the
176alternating group~$A_5$, and the modularity of~$\rho$
177is, in general, unknown. We call such a 2-dimensional representation an
178icosahedral representation''.
179The published literature contains only eight examples (up to twist)
180of odd icosahedral Galois representations that are known to satisfy Artin's
181conjecture: one of conductor $800=2^5\cdot 5^2$
182(see \cite{buhler:thesis}), and seven of conductors:
183$2083,\, 2^2\cdot 487,\, 2^2\cdot 751,\, 184 2^2 \cdot 887,\, 2^2\cdot 919,\, 185 2^5\cdot 73,\,\text{ and } 2^5\cdot 193$
186(see \cite{freyetal}).
187
188After the first draft of this paper was written, the
189preprint~\cite{bdsbt} appeared, which contains a general theorem that
190yields infinitely many (up to twist) modular icosahedral representations.
191However, we feel that our work, although much less powerful, is still
192of some worth, because it gives an effective computational approach to
193proving that certain mod~5 representations are modular, without
194computing any spaces of weight~1 forms or using effective versions of
195the Chebotarev\edit{\kevin{William, I don't have my TeXbook with me so
196don't know how to put the accent on the C.}
197\william{Kevin, How should we spell this?? I just checked the titles
198of recent papers with this word in the MathSciNet database and found:
199Chebotarev (Murty, etc.), Chebotar\"{e}v (Lenstra, however the funniness
200of the e is dropped in the AMS review),  Tschebotareff (by a Japanese author),
201Tchebotarev (Lang), Tschebotarev (van der Ploeg),
202Tchebotareff (a Frenchman, but translated to Chebotarev by the AMS),
203\v{C}ebotarev (Moshe).
204I say we go with Chebotarev,'' because it is seems to be the
205most common, and simply looks nice to me.  Also, it is the
206closest to Hendrik's without using special characters.  It
207seems strange to me to transliterate into English from Russian and use special
208characters from, I guess, French.
209}}
210density theorem.  Finally, the
211main theorem of~\cite{bdsbt} does not apply to any of the examples
212considered in the present paper.
213
214In this paper we give eight new examples that were computed
215by applying the main theorem  of~\cite{buzzard-taylor} to
216the mod~$5$ reduction of~$\rho$.
217\edit{In seven of the eight examples, we assume an unknown
218congruence bound in order to keep the computations
219manageable.\kevin{Do we? We'll have to sort this out---see
220the lemma later on in section 1.3}
221\william{The congruence bound is now known!}}
222We verify modularity mod~$5$ on a case-by-case basis. Later we shall
223explain our approach more carefully, but let us briefly summarise it here.
224By~\cite{buzzard-taylor},
225the problem is to show that the mod~5 reduction of~$\rho$ is modular.
226\edit{\william{By the way, as I'm sure you've spotted, I have added this
227para.}}
228We do this by finding a candidate mod~5 modular form at weight~5
229and then, using the table of icosahedral extensions of $\Q$ in~\cite{freyetal}
230and what we know about the 5-adic representation attached to our candidate
231form, we deduce that the mod~5 representation attached to our candidate
232form must be the reduction of~$\rho$. In particular, this paper gives
233a computational methods for checking the modularity of certain mod~5
234representations whose conductors are not too large. We now give
235more details.
236
237In each of our examples it is easy to compute a few Hecke operators
238and be morally convinced that a mod~$5$ representation should be modular;
239it is far more difficult to prove this.
240Effective variants of the Chebotarev density theorem require
241that we check vastly more traces of Frobenius than is practical.
242Instead we use the Local Langlands theorem for $\GL_2$, the
243theory of companion forms, the main theorem of \cite{buzzard-taylor},
244and Table~2 of~\cite{freyetal}, to provide proofs of modularity
245in certain cases.
246
247More precisely, let~$K$ be an icosahedral extension of~$\Q$ that is not
248totally real, and consider a minimal lift $\rho:\GQ\ra \GL_2(\C)$
249of
250   $$\GQ\ra \Gal(K/\Q)\ncisom{}A_5\subset \PGL_2(\C);$$
251the lift is minimal in the sense  that its conductor is minimal.
252Assume that~$5$ does not ramify in~$K$, and that
253a Frobenius element at~$5$ in $\Gal(K/\Q)$ does not have order~$1$ or~$5$.
254Inspired by the possibility that~$\rho$ is modular,
255we search for a mod~$5$ modular form of weight~$5$ whose existence would
256be forced by modularity of~$\rho$.  Indeed, we find
257a candidate mod~$5$ form~$f$, and then prove that the fixed field
258of the kernel of the projective mod~$5$ representation
259associated to a certain twist of~$f$  must be~$K$.
260This proves that the mod~$5$ reduction of a twist
261of~$\rho$ is modular, and the main theorem
262of \cite{buzzard-taylor} then implies
263that~$\rho$ is modular.
264We carried out this program for icosahedral representations
265of the following conductors:
266${\bf 1376} = 2^5\cdot 43$,
267${\bf 2416}=2^4\cdot 151$,
268${\bf 3184}=2^4\cdot 199$,
269${\bf 3556}=2^2\cdot 7\cdot 127$,
270${\bf 3756}=2^2\cdot 3\cdot 313$,
271${\bf 4108}=2^2\cdot 13\cdot 79$,
272${\bf 4288}=2^6\cdot 67$, and
273${\bf 5373}=3^3\cdot 199$.
274
275
276We choose an icosahedral field~$K$ and representation~$\rho$,
277then proceed as follows:
278\vspace{.5ex}
279\begin{numlist}
280\item Search for a form~$f \in S_5(N,\eps;\Fbar_5)$ whose
281      associated mod~$5$ Galois representation looks like
282      it is the mod~$5$ reduction of~$\rho$.
283\item Twist~$f$ to obtain an eigenform~$g$ with coefficients in~$\F_5$.
284\item Prove that~$\rho_g$ is unramified at~$5$ by finding a companion form.
285\item Prove that the image of $\proj\rho_g$ is~$A_5$ by ruling out all
286      other possibilities.
287\item Prove that the fixed field~$L$ of $\proj\rho_g$ has
288      root field of discriminant at most $2083^2$,
289      so~$L$ is in Table~2 of~\cite{freyetal}; deduce that~$L=K$.
290\item Apply the main theorem of~\cite{buzzard-taylor}
291      to a lift of $\rhobar=\rho_g$
292      to conclude that~$\rho$ is modular.
293\end{numlist}
294
295
296
297\section{Modularity of an icosahedral representation of
298conductor~$1376=2^5\cdot 43$}\label{sec:1376}
299In this section we prove the following theorem.
300\begin{theorem}\label{thm:1376}
301The icosahedral representations whose corresponding
302icosahedral extension
303is the splitting field of $x^5 + 2x^4+6x^3+8x^2+10x+8$
304are modular.
305\end{theorem}
306
307Let~$K$ be the splitting field of $h=x^5 + 2x^4+6x^3+8x^2+10x+8$.
308The Galois group of~$K$ is~$A_5$, so we obtain a homomorphism
309$G_\Q\ra{}A_5\subset \PGL_2(\C)$;
310let $\rho:G_\Q\ra\GL_2(\C)$ be a minimal lift, minimal
311in the sense that the Artin conductor of~$\rho$ is minimal.
312By Table~$A_5$ of~\cite{buhler:thesis}, the conductor of~$\rho$
313is $N=1376=2^5\cdot 43$.  Since
314$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5}$,
315and ${\rm disc}(h)$ is coprime to~$5$,
316any Frobenius element at~$5$ in $\Gal(K/\Q)$ has order~$2$.
317
318We use the notation of Tables 3.1 and 3.2 of~\cite[pg. 46]{buhler:thesis};
319from Table 3.2 we see that the type of~$\rho$ at~$2$
320is~$17$ and the type at~$43$ is~$2$.
321The mod~$N$ Dirichlet character~$\eps=\det(\rho)$
322factors as~$\eps=\eps_2\cdot \eps_{43}$ where~$\eps_2$ is
323a character mod~$2^5$ and~$\eps_{43}$ is a character mod~$43$.
324Corresponding to each type in Buhler's table, there is a character,
325and fortunately Buhler's level $800$ example also was of type~$17$ at~$2$
326(see the first line of~\cite[Table~3.2]{buhler:thesis}).
327By~\cite[pg.~80]{buhler:thesis} $\eps_2$ is the unique
328character\edit{
329\kevin{William: we should
330check that Buhler's example turned out to be a weight~$1$ form
331whose character at~$2$ was $\eps_2$.}
332\william{\em Yep, by Table 3.2.}
333\kevin{I didn't
334mean this---but after reading Buhler I found that in fact the answer
335to the question I meant to ask was Yep, by the beginning of chapter 6''
336(and this is probably the p80 you mention above). So we can kill
337this footnote. By the way, how long does
338it take you to check the fact'' near the beginning of Chapter 6, that
339Buhler talks philosophically about? (not that it matters)}
340\william{
341On a PII 350, it takes a total of 258 seconds total
342for my program to compute a basis of newforms
343for each space of level dividing $800$, and from this to then
344compute a Gauss-reduced basis of $q$-expansions, up to precision $360$
345for the space $S_2(800;\C)$. I didn't check how long it
346takes to verify that the $h_{i,d}$ lie in this space, but
347this should be completely trivial because my basis is
348Gauss-reduced.}}
349of conductor~$4$ and order~$2$.
350A local computation shows that the image
351of~$\eps_{43}$ has order~$3$.
352
353If~$\rho$ is modular,  then there is a weight~$1$
354newform $f_?\in S_1(N,\eps;\Qbar)$ that gives rise to~$\rho$.
355Suppose for the moment that~$\rho$ is modular, so that~$f_?$ exists.
356Choose a prime of~$\overline{\Z}$ lying
357over~$5$, and denote by~$\fbar_?$ the reduction
358of $f_?$ modulo this prime. The Eisenstein series
359$E_4\in M_4(1;\F_5)$  is congruent to~$1$ modulo~$5$, so
360$E_4\cdot{}\fbar_?\in S_5(N,\eps;\Fbar_5)$ has the same $q$-expansion
361as $\fbar_?$.  Using a computer, we can search for a
362form $f\in S_5(N,\eps,\Fbar_5)$ that has the same
363$q$-expansion as the conjectural form $E_4\cdot{}\fbar_?$.
364
365Instead of multiplying $\fbar_?$ by~$E_4$, we could have multiplied
366it by an  Eisenstein series of weight~$1$, level~$5$, and character $\eps'$.
367We used $E_4$ because the dimension of $S_5(N,\eps;\Fbar_5)$
368is~$696$ whereas the dimension of the relevant space
369$S_2(5\cdot 1376, \eps_{43})$ of weight~$2$ cusp forms is~$1040$.
370
371\subsection{Searching for the newform~$f$}
372Using modular symbols (see Section~\ref{sec:modsym}) we
373compute (at least up to semi-simplification) the space
374$S_5(1376,\eps;\F_{25})$. Note that there is injective map
375from the image of~$\eps$ into $\F_{25}^*$.  By computing
376the kernels of various Hecke operators on this space, we find~$f$.
377In the following computations, we represent nonzero elements of~$\F_{25}$
378as powers of a generator~$\alp$ of~$\F_{25}^*$, which satisfies
379$$\alp^2 + 4\alp + 2=0.$$
380Our character $\eps_{43}$ was represented
381as the map sending $(1,3)\in(\Z/2^5\Z)^*\cross(\Z/43\Z)^*$ to
382$2\alp+1$. Note that~3 is a primitive root mod~43, and that $2\alp+1$
383has order~3.
