1% conjtwist.tex
2\documentclass[11pt]{article}
3\title{\Large\sc The First Newform on $\Gamma_0(N)$ such that
4$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$}
5\author{William A. Stein}
6\usepackage{amsthm}
7\usepackage{amsmath}
8\theoremstyle{plain}
9\newtheorem{theorem}{Theorem}[section]
10\newtheorem{proposition}[theorem]{Proposition}
11\newtheorem{corollary}[theorem]{Corollary}
12\newtheorem{claim}[theorem]{Claim}
13\newtheorem{lemma}[theorem]{Lemma}
14\newtheorem{conjecture}[theorem]{Conjecture}
15\theoremstyle{definition}
16\newtheorem{definition}[theorem]{Definition}
17\newtheorem{algorithm}[theorem]{Algorithm}
18\newtheorem{question}[theorem]{Question}
19
20\theoremstyle{remark}
21\newtheorem{goal}[theorem]{Goal}
22\newtheorem{remark}[theorem]{Remark}
23\newtheorem{example}[theorem]{Example}
24\newtheorem{exercise}[theorem]{Exercise}
25
26\newcommand{\defn}[1]{{\em #1}}
27\newcommand{\e}{\mathbf{e}}
28\newcommand{\Gam}{\Gamma}
29\newcommand{\X}{\mathcal{X}}
30\newcommand{\E}{\mathcal{E}}
31\newcommand{\q}{\mathbf{q}}
32\newcommand{\cross}{\times}
33\newcommand{\ra}{\rightarrow}
34\newcommand{\la}{\leftarrow}
35\newcommand{\tensor}{\otimes}
36\newcommand{\eps}{\varepsilon}
37\newcommand{\vphi}{\varphi}
38\newcommand{\comment}[1]{}
39\newcommand{\Q}{\mathbf{Q}}
40\newcommand{\Qbar}{\overline{\Q}}
41\newcommand{\A}{\mathcal{A}}
42\newcommand{\p}{\mathfrak{p}}
43\newcommand{\m}{\mathfrak{m}}
44\renewcommand{\L}{\mathcal{L}}
45\renewcommand{\l}{\ell}
46\renewcommand{\t}{\tau}
47\renewcommand{\star}{\times}
48\renewcommand{\P}{\mathbf{P}}
49\renewcommand{\a}{\mathfrak{a}}
50\DeclareMathOperator{\Br}{Br}
51\DeclareMathOperator{\End}{End}
52\DeclareMathOperator{\new}{new}
53\DeclareMathOperator{\Aut}{Aut}
54\DeclareMathOperator{\Gal}{Gal}
55
56\begin{document}
57\maketitle
58
59\begin{abstract}
60We compute the minimum $N$ such that there
61is a newform $f=\sum a_n q^n\in{}S_2(\Gamma_0(N))$ with
62the odd property that
63$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$.\footnote{Henri
64Cohen told me that
65he discoved this example many years ago, and Elkies explained'' it to him.
66This example is not in the literature, and I found
67it indepedently of Cohen and Elkies.}
68\end{abstract}
69
70\section{The Newform}
71Using modular symbols, we computed the first few terms of each
72newform $f=\sum a_n q^n$ on $S_2(\Gamma_0(N))$ for $N\leq 512$.
73For each newform of level $<512$, we found that there
74exists an integer~$n$ such that the field $K_f=\Q(\ldots, a_m, \ldots)$
75generated by all Fourier coefficients of~$f$ equals the field
76$\Q(a_n)$ generated by a single coefficient.  This lead me
77to suspect that we always have equality.
78
79We do not always have equality.
80The characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is
81$82 (x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2), 83$
84and if we let $f=\sum a_n q^n$
85be one of the newforms in the four-dimensional kernel~$V$
86of $T_3^2-6$, then
87$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11} 88 -2\sqrt{3}q^{13}-6\sqrt{2}q^{15} +\cdots$$
89provides an example in which $K_f\neq \Q(a_n)$ for some~$n$.
90
91We will use the following theorem (see \cite[Prop.~3.64]{shimura}):
92
93\begin{theorem}[Shimura]\label{thm:shimura}
94Let $N, r, s, k>0$ be integers such that $s|N$ and let
95$M$ be the least common multiple of $N$, $r^2$, and $rs$.
96Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).