384
385If the least common multiple of the degrees of the factors of
386the polynomial~$h$ modulo an
387unramified prime~$p$ is~$2$, then $\Frob_p\in\Gal(K/\Q)$
388has order~$2$.  The minimal polynomial of $\rho(\Frob_p)\in\GL_2(\C)$
389is then $x^2-1$, so $\rho(\Frob_p)$ has trace~$0$.
390The first three primes $p \nmid 5\cdot 1376$ such
391that $\rho(\Frob_p)$ has  order~$2$ are $p= 19,31,97$.
392We computed the mod~$5$ reduction $\sS_5(1376,\eps;\F_{25})^{+}$
393of the $\Z_5[\zeta_3]$-lattice of
394modular symbols of level~$1376$ and
395character~$\tilde{\eps}$ where
396complex conjugation acts as $+1$.
397Here~$\tilde{\eps}$ denotes the Teichm\"uller lift of~$\eps$.
398
399Let~$V$ be the intersection of the kernels of $T_{19}$, $T_{31}$, and
400$T_{97}$ inside of the space $\sS_5(1376,\eps;\F_{25})^{+}$ of mod~5
401modular symbols.
402The space~$V$ is $8$-dimensional,
403and no doubt all the eigenforms in this space give rise to~$\rho$ or one
404of its twists. One of the eigenvalues of~$T_3$ on this space
405is~$\alp^{16}$, and the kernel $V_1$ of $T_3-\alp^{16}$ is $2$-dimensional
406over $\F_{25}$. The Hecke operator~$T_5$ acted as a diagonalisable matrix on
407$V_1$, with eigenvalues $\alp^{10}$ and $\alp^{22}$, so the corresponding
408two systems of eigenvalues must correspond to mod~$5$ modular eigenforms,
409and furthermore we must have found all mod~$5$ modular eigenforms
410of this level, weight and character,
411such that $a_{19}=a_{31}=0$ and $a_3=\alp^{16}$.
412
413\begin{remark}
414The careful reader might wonder how we know that the
415systems of mod~$5$ eigenvalues really do correspond to mod~$5$ modular
416forms, and not to perhaps some strange mod~$5$ torsion in the space of
417modular symbols. However, we eliminated this possibility by
418computing the dimension of the full space of mod~$5$ modular
419symbols where complex conjugation acts as~$+1$, and checking that it
420equals $696$, the dimension of $S_5(1376,\tilde{\eps},\C)$, which we
421computed using the formula in \cite{cohen-oesterle}.
422\edit{
423\kevin{I still have never seen this
424     paper, so can't give a precise reference. Did you check it all in magma?
425     My pari port gives 696 :-) :-) :-)}
427 I have carefully re-checked all of these numbers in Magma, using
428Cohen-Oesterle.  See the end of this tex file for the Magma code.}
429}
430\end{remark}
431
432Let~$f$ be the eigenform in~$V_1$ that satisfies
433$a_5=\alp^{22}$; the $q$-expansion of~$f$ begins
434$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 435 + \alp^{14}q^9 + 4q^{11}+\cdots.$$
436Further eigenvalues are given in Table~\ref{table:1376} below.
437The primes~$p$ in the table such that~$a_p=0$ are
438exactly those
439predicted by considering the splitting behavior of~$h$.
440This is strong evidence that~$\rho$ is modular,
441and also that our modular symbols algorithm have been correctly
442implemented.
443\newpage
444
445\begin{center}
446\label{table:1376}
447{\bf Table~\ref{table:1376}}
448$$\begin{array}{|rl|}\hline 4492&0\\ 4503&\alpha^{16}\\ 4515&\alpha^{22}\\ 4527&\alpha^{14}\\ 45311&4\\ 45413&\alpha^{14}\\ 45517&\alpha^{14}\\ 45619&0\\ 45723&\alpha^{16}\\ 45829&\alpha^{8}\\ 45931&0\\ 46037&\alpha^{10}\\ 46141&1\\ 46243&\alpha^{10}\\ 46347&1\\ 46453&\alpha^{22}\\ 46559&4\\ 46661&\alpha^{14}\\ 46767&\alpha^{4}\\ 46871&\alpha^{20}\\ 46973&\alpha^{2}\\ 470\hline\end{array} 471\begin{array}{|rl|}\hline 47279&\alpha^{20}\\ 47383&\alpha^{4}\\ 47489&\alpha^{10}\\ 47597&0\\ 476101&\alpha^{8}\\ 477103&\alpha^{14}\\ 478107&0\\ 479109&\alpha^{10}\\ 480113&2\\ 481127&0\\ 482131&2\\ 483137&0\\ 484139&\alpha^{22}\\ 485149&\alpha^{4}\\ 486151&1\\ 487157&\alpha^{14}\\ 488163&0\\ 489167&\alpha^{22}\\ 490173&4\\ 491179&\alpha^{2}\\ 492181&\alpha^{14}\\ 493\hline\end{array} 494\begin{array}{|rl|}\hline 495191&\alpha^{10}\\ 496193&4\\ 497197&0\\ 498199&3\\ 499211&0\\ 500223&0\\ 501227&\alpha^{10}\\ 502229&0\\ 503233&\alpha^{14}\\ 504239&0\\ 505241&\alpha^{2}\\ 506251&\alpha^{2}\\ 507257&3\\ 508263&\alpha^{16}\\ 509269&2\\ 510271&\alpha^{8}\\ 511277&0\\ 512281&\alpha^{16}\\ 513283&0\\ 514293&3\\ 515307&\alpha^{4}\\ 516\hline\end{array} 517\begin{array}{|rl|}\hline 518311&\alpha^{22}\\ 519313&0\\ 520317&0\\ 521331&\alpha^{14}\\ 522337&0\\ 523347&\alpha^{16}\\ 524349&\alpha^{4}\\ 525353&0\\ 526359&0\\ 527367&\alpha^{22}\\ 528373&0\\ 529379&3\\ 530383&3\\ 531389&1\\ 532397&\alpha^{16}\\ 533401&0\\ 534409&2\\ 535419&3\\ 536421&\alpha^{20}\\ 537431&4\\ 538433&\alpha^{4}\\ 539\hline\end{array} 540\begin{array}{|rl|}\hline 541439&\alpha^{20}\\ 542443&0\\ 543449&0\\ 544457&0\\ 545461&0\\ 546463&\alpha^{10}\\ 547467&0\\ 548479&0\\ 549487&\alpha^{8}\\ 550491&\alpha^{2}\\ 551499&\alpha^{20}\\ 552503&\alpha^{2}\\ 553509&\alpha^{8}\\ 554521&\alpha^{10}\\ 555523&\alpha^{14}\\ 556541&\alpha^{20}\\ 557547&\alpha^{22}\\ 558557&3\\ 559563&1\\ 560569&\alpha^{16}\\ 561571&\alpha^{22}\\ 562\hline\end{array} 563\begin{array}{|rl|}\hline 564577&\alpha^{14}\\ 565587&\alpha^{20}\\ 566593&0\\ 567599&\alpha^{22}\\ 568601&0\\ 569607&\alpha^{16}\\ 570613&2\\ 571617&0\\ 572619&\alpha^{20}\\ 573631&\alpha^{20}\\ 574641&4\\ 575643&1\\ 576647&4\\ 577653&1\\ 578659&\alpha^{14}\\ 579661&2\\ 580673&\alpha^{8}\\ 581677&4\\ 582683&0\\ 583691&\alpha^{16}\\ 584701&\alpha^{14}\\ 585\hline\end{array} 586\begin{array}{|rl|}\hline 587709&4\\ 588719&\alpha^{4}\\ 589727&4\\ 590733&0\\ 591739&2\\ 592743&\alpha^{22}\\ 593751&\alpha^{4}\\ 594757&\alpha^{2}\\ 595761&\alpha^{2}\\ 596769&0\\ 597773&0\\ 598787&\alpha^{20}\\ 599797&\alpha^{16}\\ 600809&3\\ 601811&\alpha^{16}\\ 602821&2\\ 603823&\alpha^{10}\\ 604827&\alpha^{10}\\ 605829&\alpha^{22}\\ 606839&0\\ 607853&\alpha^{14}\\ 608\hline\end{array} 609\comment{ 610\begin{array}{|rl|}\hline 611857&0\\ 612859&0\\ 613863&\alpha^{4}\\ 614877&\alpha^{8}\\ 615881&1\\ 616883&0\\ 617887&2\\ 618907&0\\ 619911&1\\ 620919&1\\ 621929&0\\ 622937&\alpha^{2}\\ 623941&\alpha^{4}\\ 624947&2\\ 625953&\alpha^{8}\\ 626967&4\\ 627971&\alpha^{2}\\ 628977&0\\ 629983&0\\ 630991&3\\ 631997&3\\ 632\hline\end{array}} 633$$
634\end{center}
635
636\subsection{Twisting into $\GL(2,\F_5)$}
637Although there is a representation
638$\rho_f:\GQ\ra\GL(2,\F_{25})$ attached to $f$,
639it is difficult to say anything about its image without further
640work. We use a trick to show that the image of $\rho_f$ is small.
641Firstly, for a character~$\chi:\GQ\to\Fbar_5$, let~$\tilde\chi$
642denote its Teichm\"uller lift to~$\Qbar_5$. By a result of Carayol,
643there is a characteristic 0 eigenform
644$\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$ lifting $f$.
645The twist $\tilde{g}=\tilde{f} \tensor \tilde{\eps}_{43}$ is, by
646\cite[Prop. 3.64]{shimura:intro}, an eigenform in
647$S_5(43N, \tilde{\eps}_2; \Qbar_5)$, and its reduction is
648a form $g\in S_5(43N,\eps_2,\F_{25})$.
649The eigenvalues $a_p(g) = a_p(f) \eps_{43}(p)$, for
650$p\nmid 5N$, are given in Table~\ref{table:1376twist}.
651\begin{center}\label{table:1376twist}
652{\bf Table~\ref{table:1376twist}}
653$$654\begin{array}{|rl|}\hline 6552&*\\%0\\ 6563&1\\ 6575&*\\%3\\ 6587&2\\ 65911&4\\ 66013&2\\ 66117&2\\ 66219&0\\ 66323&1\\ 66429&1\\ 66531&0\\ 66637&3\\ 66741&1\\ 66843&*\\%0\\ 66947&1\\ 67053&2\\ 671\hline\end{array} 672\begin{array}{|rl|}\hline 67359&4\\ 67461&2\\ 67567&4\\ 67671&4\\ 67773&3\\ 67879&4\\ 67983&4\\ 68089&3\\ 68197&0\\ 682101&1\\ 683103&2\\ 684107&0\\ 685109&3\\ 686113&2\\ 687127&0\\ 688131&2\\ 689\hline\end{array} 690\begin{array}{|rl|}\hline 691137&0\\ 692139&2\\ 693149&4\\ 694151&1\\ 695157&2\\ 696163&0\\ 697167&2\\ 698173&4\\ 699179&3\\ 700181&2\\ 701191&3\\ 702193&4\\ 703197&0\\ 704199&3\\ 705211&0\\ 706223&0\\ 707\hline\end{array} 708\begin{array}{|rl|}\hline 709227&3\\ 710229&0\\ 711233&2\\ 712239&0\\ 713241&3\\ 714251&3\\ 715257&3\\ 716263&1\\ 717269&2\\ 718271&1\\ 719277&0\\ 720281&1\\ 721283&0\\ 722293&3\\ 723307&4\\ 724311&2\\ 725\hline\end{array} 726\begin{array}{|rl|}\hline 727313&0\\ 728317&0\\ 729331&2\\ 730337&0\\ 731347&1\\ 732349&4\\ 733353&0\\ 734359&0\\ 735367&2\\ 736373&0\\ 737379&3\\ 738383&3\\ 739389&1\\ 740397&1\\ 741401&0\\ 742409&2\\ 743\hline\end{array} 744\begin{array}{|rl|}\hline 745419&3\\ 746421&4\\ 747431&4\\ 748433&4\\ 749439&4\\ 750443&0\\ 751449&0\\ 752457&0\\ 753461&0\\ 754463&3\\ 755467&0\\ 756479&0\\ 757487&1\\ 758491&3\\ 759499&4\\ 760503&3\\ 761\hline\end{array} 762\begin{array}{|rl|}\hline 763509&1\\ 764521&3\\ 765523&2\\ 766541&4\\ 767547&2\\ 768557&3\\ 769563&1\\ 770569&1\\ 771571&2\\ 772577&2\\ 773587&4\\ 774593&0\\ 775599&2\\ 776601&0\\ 777607&1\\ 778613&2\\ 779\hline\end{array} 780\begin{array}{|rl|}\hline 781617&0\\ 782619&4\\ 783631&4\\ 784641&4\\ 785643&1\\ 786647&4\\ 787653&1\\ 788659&2\\ 789661&2\\ 790673&1\\ 791677&4\\ 792683&0\\ 793691&1\\ 794701&2\\ 795709&4\\ 796719&4\\ 797\hline\end{array} 798\comment{ 799\begin{array}{|rl|}\hline 800727&4\\ 801733&0\\ 802739&2\\ 803743&2\\ 804751&4\\ 805757&3\\ 806761&3\\ 807769&0\\ 808773&0\\ 809787&4\\ 810797&1\\ 811809&3\\ 812811&1\\ 813821&2\\ 814823&3\\ 815827&3\\ 816\hline\end{array} 817\begin{array}{|rl|}\hline 818829&2\\ 819839&0\\ 820853&2\\ 821857&0\\ 822859&0\\ 823863&4\\ 824877&1\\ 825881&1\\ 826883&0\\ 827887&2\\ 828907&0\\ 829911&1\\ 830919&1\\ 831929&0\\ 832937&3\\ 833941&4\\ 834\hline\end{array} 835\begin{array}{|rl|}\hline 836947&2\\ 837953&1\\ 838967&4\\ 839971&3\\ 840977&0\\ 841983&0\\ 842991&3\\ 843997&3\\ 844&\\ 845&\\ 846&\\ 847&\\ 848&\\ 849&\\ 850&\\ 851&\\ 852\hline\end{array}} 853$$
854\end{center}
855
856\begin{proposition}\label{prop:1376-g}
857Let $g=f\tensor \eps_{43}$.  Then $a_p(g)\in \F_5$
858for all~$p\nmid \ell N$.
859\end{proposition}
860\begin{proof}
861Consider an eigenform $\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$
862lifting~$f$ as above.
863Associated to~$\tilde{f}$ there is an automorphic
864representation~$\pi=\tensor_v'\pi_v$ of $\GL(2,\bA)$, where~$\bA$
865is the ad\{e}le ring of~$\Q$.
866Because $43\mid\mid N$, and~$43$ divides the conductor
867of $\eps$, we see that the local component $\pi_{43}$ of $\pi$ at
868$43$ must be ramified principal series. By Carayol's theorem,
869$\rho_{\tilde{f}}|_{D_{43}} \sim 870 \abcd{\Psi_1}{0}{0}{\Psi_2}$
871with, without loss of generality,~$\Psi_2$ unramified.  We have
872$(\Psi_1\cdot \Psi_2)|_{I_{43}}=\tilde{\eps}|_{I_{43}}=\tilde{\eps}_{43}$,
873therefore, $\rho_{\tilde{f}}|_{I_{43}} \sim 874 \abcd{\tilde{\eps}_{43}}{0}{0}{1}$.
875
876Now twist~$\tilde{f}$ by $\tilde{\eps}_{43}^{-1}$; we find that
877$\rho_{\tilde{f}\tensor\tilde{\eps}_{43}^{-1}}|_{I_{43}} \sim 878 \abcd{1}{0}{0}{\tilde{\eps}^{-1}_{43}}$.
879In particular, there is an
880eigenform~$\tilde{f}'\in S_5(N,\tilde{\eps}_2\tilde{\eps}^{-1}_{43},\Qbar_5)$
881whose associated Galois representation is the twist by $\tilde{\eps}^{-1}_{43}$
882of that of $\tilde{f}$ (recall that $N=1376$ and so~$43$ divides~$N$
883exactly once). Let~$f'$ denote the mod~$5$ reduction of~$\tilde{f}'$. Then
884one checks easily that $f'\in S_5(N,\eps_2\eps^{-1}_{43},\F_{25})=S_5(N,\eps^5,\F_{25})$.
885
886For all primes $p\nmid5N$ we have $a_p(f')=\eps_{43}(p)^{-1}a_p(f)$.
887In particular, we have $a_p(f')=0$ for
888$p=19,31$.
889Also, $\eps_{43}(3)=\alp^8$ and $\eps_{43}(5) =\alp^8$, so
890$$a_3(f')=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
891$$a_5(f')=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5.$$
892Now if $\sigma$ is the non-trivial automorphism of $\F_{25}$,
893then $\sigma(f')$ and $f$ both lie in
894$S_5(1376,\eps;\F_{25})$ and have same~$a_p$ for
895$p=3,5,19,31$, so they are equal because we found~$f$
896by computing the unique eigenform with given~$a_p$ for $p=3,5,19,31$.
897So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
898Thus for all $p\nmid 5N$, we see that
899$a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
900$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10} 901 = a_p(f) \eps_{43} = a_p(g)$.
902\end{proof}
903
904\subsection{Proof that~$\rho_g$ is unramified at~$5$}
905We begin with a generalisation of~\cite{sturm:cong}.\edit{\kevin{Are you happy
906to put this lemma here? I expect you're just relieved to see the
907lemma at all :-)}
908\william{It fits naturally here.  And {\em yes}, I am very relieved
909to see the lemma at all!}}
910Let $M>4$ be an integer, and let $h=\sum_{n\geq1}c_nq^n$ be a
911normalised cuspidal eigenform
912of some weight~$k\geq1$, level~$M$ and character~$\chi$, defined over some
913field of characteristic not dividing~$M$.
914Let~$I$ be the set of primes~$p$
915dividing~$M$ such that~$h$ is $p$-new, and
916such that either~$p$ does not divide $M/\cond(\chi)$, or~$p$
917divides~$M$ exactly once. Let~$C$ denote
918the orbit of the cusp~$\infty$ in $X_1(M)$
919under the action of the group generated by $w_p$ for $p\in I$, and
920the diamond operators $\langle d\rangle_M$. By Corollary~4.6.18
921of \cite{miyake}\edit{\kevin{don't know the publisher/year and don't have
922it with me, sorry. Working offline and at home is hard!}\william{Fixed.}},
923we see that the first~$t$ terms of the $q$-expansion
924of~$h$ at any cusp in~$C$ are determined by~$M$,~$k$, $\chi$, $c_p$
925for~$p$ in~$I$, and $c_n$ for $1\leq n\leq t$. The size of~$C$ is
926$\phi(M).2^{|I|-1}.$ The usefulness of this result is that
927if $h_1$ and $h_2$ are two normalised eigenforms of the same level,
928weight and character as above, both new at all primes in~$I$,
929and the coefficients of $q^n$ in the
930$q$-expansions of $h_1$ and $h_2$ agree for $n\in I$ and $n\leq t$,
931then $h_1-h_2$ has a zero of order at least $t+1$ at all cusps in $C$,
932and in particular if $\phi(M).2^{|I|-1}(t+1)>k/12[\SL_2(\Z):\Gamma_1(M)]$
933then $h_1=h_2$. Using the fact that $[\Gamma_0(M):\Gamma_1(M)]=\phi(M)/2$,
934we deduce
935\begin{lemma}\label{lem:bound}
936Let $h_1$ and $h_2$ be two normalised eigenforms as above.
937If the coefficients of $q^n$ in the $q$-expansions of $h_1$ and $h_2$ agree
938for all primes in $I$ and for all
939$n\leq\frac{k}{12}[\SL_2(\Z):\Gamma_0(M)]/2^{|I|}$ then $h_1=h_2$.
940\end{lemma}\edit{\kevin{\bf William: I have been a
941bit vague here---but tell me whether this
942result (a) looks right and (b) looks useful, and then we can go from
943there. I could strengthen it a little in the case when $p$ is in $I$
944and $p^2$ divides $M$ but I don't know whether this is necessary. Note that
945it gives exactly what Sturm says in the squarefree case. Note also that
946it's easy to check that e.g. our first form is new at 43 because the
947char is non-trivial! I could give details of Miyake's argument (which
948is in fact over $\C$ but which I believe I can generalise to an arbitrary
949field if necessary) if you want.}
950\william{
951I think that both (a) and (b) are true.
952I would like to see details of Miyake's argument over an arbitrary field.
953Even if we don't put them in the paper, I need to understand them
955also use this result to improve my MAGMA software.
956
957Am I mistaken in believing that multiplicity one'' implies that
958the forms we consider are all new?
959}}
960
961We now go back to the explicit situation we are concerned with.
962Although~$g$ is an eigenform of level $59168=2^5\cdot 43^2$,
963we can still consider the corresponding representation
964$\rho_g :\GQ\ra \GL(2,\F_5)$, and then directly analyze
965its ramification.
966\begin{proposition}
967The representation~$\rho_g$ is unramified at~$5$.
968\end{proposition}
969\begin{proof}
970Continuing the modular symbols computations as above,
971we find that~$V_1$ is spanned by the two eigenforms
972\begin{align*}
973f\,\,&=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
974 + \alp^{14}q^9 + 4q^{11}+\cdots\\
975f_1&=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7
976 + \alp^{14}q^9 + 4q^{11}+\cdots.
977\end{align*}
978For $p\neq 5$ and $p\leq 997$, we have $a_p(f_1)=a_p(f)$.
979To check that $a_p(f) = a_p(f_1)$ for all $p\neq 5$,
980it suffices to show that the difference~$f-f_1$ has
981$q$-expansion involving only powers of~$q^5$;
982for this we use the $\theta$-operator
983$q\frac{d}{dq}:S_5(1376,\eps,\F_{25})\ra S_{11}(1376,\eps;\F_{25})$.
984Since~$\theta$ sends normalized eigenforms to normalized eigenforms,
985it suffices to check that the subspace of
986$S_{11}(1376,\eps;\F_{25})$ generated by~$\theta(f)$
987and~$\theta(f_1)$ has dimension~$1$.
988Lemma~\ref{lem:bound} implies that it suffices to verify that the
989coefficients $a_p(\theta(f))$ and $a_p(\theta(f_1))$ are equal for all
990$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(1376)]\cdot \frac{1}{4} 991 = 484.$$
992The eigenform~$f$ must be new because we computed it by finding
993the intersections of the kernels of Hecke operators $T_p$ with
994$p\nmid 1376$; if~$f$ were an oldform then the intersection of the
995kernels of these Hecke operators
996would necessarily have dimension greater than~$1$.