97If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then
98$$\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$
99\end{theorem}
100
101\begin{lemma}\label{preserves}
102Suppose $n\geq 6$ and let~$\eps$ be a primitive
103Dirichlet character of conductor dividing~$8$.
104Then the map $f\mapsto f\tensor\eps$
105preserves $S_k(\Gamma_0(2^n))$.
106\end{lemma}
107\begin{proof}
108Set $r=8$ in Theorem~\ref{thm:shimura}, and note that $\eps^2=1$.
109\end{proof}
110
111
112It is clear from the $q$-expansion of~$f$ that
113$\Q(\sqrt{2},\sqrt{3})\subset K_f$.
114The newform~$f$ and its companions lie inside
115of the kernel~$V$ of $T_3^2-6$.  A modular symbols computation
116shows that this kernel has dimension~$4$, which proves that
117$K_f=\Q(\sqrt{2},\sqrt{3})$.
118
119
120\begin{proposition}
121There is no integer~$n$ such that $K_f\neq \Q(a_n)$.
122\end{proposition}
123
124\begin{proof}
125Let~$\chi$ be a character of conductor dividing~$8$.  By
126Lemma~\ref{preserves}, $f\tensor\chi$ lies in $S_2(\Gamma_0(512))$.
127We check computationally that $f\tensor\chi$ is one of the four
128Galois-conjugates of~$f$, and that each conjugate is of the form
129$f\tensor\chi$ for some~$\chi$.  Let~$n>2$ be a positive integer and
130let~$\sigma$ be an automorphism.  Then we have just showed that
131$\sigma(a_n) = \chi(n) a_n$ for some character~$\chi$ of order at
132most~$2$, so $\chi(n)\in\{\pm 1\}$.  Thus the action of
133$\Gal(\Qbar/\Q)$ on $a_n$ factors through $\{ \pm 1\}$, so
134$\Q(a_n)$ has degree at most~$2$.
135\end{proof}
136
137
138\vspace{2ex}
139
140\noindent{\bf Acknowledgment:} It is a pleasure to thank Kevin Buzzard and
142
143\begin{thebibliography}{HHHHHHH}
144%\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
145\bibitem[1]{shimura} G. Shimura, {\em Introduction to the
146Arithmetic Theory of Automorphic Functions}, Princeton
147University Press, (1994).
148\end{thebibliography} \normalsize\vspace*{1 cm}
149
150\end{document}
151
152\section{Don't read this.  It contains a mistake.}
153
154We introduce some notation in order to recall Ribet's Theorem.
155Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$.  Let
156$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f= 157 f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$
158It is known that $\Gamma$ is an abelian subgroup
159of $\Aut(E)$.  Let $F$ be the subfield of $E$ fixed by $\Gamma$.
160
161\begin{remark}
162What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?
163Embed $E$ in its normal closure $K$.
164Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms
165which preserve $E$.  Since every automorphism extends,
166$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup
167of $G$ corresponding to $E$ via Galois theory.
168Note that $J$ is normal in $H$ but need not be normal in $G$.
169The inverse image $\Gamma'$ of
170$\Gamma$ in $H$ is a subgroup of the same index as the index
171of $\Gamma$ in $\Aut(E)$.
172Unless I make a mistake computing indices, we get
173$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]} 174 = \frac{[E:\Q]}{\#\Gamma}$$
175which is just what we would expect.
176\end{remark}
177
178For a primitive character $\vphi$ of conductor $c$ define
179$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$
180For $\gamma,\delta\in\Gamma$, let
181$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1}) 182 g(\gamma(\chi_{\delta}^{-1}))} 183 {g(\chi_{\gamma\delta}^{-1})}.$$
184
185In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$
186in terms of generators and relations.
187Let $\X$ be the $E$-vector space
188$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$
189where the $X_{\gamma}$ are formal symbols.  By imposing
190on the $X_{\gamma}$ the rules
191\begin{eqnarray*}
192X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},
193         \quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\
194X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}
195\end{eqnarray*}
196we make $\X$ into an associative algebra.
197
198\begin{theorem}[Ribet\cite{ribet}]\label{ribet}
199There algebra $\X$ is a central simple
200algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.
201Furthermore, the $2$-cocycle $c$ corresponds to the class of
202$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.