997Because it takes less than a second
998to compute each $a_p(\theta(f))$, we were easily able to verify that the
999space generated by $\theta(f)$ and $\theta(f_1)$ has dimension~$1$.
1000
1001\begin{remark}
1002It is possible to avoid appealing to Lemma~\ref{lem:bound} by using
1003one of the following two alternative methods:
1004\begin{enumerate}
1005\item Define~$\theta$ directly on modular symbols and compute it.
1006\item Compute the intersection
1007  $$\bigcap_{p\geq 2} \ker(T_p - pa_p(f)) 1008 \subset S_{11}(1376,\eps;\F_{25}).$$
1009  Since~$\theta(f)$ and~$\theta(f_1)$ both lie
1010  in the intersection, the moment the dimension
1011  of a partial intersection is~$1$, it follows
1012  that $\theta(f-f_1)=0$.
1013\end{enumerate}
1014We successfully carried out both alternatives.
1015For the first, we showed that~$\theta$ on modular symbols is
1016induced by multiplication by
1017$X^5Y - Y^5X$.
1018For the second, we find that after intersecting
1019kernels for $p\leq 11$, the dimension is already~$1$.
1020The first of these two methods took much less
1021time than the second.
1022\end{remark}
1023\edit{\kevin{Have I done this? That is, is my result strong enough?
1024The point is that for arbitrary forms on $\Gamma_1(N)$ you have to
1025check a long way but for 2 forms which are eigenforms for the diamond
1026ops with the same character you can check
1027they're the same just by using the $\Gamma_0(N)$ bounds, because if the
1028q-exps agree at one cusp then they agree at lots of cusps
1029($\phi(N)/2$ cusps, to be precise, the orbit of infinity under the
1030diamond operators) Is this all you needed? It's kind
1031of like why you don't need to deal with a huge huge space when working
1032with modular symbols for $\Gamma_1(N)$ with a character. Anyway, if my
1033result is strong enough then perhaps this section needs mild rewriting.}
1034}
1035
1036Next we use that $\theta(f-f_1)=0$ to show that $\rho_g$ is unramified,
1037thus finishing the proof of the proposition.
1038Since~$f$ is ordinary, Deligne's theorem (see~\cite[\S12]{gross:tameness})
1039implies that
1040$$\rho_f|_{D_5}\sim 1041 \mtwo{\alp}{*}{0}{\beta}\qquad\text{over \Fbar_5}$$
1042with both~$\alp, \beta$ unramified,
1043$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$, and
1044$\beta(\Frob_5)=\alp^{22}$.
1045Since $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
1046$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
1047with
1048$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
1049$\beta'(\Frob_5)=\alp^{10}$;
1050in particular, $\alp'=\beta$.
1051Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so
1052$\rho_f|_{D_5}\sim\alp\oplus\beta$ and hence there is a choice
1053of basis so that $*=0$.
1054
1055\end{proof}
1056
1057
1058
1059\subsection{The image of $\proj \rho_g$}
1060\begin{proposition}
1061The image of $\proj \rho_g$ is $A_5$.
1062\end{proposition}
1063\begin{proof}
1064The image~$H$ of $\proj \rho_g$ in $\PGL_2(\F_5)$ is easily checked to
1065lie in $\PSL_2(\F_5)\cong A_5$ because of what we know about the determinant
1066of $\rho_g$. Hence $H$ is a subgroup of $A_5$ that
1067contains an element of order~$2$ (complex conjugation)
1068and an element of order~$3$ (for example, $\rho_g(\Frob_7)$ has
1069characteristic polynomial $x^2-2x-1$).
1070This proves that~$H$ is isomorphic to either~$S_3$,~$A_4$, or~$A_5$.
1071Let $L$ be the number field cut out by~$H$.
1072If~$L$ were an $S_3$ extension, then there would be a
1073quadratic extension contained in it which is unramified
1074outside $2\cdot 5\cdot 43$; it is furthermore unramified at~$5$ by
1075the previous section and unramified at $43$ because $I_{43}$
1076has order~$3$.  Thus it is one of the three quadratic
1077fields unramified outside~$2$. In particular, the trace of $\Frob_p$
1078would be zero for all primes in a certain congruence class
1079modulo~8.
1080However, there are primes~$p$ congruent to $3$, $5$, and $7$
1081mod $8$ such that $a_p(g)\neq 0$, e.g., $3$, $7$, and $13$.
1082
1083If $H$ were isomorphic to $A_4$,
1084then let~$M$ denote the cyclic extension of degree~3 over~$\Q$ contained
1085in~$L$. Now~$M$ is unramfied at~2 and~5, and hence is the subfield
1086of $\Q(\zeta_{43})$ of degree~3.
1087Choose $p\nmid 1376\cdot 5$ that is inert in~$M$.  The
1088order of $\rho_g(\Frob_p)$ in $\GL_2(\F_5)$ must be divisible
1089by~$3$.  However, a quick check using
1090Table~\ref{table:1376twist} shows that this is
1091usually not case, even for $p=3$.
1092\end{proof}
1093
1094
1095\subsection{Bounding the ramification at~$2$ and~$43$}
1096Let~$L$ be the fixed field of $\ker(\proj(\rho_g))$. We have just
1097shown that $\Gal(L/\Q)$ is isomorphic to $A_5$.
1098By a root field for~$L$, we mean
1099a non-Galois extension of $\Q$ of degree~5 whose Galois closure is~$L$.
1100\begin{proposition}
1101The discriminant of a root
1102field for~$L$ divides $(43\cdot 8)^2=344^2$, and
1103in particular,~$L$ must be mentioned in Table~1
1104of \cite[pg 122]{freyetal}.\end{proposition}
1105\begin{proof}
1106The analysis of the local behavior of~$\rho_f$ at~$43$ given in
1107Proposition~\ref{prop:1376-g}
1108shows that the inertia group at~$43$ in $\Gal(L/\Q)$ has order~$3$. Using
1109Table~3.1 of~\cite{buhler:thesis}, we see that if
1110$\Gal(L/\Q)\isom A_5$
1111then it must be type~$2$'' at 43, and hence the discriminant of a root
1112field of~$L$, that is, of a non-Galois extension of~$\Q$ of degree~$5$
1113whose Galois closure is~$L$, must be $43^2$ at~$43$.
1114
1115At~$2$ the behavior of~$\rho$ is more subtle and we shall not analyze
1116it fully. But we can say that, because~$\rho$ has arisen from
1117a form of level $1376=2^5.43$, we must be either of type~$5$
1118or one of types~$14$--$17$. In particular, the discriminant at~$2$ of a root
1119field for~$L$ will be at most~$2^6$.
1120
1121Finally,~$L$ is unramified at all other primes, because~$\rho$ is.
1122Hence the discriminant of a root field for~$L$, assuming that
1123$\Gal(L/\Q)\cong A_5$, divides $(43.8)^2=344^2$.
1124\end{proof}
1125
1126We know that~$L$ is an icosahedral extension of~$\Q$ with
1127discriminant dividing $43^2\cdot 2^6$.  Table~1 of \cite[pg 122]{freyetal}
1128contains all icosahedral extensions, such that the discriminant
1129of a root field is bounded by $2083^2$.  The table
1130must contain~$L$; there is only one icosahedral extension with
1131discriminant dividing $43^2\cdot 2^6$, so $L=K$.
1132
1133\subsection{Obtaining a classical weight one form}
1134We have shown that a twist of the icosahedral
1135representation $\rho:\GQ\ra\GL(2,\C)$,
1136obtained by lifting $\GQ\ra \Gal(K/\Q)\ncisom A_5$,
1137has a mod~$5$ reduction $\rho_g:\GQ\ra \GL_2(\F_5)$ that
1138is modular.  Since~$\rho$ ramifies at only finitely many primes,
1139and~$\rho$ can be chosen to be unramified at~$5$ with distinct eigenvalues,
1140\cite{buzzard-taylor} implies that~$\rho$ arises from
1141a classical weight~$1$ newform.\edit{\kevin{You write say something
1142about choice of iso $\overline{\Q}_5=\C$'' but I think that we can just
1143say nothing.}\william{I'm happy with saying nothing, as the reader
1144will know what needs to be said upon looking at [BT].}}
1145
1146
1147
1148
1149\section{More examples}
1150\subsection{Conductor~$2416=2^4\cdot 151$}
1151\label{sec:2416}
1152Consider the icosahedral extension~$K$ defined by $h=x^5-2x^3+2x^2+5x+6$.
1153A Frobenius element at~$5$ has order~$2$.
1154The first three primes~$p$ such that $a_p=0$ are
1155$53$, $97$, and $127$.
1156The type at~$2$ is~16, and the type at~$151$ is~2.
1157The order of $\eps_{151}$ is~$3$, and a local computation would tell us
1158what $\eps_2$ was but we preferred to guess---if we guessed wrong then
1159we would almost certainly find no forms and we could just guess again.
1160We guessed that $\eps_2$ was the character of conductor~$4$, and this
1161turned out to be correct.
1162
1163Using \cite{cohen-oesterle}\edit{\kevin{Computing in general is currently
1164a real pain for me. I am currently spending as much time as possible
1165at home (Joel is too young to be put in a creche at the minute)
1166and hence am (a) using Windows 98 and (b) usually offline (it's
1168to this is the intense irritation caused by the fact that my office in
1169London is being used by someone who has rebooted the computer into
1170Windows so I can't telnet into it, and you can see why doing any computations
1171at all is a real pain! Fortunately I have gp and I have ported all the
1172DimensionCuspForms stuff (it's a pain because evaluate'' typically gives
1173complex numbers to 30dps!!) and indeed it's 1210. What's the precise
1174reference to [CO] by the way?}} we find $\dim S_5(2416,\eps)=1210$; this
1175is the same as the dimension obtained using modular symbols, so
1176there is no spurious $5$-torsion.
1177We compute the space of modular symbols corresponding to
1178$S_5(2416,\eps;\F_{25})$, and then the $8$-dimensional
1179intersection~$V$ of the kernels of~$T_{53}$, $T_{97}$, and~$T_{127}$.
1180The characteristic polynomial of~$T_3$ is
1181$(x+2)^4(x+3)^4$.  The kernel~$V_1$ of $T_3+2$ has dimension~$4$.
1182The characteristic polynomial of~$T_7$ on~$V_1$
1183is $(x+\alp^4)^4$; the kernel~$V_2\subset V_1$ of $T_7+\alp^4$ has
1184dimension~$2$.
1185The characteristic polynomial of~$T_5$ on~$V_2$ is
1186$(x+\alp^{10})(x+\alp^{22})$.
1187Let~$f$ be the newform such that $a_5+\alp^{10}=0$.
1188We have
1189$$f = q +3q^3 + \alp^{22}q^5 + \alp^{16}q^7 + \alp^{4} q^{11} + \cdots.$$
1190
1191There is one other newform $f'$ in~$V_2$, and it is a companion of~$f$.
1192We verified this using Lemma~\ref{lem:bound} by
1193checking that $a_p(f)=a_p(f')$ for those primes~$p\neq 5$ such that
1194$$p \leq 1195\frac{11}{12}\cdot 1196 [\SL_2(\Z):\Gamma_0(2416)]\cdot \frac{1}{4} = 836.$$
1197
1198
1199Let $g=f\tensor\eps_{151}$.
1200We have $a_p(f\tensor\eps_{151}^{-1})=0$ for $p=53,97,127$.
1201Also,
1202$\eps_{151}(3)=1$,
1203$\eps_{151}(5)=\alp^8$,
1204$\eps_{151}(7) =\alp^{20}$, so
1205$a_3(f\tensor\eps_{151}^{-1})=3/1=3=3^5$,
1206$a_5(f\tensor\eps_{151}^{-1})= \alp^{22}/\alp^8 = \alp^{14}=(\alp^{22})^5$,
1207$a_7(f\tensor\eps_{151}^{-1})= \alp^{16}/\alp^{8} = \alp^{8} = (\alp^{16})^5$.
1208Since $f\tensor\eps_{151}^{-1}$ and~$\sigma(f)$ both lie
1209in $S_5(2416,\eps_2\eps_{151}^2;\F_{25})$ and have same~$a_p$ for
1210$p=3,5,7,53,97,127$, they are equal.
1211Exactly as in Proposition~\ref{prop:1376-g} we see
1212that $g\in S_5(151N,\eps_2;\F_5)$.
1213
1214We have not yet bounded ramification or checked that the image
1215of~$\rho_g$ is~$A_5$.\edit{\kevin{This shouldn't take too long but I didn't
1216want to waste time on it until you confirmed that my bound was good enough.
1217By the way, do you want to move the lemma to a different section or
1218anything?}\william{{\bf If I understand it correctly,
1219you're bound is definitely good enough.}}}
1220This computation can be done with the
1221help of the following table:
1222{\arraycolsep=.36em
1223$$\begin{array}{|rl|}\hline 12242&0\\ 12253&3\\ 12265&\alp^{22}\\ 12277&\alp^{16}\\ 122811&\alp^{4}\\ 122913&\alp^{2}\\ 123017&\alp^{22}\\ 123119&3\\ 1232\hline\end{array} 1233\begin{array}{|rl|}\hline 123423&\alp^{22}\\ 123529&3\\ 123631&\alp^{16}\\ 123737&\alp^{22}\\ 123841&2\\ 123943&\alp^{8}\\ 124047&\alp^{8}\\ 124153&0\\ 1242\hline\end{array} 1243\begin{array}{|rl|}\hline 124459&1\\ 124561&\alp^{8}\\ 124667&3\\ 124771&\alp^{8}\\ 124873&2\\ 124979&2\\ 125083&2\\ 125189&\alp^{20}\\ 1252\hline\end{array} 1253\begin{array}{|rl|}\hline 125497&0\\ 1255101&2\\ 1256103&\alp^{16}\\ 1257107&1\\ 1258109&\alp^{20}\\ 1259113&\alp^{22}\\ 1260127&0\\ 1261131&4\\ 1262\hline\end{array} 1263\begin{array}{|rl|}\hline 1264137&\alp^{20}\\ 1265139&0\\ 1266149&\alp^{22}\\ 1267151&2\\ 1268157&\alp^{22}\\ 1269163&\alp^{8}\\ 1270167&\alp^{10}\\ 1271173&0\\ 1272\hline\end{array} 1273\begin{array}{|rl|}\hline 1274179&2\\ 1275181&\alp^{2}\\ 1276191&\alp^{10}\\ 1277193&0\\ 1278197&\alp^{14}\\ 1279199&\alp^{16}\\ 1280211&0\\ 1281223&0\\ 1282\hline\end{array} 1283\begin{array}{|rl|}\hline 1284227&\alp^{20}\\ 1285229&3\\ 1286233&\alp^{22}\\ 1287239&\alp^{10}\\ 1288241&\alp^{2}\\ 1289251&\alp^{4}\\ 1290257&0\\ 1291263&\alp^{2}\\ 1292\hline\end{array} 1293\begin{array}{|rl|}\hline 1294269&0\\ 1295271&\alp^{10}\\ 1296277&\alp^{20}\\ 1297281&\alp^{16}\\ 1298283&0\\ 1299293&2\\ 1300307&\alp^{4}\\ 1301311&2\\ 1302\hline\end{array}$$}
1303
1304
1305\subsection{Conductor~$3184=2^4\cdot 199$}
1306Consider the icosahedral extension~$K$ defined by
1307$h=x^5+5x^4+8x^3-20x^2-21x-5$.
1308A Frobenius element at~$5$ has order~$2$.
1309The first~$3$ primes~$p$ such that $a_p=0$ are
1310$31$, $89$, and $97$.
1311The type at~$2$ is~$16$ and the type at~$199$ is~$2$.
1312As in Section~\ref{sec:2416}, $\eps_2$ is of
1313conductor~$4$ and order~$2$, and the order of~$\eps_{199}$ is~$3$.
1314\edit{\kevin{We don't need to
1315guess any more---I think that the calculation in the last section
1316has convinced us that type 16 is associated to char non-trivial of
1317conductor~4. Probably I should think of some way of explaining this
1318that makes us look less like idiots.}\william{I fixed this.}}
1319
1320Using \cite{cohen-oesterle}
1321we find $\dim S_5(3184,\eps)=1594$,
1322which agrees with the dimension computed using modular symbols.
1323We compute $S_5(3184,\eps;\F_{25})$ and then the
1324$8$-dimensional intersection~$V$ of the kernels of
1325$T_{31}$, $T_{89}$, and $T_{97}$.
1326The characteristic polynomial of~$T_3$ on~$V$ is
1327$(x+\alp^4)^4(x+\alp^{16})^4$.  The dimension of
1328$V_1=\ker(T_3+\alp^4)$ is~$2$. It appears that all
1329Hecke operators~$T_p$ act diagonally on $V_1$, except~$T_5$
1330which has the distinct eigenvalues~$2$ and~$3$.
1331We thus isolate a one-dimensional eigenspace, spanned by
1332a form~$f$ whose Hecke eigenvalues are:
1333{\arraycolsep=.4em
1334$$1335\begin{array}{|rl|}\hline 13362&0\\ 13373&\alpha^{16}\\ 13385&3\\ 13397&\alpha^{22}\\ 134011&3\\ 134113&\alpha^{22}\\ 134217&3\\ 134319&\alpha^{16}\\ 1344\hline\end{array} 1345\begin{array}{|rl|}\hline 134623&\alpha^{4}\\ 134729&\alpha^{16}\\ 134831&0\\ 134937&\alpha^{8}\\ 135041&\alpha^{2}\\ 135143&\alpha^{10}\\ 135247&\alpha^{4}\\ 135353&\alpha^{14}\\ 1354\hline\end{array} 1355\begin{array}{|rl|}\hline 135659&3\\ 135761&1\\ 135867&4\\ 135971&\alpha^{22}\\ 136073&\alpha^{10}\\ 136179&\alpha^{16}\\ 136283&3\\ 136389&0\\ 1364\hline\end{array} 1365\begin{array}{|rl|}\hline 136697&0\\ 1367101&2\\ 1368103&2\\ 1369107&3\\ 1370109&0\\ 1371113&\alpha^{2}\\ 1372127&0\\ 1373131&\alpha^{14}\\ 1374\hline\end{array} 1375\begin{array}{|rl|}\hline 1376137&1\\ 1377139&1\\ 1378149&\alpha^{22}\\ 1379151&\alpha^{2}\\ 1380157&0\\ 1381163&\alpha^{16}\\ 1382167&0\\ 1383173&0\\ 1384\hline\end{array} 1385\begin{array}{|rl|}\hline 1386179&\alpha^{8}\\ 1387181&2\\ 1388191&3\\ 1389193&\alpha^{14}\\ 1390197&0\\ 1391199&2\\ 1392211&4\\ 1393223&0\\ 1394\hline\end{array} 1395\begin{array}{|rl|}\hline 1396227&1\\ 1397229&\alpha^{2}\\ 1398233&0\\ 1399239&3\\ 1400241&2\\ 1401251&0\\ 1402257&0\\ 1403263&1\\ 1404\hline\end{array} 1405\begin{array}{|rl|}\hline 1406269&\alpha^{2}\\ 1407271&\alpha^{2}\\ 1408277&0\\ 1409281&1\\ 1410283&\alpha^{10}\\ 1411293&\alpha^{8}\\ 1412307&\alpha^{20}\\ 1413311&\alpha^{20}\\ 1414\hline\end{array} 1415$$}
1416
1417The bound from Lemma~\ref{lem:bound} is
1418$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3184)]\cdot 1419 \frac{1}{4} = 1100.$$
1420
1421
1422\subsection{Conductor~$3556=2^2\cdot 7\cdot 127$}
1423Consider the icosahedral extension~$K$ defined by
1424$h=x^5+3x^4+9x^3-6x^2-4x-40$.
1425A Frobenius element at~$5$ has order~$3$.
1426The first~$3$ unramified~$p$ such that $a_p=0$ are~$19$,~$29$, and~$89$.
1427The type at~$2$ is~5; the type at~$7$ is~3; the type at~$127$ is~2.
1428The order of~$\eps_{127}$ is~$3$.
1429A local analysis shows that the order
1430of~$\eps_{2}$ is~$1$ and of~$\eps_7$ is~$2$.\edit{\kevin{William: I took
1431out some bad guesses here! If the type is $\leq15$ and not 6 then
1432I think I can work out the character, and indeed I can verify that your
1433first guess is wrong!}}
1434The formula of~\cite{cohen-oesterle}
1435gives $\dim S_5(3556,\eps)=2042$, which agrees
1436with the dimension of the corresponding space of modular symbols.
1437
1438The intersection~$V$ of the kernels of~$T_{19}$,~$T_{29}$, and~$T_{89}$
1439has dimension~$8$.
1440The characteristic  polynomial of~$T_{3}$ on~$V$
1441is $(x+\alp^4)^4(x+\alp^{16})^4$,
1442and the kernel~$V_1$ of $T_3+\alp^4$ is of dimension~$2$.
1443It appears that all Hecke operators~$T_p$ act
1444diagonally on~$V_1$ except~$T_5$, whose eigenvalues
1445\edit{\kevin{How come it says $\alpha^{14}$ in
1446the table then? Are we out by a sign here?}\william{I gave the
1447signs incorrectly in this particular sentence.  The table is fine,
1448and the sentence is now fixed.}}
1449are~$\alp^{14}$ and~$\alp^{22}$. The eigenvalues of the
1450newform with $a_5 = \alp^{14}$ are given in the following table.
1451{\arraycolsep=.4em
1452$$1453\begin{array}{|rl|}\hline 14542&0\\ 14553&\alpha^{16}\\ 14565&\alpha^{14}\\ 14577&\alpha^{10}\\ 145811&\alpha^{2}\\ 145913&\alpha^{22}\\ 146017&\alpha^{14}\\ 146119&0\\ 1462\hline\end{array} 1463\begin{array}{|rl|}\hline 146423&\alpha^{10}\\ 146529&0\\ 146631&\alpha^{16}\\ 146737&\alpha^{20}\\ 146841&\alpha^{14}\\ 146943&\alpha^{2}\\ 147047&1\\ 147153&\alpha^{2}\\ 1472\hline\end{array} 1473\begin{array}{|rl|}\hline 147459&\alpha^{8}\\ 147561&4\\ 147667&\alpha^{10}\\ 147771&\alpha^{4}\\ 147873&4\\ 147979&\alpha^{16}\\ 148083&\alpha^{8}\\ 148189&0\\ 1482\hline\end{array} 1483\begin{array}{|rl|}\hline 148497&\alpha^{16}\\ 1485101&\alpha^{4}\\ 1486103&\alpha^{4}\\ 1487107&0\\ 1488109&\alpha^{14}\\ 1489113&\alpha^{4}\\ 1490127&\alpha^{20}\\ 1491131&1\\ 1492\hline\end{array} 1493\begin{array}{|rl|}\hline 1494137&2\\ 1495139&\alpha^{22}\\ 1496149&\alpha^{2}\\ 1497151&0\\ 1498157&0\\ 1499163&\alpha^{14}\\ 1500167&0\\ 1501173&\alpha^{10}\\ 1502\hline\end{array} 1503\begin{array}{|rl|}\hline 1504179&\alpha^{4}\\ 1505181&1\\ 1506191&1\\ 1507193&0\\ 1508197&\alpha^{22}\\ 1509199&\alpha^{2}\\ 1510211&\alpha^{8}\\ 1511223&0\\ 1512\hline\end{array} 1513\begin{array}{|rl|}\hline 1514227&3\\ 1515229&0\\ 1516233&0\\ 1517239&0\\ 1518241&\alpha^{10}\\ 1519251&0\\ 1520257&\alpha^{16}\\ 1521263&\alpha^{14}\\ 1522\hline\end{array} 1523\begin{array}{|rl|}\hline 1524269&\alpha^{4}\\ 1525271&\alpha^{20}\\ 1526277&\alpha^{16}\\ 1527281&0\\ 1528283&\alpha^{20}\\ 1529293&\alpha^{14}\\ 1530307&\alpha^{8}\\ 1531311&0\\ 1532\hline\end{array} 1533$$
1534}
1535By checking that the Hecke operators $T_p$ act diagonally
1536on $V_1$, for $p\neq 5$ with
1537$$p\leq\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3556)]\cdot 1538 \frac{1}{8} = 704,$$
1539we see that the other newform in~$V_1$ is a companion of~$f$.
1540
1541Let $g=f\tensor\eps_{127}$.
1542We have $a_p(f\tensor\eps_{127}^{-1})=0$ for $p=19, 29, 89.$
1543We have $\eps_{127}(3)=\alp^8$ and $\eps_{127}(5)=1$, so
1544$a_3(f\tensor\eps_{127}^{-1})= \alp^{16}/\alp^8 = \alp^{8}$,
1545and $a_{5}(f\tensor\eps_{127}^{-1})= \alp^{14}/1=\alp^{14}$.
1546At first glance this form looks like the Galois conjugate
1547of~$f$; and if it were the Galois conjugate then the form
1548twisted by the character at~$127$ will, as in
1549Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
1550for $p \nmid \ell N$ and hence~$\rho_g$
1551will go into $\GL(2,\F_5)$.  Unfortunately the
1552newform associated to the twisted form is not
1553quite the conjugate of our original form; the coefficient
1554of~$a_5$ is wrong!  The conjugate form has
1555$a_5=(\alp^{14})^5 = \alp^{22}$.
1556The other newform $f_1\in V_1$, which is a companion
1557of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
1558This proves that~$g=\sigma(f_1)$.
1559An argument as in
1560Proposition~\ref{prop:1376-g} shows that $a_p(g)\in\F_5$
1561for all $p\nmid \ell N$.
1562
1563\edit{\william{[[Kevin has an argument that
1564uses that~$\Frob_5$ has order~$3$
1565in the~$A_5$ field to show that~$\rho_f$ is unramified at~$5$.
1566Unfortunately his argument uses that the $a_p(g)$ are in~$\F_5$,
1567for $p\nmid \ell N$!
1568It might be possible to appeal to the Gross and Coleman-Voloch theorems
1569to prove that~$f_1$ is indeed a companion form.  This would
1570avoid any further computation.]]}
1571\kevin{I'm not sure I believe you.
1572You see: if we want to go down this path, we have to check that the
1573repn is unram at 5. But we don't know a darn thing about the repn yet.
1574Am I right in thinking that we are stuck in this case?}
1575\william{Given the lemma 1.3, I think we can just use the standard
1576technique and not go down this path!?}}
1577\comment{
1578\edit{\william{This shouldn't be needed.}}
1579Now $\Frob_5$ has order~$3$ in the~$A_5$
1580extension in Buhler so I think what should happen in your computations is
1581that the form of level $3556\cdot 127$
1582given by twisting the form of level~$3556$
1583will have all~$a_p$ in $\Z/5\Z$ apart from $a_5$ which will be
1584in~$\F_{25}$ and nonzero. The image of a
1585decomposition group at~$5$ can then be shown
1586to be upper-triangularisable (Deligne) and the chars on the diagonal
1587are unramified of order~$3$. So the image of a decomp group at~$5$
1588will have order $3\cdot\text{(power of$5$)}$. But there
1589are no subgroups of the subgroup of $\GL(2,\Z/5\Z)$ consisting of elements
1590of det $\pm 1$'' of order~$15$ because their image in
1591$\PSL(2,\Z/5\Z)$ would also
1592have order~$15$ and~$A_5$ has no such subgroups. So no companion form
1593argument is needed---the repn will be unramified at~$5$.
1594}
1595
1596
1597\subsection{Conductor~$3756=2^2\cdot 3\cdot 313$}
1598Consider the icosahedral extension~$K$ defined by
1599$h=x^5 - 3x^3 + 10x^2 + 30x - 18$.
1600A Frobenius element at~$5$ has order~$3$.
1601The first~$3$ unramified~$p$ such that $a_p=0$ are $17, 61, 67$.
1602The type at~$2$ is~$5$; the type at~$3$ is~$3$; the type
1603at~$127$ is~$2$.
1604The order of~$\eps_2$ is~$1$,
1605the order of~$\eps_3$ is~$2$, and
1606the order of~$\eps_{127}$ is~$3$;
1607
1608Using \cite{cohen-oesterle} we find $\dim S_5(3756,\eps)=2506$;
1609this is the same as the dimension computed using modular symbols,
1610so there is no spurious $5$-torsion.
1611We find that~$V$ has dimension~$8$,
1612and the characteristic polynomial of~$T_{11}$ is
1613$(x+\alp^{4})^4(x+\alp^{16})^4$
1614(we do not use $T_3$ since $3\mid N$).
1615The kernel~$V_1$ of $T_{11}+\alp^{4}$ has dimension~$2$.
1616The characteristic polynomial of~$T_5$ on~$V_1$
1617is $(x+\alp^2)(x+\alp^{10})$.
1618The first few~$a_p$ of the newform~$f$
1619in~$V_2$ such that $a_5+\alp^2=0$ are given in
1620the following table.
1621{\arraycolsep=.4em
1622$$\begin{array}{|rl|}\hline 16232&0\\ 16243&\alp^{14}\\ 16255&\alp^{14}\\ 16267&3\\ 162711&\alp^{16}\\ 162813&\alp^{10}\\ 162917&0\\ 163019&3\\ 1631\hline\end{array} 1632\begin{array}{|rl|}\hline 163323&3\\ 163429&\alp^{2}\\ 163531&\alp^{22}\\ 163637&\alp^{22}\\ 163741&\alp^{20}\\ 163843&\alp^{16}\\ 163947&\alp^{4}\\ 164053&4\\ 1641\hline\end{array} 1642\begin{array}{|rl|}\hline 164359&\alp^{8}\\ 164461&0\\ 164567&0\\ 164671&0\\ 164773&0\\ 164879&3\\ 164983&\alp^{10}\\ 165089&\alp^{20}\\ 1651\hline\end{array} 1652\begin{array}{|rl|}\hline 165397&4\\ 1654101&0\\ 1655103&2\\ 1656107&\alp^{4}\\ 1657109&\alp^{20}\\ 1658113&2\\ 1659127&\alp^{10}\\ 1660131&\alp^{14}\\ 1661\hline\end{array} 1662\begin{array}{|rl|}\hline 1663137&0\\ 1664139&0\\ 1665149&0\\ 1666151&2\\ 1667157&\alp^{20}\\ 1668163&1\\ 1669167&0\\ 1670173&\alp^{20}\\ 1671\hline\end{array} 1672\begin{array}{|rl|}\hline 1673179&0\\ 1674181&\alp^{14}\\ 1675191&0\\ 1676193&\alp^{20}\\ 1677197&\alp^{22}\\ 1678199&0\\ 1679211&\alp^{10}\\ 1680223&0\\ 1681\hline\end{array} 1682\begin{array}{|rl|}\hline 1683227&\alp^{8}\\ 1684229&\alp^{16}\\ 1685233&\alp^{4}\\ 1686239&\alp^{20}\\ 1687241&\alp^{10}\\ 1688251&\alp^{20}\\ 1689257&1\\ 1690263&\alp^{10}\\ 1691\hline\end{array} 1692\begin{array}{|rl|}\hline 1693269&2\\ 1694271&1\\ 1695277&3\\ 1696281&\alp^{20}\\ 1697283&2\\ 1698293&\alp^{20}\\ 1699307&1\\ 1700311&\alp^{16}\\ 1701\hline\end{array} 1702$$}
1703
1704It appears that the other newform in~$V_1$ is a companion of~$f$,
1705but we have not proved this.
1706
1707Let $g=f\tensor\eps_{313}$.
1708We have $a_p(f\tensor\eps_{313}^{-1})=0$ for $p=17, 61, 67.$
1709We have $\eps_{313}(5)=1$ and $\eps_{313}(11)=\alp^8$, so
1710$a_5(f\tensor\eps_{313}^{-1})= \alp^{14}/1 = \alp^{14}$,
1711and $a_{11}(f\tensor\eps_{313}^{-1})= \alp^{16}/\alp^{8}=\alp^8$.
1712At first glance this form looks like the Galois conjugate
1713of~$f$; and if it were the Galois conjugate then the form
1714twisted by the character at~$313$ will, as in
1715Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
1716for $p \nmid \ell N$ and hence~$\rho_g$
1717will go into $\GL(2,\F_5)$.  Unfortunately the
1718newform associated to the twisted form is not
1719quite the conjugate of our original form; the coefficient
1720of~$a_5$ is wrong!  The conjugate form has
1721$a_5=(\alp^{14})^5 = \alp^{22}$.
1722The other newform $f_1\in V_1$, which appears to be a companion
1723of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
1724This proves that $g=\sigma(f_1)$.
1725The bound from Lemma~\ref{lem:bound} is
1726$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3756)]\cdot 1727 \frac{1}{8} = 863.5,$$
1728and we are easily able to verify that $f_1$ is a companion.
1729An argument as in
1730Proposition~\ref{prop:1376-g} then shows that $a_p(g)\in\F_5$
1731for all $p\nmid \ell N$.
1732
1733For this example it remains to check
1734the discriminant bound, and surjectivity onto~$A_5$.
1735
1736\subsection{Conductor~$4108=2^2\cdot 13\cdot 79$}
1737Consider the icosahedral extension~$K$ defined by
1738$h=x^5+4x^4+3x^3+9x^2+4x+5$.
1739A Frobenius element at~$5$ has order~$3$.
1740The first~$4$ unramified~$p$ such that
1741$a_p=0$ are $17$, $23$, $31$, and $89$.
1742The type at~$2$ is~5, the type at~$13$ is~2,
1743and the type at~$79$ is~3.
1744The order of~$\eps_2$ is~$1$,
1745the order of~$\eps_{13}$ is~$3$,
1746and the order of~$\eps_{79}$ is~$2$.
1747
1748Using~\cite{cohen-oesterle} we find $\dim S_5(4108,\eps)= 2234$;
1749
1750The intersection~$V$ of the kernels of $T_{17}$, $T_{23}$,
1751$T_{31}$, and $T_{89}$ has dimension~$8$.
1752The characteristic polynomial of~$T_3$ on~$V$ is
1753$(x+\alp^4)^4(x+\alp^{16})^4$.
1754The dimension of $V_1=\ker(T_3+\alp^4)$ is~$2$.
1755The characteristic polynomial of~$T_5$ on~$V_1$ is
1756$(x+\alp^4)(x+\alp^{20})$ and all of the other~$T_p$ appear
1757to act as scalars.
1758
1759{\arraycolsep=.4em
1760$$1761\begin{array}{|rl|}\hline 17622&0\\ 17633&\alpha^{16}\\ 17645&\alpha^{11}\\ 17657&\alpha^{20}\\ 176611&\alpha^{10}\\ 176713&4\\ 176817&0\\ 176919&\alpha^{14}\\ 1770\hline\end{array} 1771\begin{array}{|rl|}\hline 177223&0\\ 177329&\alpha^{22}\\ 177431&0\\ 177537&\alpha^{22}\\ 177641&\alpha^{10}\\ 177743&\alpha^{2}\\ 177847&3\\ 177953&4\\ 1780\hline\end{array} 1781\begin{array}{|rl|}\hline 178259&\alpha^{14}\\ 178361&\alpha^{2}\\ 178467&\alpha^{10}\\ 178571&\alpha^{20}\\ 178673&2\\ 178779&3\\ 178883&3\\ 178989&0\\ 1790\hline\end{array} 1791\begin{array}{|rl|}\hline 179297&\alpha^{2}\\ 1793101&\alpha^{22}\\ 1794103&2\\ 1795107&\alpha^{16}\\ 1796109&1\\ 1797113&\alpha^{8}\\ 1798127&0\\ 1799131&4\\ 1800\hline\end{array} 1801\begin{array}{|rl|}\hline 1802137&\alpha^{8}\\ 1803139&0\\ 1804149&\alpha^{8}\\ 1805151&0\\ 1806157&4\\ 1807163&\alpha^{20}\\ 1808167&0\\ 1809173&\alpha^{14}\\ 1810\hline\end{array} 1811\begin{array}{|rl|}\hline 1812179&\alpha^{4}\\ 1813181&2\\ 1814191&\alpha^{20}\\ 1815193&0\\ 1816197&0\\ 1817199&\alpha^{20}\\ 1818211&\alpha^{4}\\ 1819223&\alpha^{4}\\ 1820\hline\end{array} 1821\begin{array}{|rl|}\hline 1822227&0\\ 1823229&0\\ 1824233&3\\ 1825239&2\\ 1826241&\alpha^{8}\\ 1827251&0\\ 1828257&\alpha^{10}\\ 1829263&\alpha^{10}\\ 1830\hline\end{array} 1831\begin{array}{|rl|}\hline 1832269&0\\ 1833271&\alpha^{10}\\ 1834277&0\\ 1835281&2\\ 1836283&\alpha^{4}\\ 1837293&\alpha^{14}\\ 1838307&4\\ 1839311&0\\ 1840\hline\end{array} 1841$$}
1842
1843The bound of Lemma~\ref{lem:bound} is
1844\edit{\kevin{Before we go any further, which forms can
1845we deal with rigorously now?}\william{All of them?}}
1846$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4108)]\cdot 1847 \frac{1}{8} = 770.$$
1848
1849\subsection{Conductor~$4288=2^6\cdot 67$}
1850Consider the icosahedral extension~$K$ defined by
1851$h=x^5+4x^4+5x^3+8x^2+3x+2$.
1852A Frobenius element at~$5$ has order~$3$.
1853The first~$3$ unramified~$p$ such that $a_p=0$ are $19$,
1854$23$, $47$.
1855The type at~$2$ is~18, and the type at~$67$ is~2.
1856The order of~$\eps_{67}$ is~$3$, so in order
1857for~$\eps$ to be odd, the order of~$\eps_2$
1858must be~$2$.  First we try letting $\eps_2$ be the character
1859of conductor~$4$.\edit{\kevin{I don't know anything
1860about type 18'' and your statement order 2'' is not enough
1861to tell me the character, so you should perhaps say what it is.
1862Oh! I've just noticed that this is wrong anyway!}}
1863
1864Using \cite{cohen-oesterle} we find that\edit{\kevin{These
1865sections are getting really tedious, aren't they. I think we should
1866perhaps consider instead summarising things more succinctly,
1867e.g. saying we tried our approach for 4 other forms and it worked/didn't
1868work.}
1869\william{
1870You may be right.  However, there is a lot of information in each
1871situation.  One wants to know how many Hecke operators are needed
1872to cut out the space, what the character is, and so on.  All of
1873this information would have to be summarized.}}
1874$\dim S_5(4288,\eps)=2164$.
1875the intersection~$V$ of the kernels of $T_{19}$, $T_{23}$, and
1876$T_{47}$ has dimension~$0$ --- we must have chosen~$\eps_2$
1877incorrectly.  Next we try with $\eps_2$ one of the characters
1878of conductor~$8$.\edit{\william{The
1879character $\eps$ is [1,2,2].}}  \edit{\kevin{William: this character
1880also has order 2!}}  The dimension is also~$2164$.  This time we are
1881able to compute the intersection~$V$, and it has dimension~$16$.
1882%There is a quadratic twist of
1883%conductor~$8$ that preserves the level, so the dimension should
1884%be~$16$.
1885\edit{\kevin{William---this is a very optimistic statement!
1886All these spaces have Hecke acting non-semi-simply so I wouldn't dare
1887to guess the dimension of the space, just the number of
1888eigenforms\ldots}\william{I removed the remark about the dimension
1889being~$16$.}}
1890
1891The characteristic polynomial of~$T_3$ on~$V$ is
1892$(x+2)^8(x+3)^8$.  The dimension of the kernel~$V_1$
1893of $T_3|_V+2$ is~$4$.  The characteristic polynomial
1894of~$T_7$ on~$V_1$ is $(x+\alp^8)^2(x+\alp^{20})^2$.
1895The dimension of the kernel~$V_2$
1896of $T_3|_{V_1}+\alp^8$ is~$2$.
1897The characteristic polynomial of~$T_5$ on~$V_2$ is
1898$(x+\alp^2)(x+\alp^{10})$.
1899By checking up to the bound
1900$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4288)]\cdot 1901 \frac{1}{4} = 1496$$
1902of Lemma~\ref{lem:bound}
1903we see that all $T_p$, with $p\neq 5$,
1904act as scalars on~$V_2$.
1905
1906{\arraycolsep=.4em
1907$$1908\begin{array}{|rl|}\hline 19092&0\\ 19103&3\\ 19115&\alpha^{14}\\ 19127&\alpha^{20}\\ 191311&\alpha^{20}\\ 191413&\alpha^{16}\\ 191517&\alpha^{16}\\ 191619&0\\ 1917\hline\end{array} 1918\begin{array}{|rl|}\hline 191923&0\\ 192029&\alpha^{14}\\ 192131&\alpha^{20}\\ 192237&\alpha^{4}\\ 192341&\alpha^{8}\\ 192443&2\\ 192547&0\\ 192653&4\\ 1927\hline\end{array} 1928\begin{array}{|rl|}\hline 192959&1\\ 193061&\alpha^{16}\\ 193167&\alpha^{4}\\ 193271&\alpha^{20}\\ 193373&0\\ 193479&\alpha^{20}\\ 193583&0\\ 193689&2\\ 1937\hline\end{array} 1938\begin{array}{|rl|}\hline 193997&\alpha^{10}\\ 1940101&\alpha^{14}\\ 1941103&\alpha^{20}\\ 1942107&2\\ 1943109&0\\ 1944113&\alpha^{2}\\ 1945127&\alpha^{2}\\ 1946131&1\\ 1947\hline\end{array} 1948\begin{array}{|rl|}\hline 1949137&1\\ 1950139&2\\ 1951149&4\\ 1952151&\alpha^{4}\\ 1953157&\alpha^{22}\\ 1954163&0\\ 1955167&\alpha^{14}\\ 1956173&0\\ 1957\hline\end{array} 1958\begin{array}{|rl|}\hline 1959179&1\\ 1960181&\alpha^{14}\\ 1961191&\alpha^{4}\\ 1962193&3\\ 1963197&0\\ 1964199&0\\ 1965211&\alpha^{16}\\ 1966223&2\\ 1967\hline\end{array} 1968\begin{array}{|rl|}\hline 1969227&\alpha^{14}\\ 1970229&\alpha^{4}\\ 1971233&\alpha^{2}\\ 1972239&\alpha^{20}\\ 1973241&3\\ 1974251&\alpha^{4}\\ 1975257&\alpha^{2}\\ 1976263&4\\ 1977\hline\end{array} 1978\begin{array}{|rl|}\hline 1979269&0\\ 1980271&0\\ 1981277&2\\ 1982281&\alpha^{10}\\ 1983283&3\\ 1984293&1\\ 1985307&\alpha^{22}\\ 1986311&3\\ 1987\hline\end{array} 1988$$}
1989
1990
1991\subsection{Conductor~$5373=3^3\cdot 199$}
1992There are two $A_5$ entries in Frey's
1993Table~2 of conductor 5373.  Though neither 2 nor 5 ramify,
1994the hypothesis of~\cite{bdsbt} do not apply
1995to the example below because $\Frob_2$ has order~$5$.
1996This is the only example of odd level treated in this paper.
1997
1998Consider the icosahedral extension~$K$ defined by
1999$h=x^5+2x^4+x^3+7x^2+23x-11$.
2000A Frobenius element at~$5$ has order~$2$.
2001The first few $p\nmid \ell N$ such that
2002$a_p=0$ are $7$, $23$,  $37$, $79$, $89$.
2003The type at~$3$ is~11, and the type at~$199$ is~2.
2004The order of~$\eps_3$ is~$2$, and
2005the order of~$\eps_{199}$ is~$3$.
2006
2007Using~\cite{cohen-oesterle} we find $\dim S_5(5373,\eps)= 2394$;
2008this agrees with the dimension of the space of modular symbols.
2009
2010The intersection~$V$ of the kernels of
2011$T_7$, $T_{23}$, $T_{37}$, $T_{79}$, and $T_{89}$
2012is of dimension~$8$.
2013The characteristic polynomial of~$T_2$ on~$V$ is
2014$(x+\alp^4)^4(x+\alp^{16})^4$.
2015The kernel~$V_1$ of $T_2|_V+\alp^4$
2016is of dimension~$2$. It appears that all~$T_p$ act
2017as scalars on~$V_1$ except~$T_5$ which has
2018characteristic polynomial $(x+1)(x+4)$.
2019Let~$f$ be the form with $a_5=-1$; some eigenvalues of~$f$ are
2020given in the following table.
2021{\arraycolsep=.4em
2022$$\begin{array}{|rl|}\hline 20232&\alpha^{16}\\ 20243&0\\ 20255&4\\ 20267&0\\ 202711&2\\ 202813&\alpha^{22}\\ 202917&1\\ 203019&\alpha^{16}\\ 2031\hline\end{array} 2032\begin{array}{|rl|}\hline 203323&0\\ 203429&\alpha^{4}\\ 203531&\alpha^{14}\\ 203637&0\\ 203741&\alpha^{14}\\ 203843&\alpha^{10}\\ 203947&\alpha^{4}\\ 204053&\alpha^{8}\\ 2041\hline\end{array} 2042\begin{array}{|rl|}\hline 204359&3\\ 204461&3\\ 204567&4\\ 204671&\alpha^{22}\\ 204773&\alpha^{10}\\ 204879&0\\ 204983&3\\ 205089&0\\ 2051\hline\end{array} 2052\begin{array}{|rl|}\hline 205397&\alpha^{14}\\ 2054101&3\\ 2055103&0\\ 2056107&2\\ 2057109&3\\ 2058113&\alpha^{20}\\ 2059127&\alpha^{14}\\ 2060131&\alpha^{20}\\ 2061\hline\end{array} 2062\begin{array}{|rl|}\hline 2063137&2\\ 2064139&2\\ 2065149&\alpha^{4}\\ 2066151&0\\ 2067157&4\\ 2068163&\alpha^{16}\\ 2069167&\alpha^{8}\\ 2070173&\alpha^{8}\\ 2071\hline\end{array} 2072\begin{array}{|rl|}\hline 2073179&\alpha^{8}\\ 2074181&1\\ 2075191&3\\ 2076193&\alpha^{20}\\ 2077197&0\\ 2078199&1\\ 2079211&2\\ 2080223&0\\ 2081\hline\end{array} 2082\begin{array}{|rl|}\hline 2083227&1\\ 2084229&\alpha^{20}\\ 2085233&\alpha^{4}\\ 2086239&2\\ 2087241&1\\ 2088251&4\\ 2089257&\alpha^{20}\\ 2090263&4\\ 2091\hline\end{array} 2092\begin{array}{|rl|}\hline 2093269&\alpha^{2}\\ 2094271&\alpha^{20}\\ 2095277&3\\ 2096281&1\\ 2097283&\alpha^{22}\\ 2098293&\alpha^{14}\\ 2099307&0\\ 2100311&\alpha^{2}\\ 2101\hline\end{array} 2102$$
2103}
2104
2105The bound of Lemma~\ref{lem:bound} is
2106$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(5373)]\cdot 2107 \frac{1}{4} = 1650.$$
2108
2109\comment{
2110\subsection{Conductor $N=1825=5^2\cdot 73$ and $\ell=7$}
2111In this example, the method does not apply because $5\mid N$.
2112Nonetheless, we proceed using~$\ell=7$ instead.\edit{\kevin{We can't
2113\emph{hope} to finish though, because we'll never show the thing is $A_5$
2114and not something like $GL_2(\F_7)$. So probably we should kill this
2115example completely. Somehow our paper gives a method for dealing
2116with $\GL_2(\F_5)$ (and it would also perhaps work with $\GL_2(\F_4)$)
2117but doesn't have a chance of working wiht $GL_2(\F_7)$, I shouldn't think.}
2119
2120Consider the icosahedral extension~$K$ defined by
2121$h=x^5+4x^4-x^3-21x^2-x-7$.
2122A Frobenius element at~$7$ has order~$3$, so
2123distinct eigenvalues mod~$7$ (I think).
2124The first three unramified~$p$ such that
2125$a_p=0$ are $17$, $59$, $67$, $89$.
2126The type at~$5$ is~6, and the type at~$73$ is~2.
2127The order of~$\eps_5$ is~$4$ (I think), and
2128the order of~$\eps_{73}$ is~$3$.
2129
2130Using~\cite{cohen-oesterle} we find $\dim S_8(1825,\eps)= 1104$, and
2131this agrees with modular symbols.
2132The intersection~$V$ of the kernels of $T_{17}$, $T_{59}$,
2133and~$T_{67}$ has dimension~$8$.
2134The operator~$T_{89}$ also vanishes on~$V$.
2135The character takes values in $\F_{7^2}$ which we represent
2136using a generator~$\alp$ of the multiplicative group such that
2137$\alp^2-\alp+3=0$.
2138The characteristic polynomial of~$T_2$ on~$V$ is
2139$(x+\alp)^2(x+\alp^{19})^2(x+\alp^{25})^2(x+\alp^{43})^2$.
2140Let $V_1$ be the kernel of $T_2+\alp$; then $V_1$ is spanned
2141by two newforms~$f$ and~$f_1$:
2142\begin{align*}
2143   f   &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{46}q^7+\cdots\\
2144   f_1 &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{14}q^7+\cdots
2145  \end{align*}
2146It appears that~$f$ and~$f_1$ are companions; and undoubtedly this can
2147be checked.
2148
2149It doesn't seem possible to twist and obtain a representation
2150into $\GL(2,\F_7)$. The $\eps_{73}$ part of the character
2151is $\F_7$-rational.  The $\eps_5$ part has order~$4$, and is
2152hence not $\F_7$-rational.  Twisting by~$\eps_5$ (or its inverse)
2153only replaces $\eps_5$ by $\eps_5^3$, which is still not $\F_7$-rational.
2154
2155\begin{dashlist}
2156\item Even if we could twist into $\GL(2,\F_7)$,
2157the order~$60$ of $A_5$ doesn't divide the order of $\GL(2,\F_7)$.
2158\item  In this $N=1825$ example, we found a modular representation
2159       $\rho_g : \GQ \ra \GL(2,\F_{7^2})$
2160that (morally) is the modulo a prime-over-$7$ reduction of
2161the Artin representation  $\rho : \GQ \ra \GL(2,\cO)$.
2162Here~$\cO$ is the ring of integers in a finite extension of~$\Q_7$.
2163If someone ever proves modularity of~$\rho$ they will
2164have proved that~$\rho_g$ admits a lift to a characteristic zero
2165representation whose image is an extension of~$A_5$;
2166perhaps we can prove this.  We could then try and analyze the
2167ramification and conclude that the lift of~$\rho_g$ is (a twist) of~$\rho$.
2168Then~\cite{buzzard-taylor} implies modularity of~$\rho$.
2169\end{dashlist}
2170}
2171
2172
2173\section{Computations}
2174\subsection{Higher weight modular symbols}
2175\label{sec:modsym}
2176The second author developed software that computes the space of
2177weight~$k$ modular symbols $\sS_k(N,\eps)$, for $k\geq 2$ and
2178arbitrary~$\eps$.
2179See~\cite{merel:1585} for the standard facts about higher weight
2180modular symbols, and~\cite{stein:phd} for a description of
2181how to efficiently compute with them.
2182
2183Let $K=\Q(\eps)$ be the field generated by the values of~$\eps$.
2184 The cuspidal modular symbols $\sS_k(N,\eps)$ are a
2185finite dimensional vector space over~$K$, which is generated by all
2186linear combinations of  higher weight modular symbols
2187    $$X^i Y^{k-2-i}\{\alp,\beta\}$$
2188that lie in the kernel of an appropriate boundary map.  There is an
2189involution~$*$ that acts on $\sS_k(N,\eps)$, and
2190$\sS_k(N,\eps)^+\tensor_K\C$ is isomorphic, as a module over the Hecke
2191algebra, to the space $S_k(N,\eps;\C)$ of cusp forms.
2192
2193Fix $k=5$.  In each case considered in this paper,
2194there is a prime ideal~$\lambda$
2195of the ring of integers $\mathcal{O}$ of~$K$
2196such that $\mathcal{O}/\lambda\isom \F_{25}$.
2197Let~$\cL$ be the $\mathcal{O}$-module generated by all modular
2198symbols of the form $X^iY^{3-i}\{\alp,\beta\}$,
2199and let
2200 $$\sS_5(N,\eps;\F_{25})=(\cL\tensor_{\mathcal{O}}\F_{25})\cap \sS_5(N,\eps).$$
2201This is the space that we computed.
2202The Hecke algebra acts on $\sS_5(N,\eps;\F_{25})$, so when
2203we find an eigenform we find a maximal ideal of the Hecke algebra.
2204
2205As an extra check on our computation of
2206$\sS_5(N,\eps;\F_{25})$, we computed the dimension
2207of $S_5(N,\eps;\C)$ using both the formula of~\cite{cohen-oesterle}
2208and the Hijikata trace formula (see~\cite{hijikata:trace})
2209applied to the identity Hecke operator.
2210
2211
2212\comment{%it's all in my thesis and it's not that relevant.
2213The Manin symbols are
2214$[i, (c,d)]$ where $0\leq i\leq k-2=3$ and
2215$(c,d)$ vary over points in the projective plane.
2216The Manin symbol $[i,(c',d')]$ corresponds to the
2217modular symbol $(g.X^iY^{3-i})\{g(0),g(\infty)\}$
2218where $g=\abcd{a}{b}{c}{d}\in\SL_2(\Z)$ is a matrix whose lower
2219two entries are congruent to $(c',d')$ modulo $N$,
2220and $g.X^iY^{3-i} := (dX-bY)^i(-cX+aY)^{3-i}$.
2221Let $\sigma=\abcd{0}{-1}{1}{0}$, $\tau={0}{-1}{1}{-1}$
2222and for $\gamma\in\SL_2(\Z)$, let
2223$[i,(c,d)]\gamma = [\gamma.X^iY^{3-i}, (c,d)\gamma]$.
2224Since there are only finitely many
2225Manin symbols, we  can
2226compute $\sS_5(N,\eps)$ as the quotient of the $\F$-vector
2227space generated by Manin symbols modulo
2228the following relations:
2229\begin{align*}
2230    {[i,(c,d)] + [i,(c,d)]\sigma} &= 0\\
2231    {[i,(c,d)] + [i,(c,d)]\tau + [i,(c,d)]\tau^2} &= 0\\
2232    {[i,(n c,n d)]}&=\eps(n)[i,(c,d)]
2234\end{align*}
2235The quotient was computed by using a fast hashing'' function
2236to quotient out by the $2$-term relations.  The quotient
2237by the $3$-term relations was then computed using sparse
2238Gauss elimination.  One important subtlety is that, e.g., $\sigma$
2239and~$\tau$ do not commute so, after modding out by
2240the~$\sigma$ relations, it is important to mod out by~$3$
2241term relations coming both from~$\tau$ and~$\sigma\tau$.
2242
2243The main result of~\cite{merel:1585} gives
2244a way to compute the action of $T_p$ directly
2245on the Manin symbols.
2246Suppose $f\in\sS_5(N,\eps;\F_{25})$ is an eigenvector; to
2247naively compute the action of~$T_p$ on~$f$ requires computing
2248the action of~$T_p$ on each Manin symbol involved in~$f$,
2249and then summing the result. This requires roughly
2250$\dim\sS$ times as long as computing~$T_p$ on a single
2251Manin symbol.
2252In order to quickly compute a large number of
2253Hecke eigenvalues we use the following projection trick.
2254Let $\vphi\in\Hom(\sS_5(N,\eps;\F_{25}),\F_{25})$ be a (left) eigenvector for all
2255Hecke operators~$T_p$ having the same eigenvalues as~$f$.
2256Choose a Manin symbol $x=[i,(c,d)]$ such
2257that $\vphi(x)\neq 0$.  Since~$x$ is of a very simple form,
2258it is easy to compute~$T_p(x)$ quickly.  We have
2259 $\vphi(T_p(x)) = (T_p(\vphi))(x) = a_p \vphi(x)$,
2260so since $\vphi(x)\neq 0$ we divide and find
2261$a_p = \vphi(T_p(x))/\vphi(x)$.
2262In fact, we use a generalization of this trick to
2263quickly compute the action of~$T_p$ on any Hecke stable subspace
2264$V\subset \Hom(\sS(N,\eps;\F_{25}),\F_{25})$.
2265}
2266
2267
2268\subsection{Complexity}
<`