203\end{theorem}
204
205
206\section{Example in which the endomorphism ring has order $2$ in the Brauer group}
207
208\comment{%PARI code
209? nf=nfinit(subst(E[1],x,t));
210? nffactor(nf,x^2-2)
211? a=-1/12*t^2 + 3/2;
212? nffactor(nf,x^2-3)
213? b=-1/24*t^3 + 7/4*t;
214? c=lift(Mod(a*b,t^4-36*t^2+36))
215? d=1;
216? v=subst(E[2],x,t);
217? N=mattranspose([3/2,0,-1/12,0;0,7/4,0,-1/24;0,5/2,0,-1/12;1,0,0,0])^(-1);
218? w(f)=vector(4,i,polcoeff(f,i-1))~;;
219? N*w(v[5])
220? N*w(v[7])
221}
222
223
224\begin{theorem}
225The endomorphism ring of $A_f$ is $M_2(\Q)$.
226\end{theorem}
227\begin{proof}
228
229{\bf NOTE, added July 2002: } The proof below is wrong, as Pete
230Clark has pointed out.  I recommend ignoring it, and looking at
231\cite{ghate:endo} instead (see their Example 4.3.1).
232
233Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$
234where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.
235Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of
236conductor dividing $8$, where $\chi_d$ corresponds to the field
237$\Q(\sqrt{d})$.
238We have
239\begin{eqnarray*}
240  f & = & f\tensor\chi_1\\
241  \gamma_2 f  = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\
242  \gamma_3 f  = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
243     &=& f\tensor\chi_{-1}\\
244  \gamma_6 f  = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
245     &=& f\tensor\chi_{-2}
246\end{eqnarray*}
247
248The sums are
249\begin{eqnarray*}
250  g(\chi_1) &=& 1\\
251  g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\
252  g(\chi_{2}) &=&
253    e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}
254     = 2\sqrt{2}\\
255  g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}
256      - e^{2\pi i 5/8} - e^{2\pi i 7/8}
257     = 2i\sqrt{2}.
258\end{eqnarray*}
259Thus we can compute
260$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})} 261 {g(\chi_{\gamma_i\gamma_j})}.$$
262For example,
263$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$
264By Theorem~\ref{ribet} we obtain a presentation of
265$(\End A)\tensor\Q$.
266Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so
267that the generators $X_{\gamma}$ commute.  The endomorphism
268ring is not commutative as $E$ does not commute with $X_{\gamma}$
269for nontrivial $\gamma$.
270
271The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$
272corresponding to $(\End A)\tensor\Q$.  We ask, does this element have
273order $1$ or order $2$?  Let $K=\Q_2(\sqrt{2},\sqrt{3})$.
274Then, by inf-res, $c$ arises from an
275element of
276  $$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$
277Note that inf {\em is} injective because of Hilbert's
278Theorem 90; see Proposition 6 on page 156 of \cite{serre}.
279
280I think that $c$ must have order dividing $2$ and
281$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$
282is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.
283Using the Local symbol in Chapter XIV of we can write down the
284nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.
285
286Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$.  What is
287$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?
288We have:
289$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad 290 c(\tau,\tau) = -4.$$
291
292This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;
293see Washington's article \cite{washington}.  Thus this cocycle is
294trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.
295If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.
296But this is impossible as the ramification degree of
297$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$
298is $4$.  Thus we finally conclude that $c$ is nontrivial and hence
299obtain the theorem.  {\bf No, it is trivial.}
300\end{proof}
301
302
303\begin{thebibliography}{HHHHHHH}
304\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
305\bibitem[1]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators
306for $\Gamma_0(N)$ and class fields over quadratic number fields},
307J. Math. Soc. Japan, {\bf 25} (1973), 629--643.
308\bibitem[2]{koike} M. Koike, {\em On certain abelian varieties
309obtained from new forms of weight $2$ on
310$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,
311{\bf 62} (1976), 29--39.
312\bibitem[3]{ribet} K. Ribet, {\em Twists of Modular Forms and
313Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},
314(1980), 43--62.
315\bibitem[4]{serre} J-P. Serre, {\em Local Fields}, Springer-Verlag,
316(1979).
317\bibitem[5]{shimura} G. Shimura, {\em Introduction to the
318Arithmetic Theory of Automorphic Functions}, Princeton
319University Press, (1994).
320\bibitem[6]{washington} L.C. Washington, {\em Galois Cohomology},
321In Modular Forms and Fermat's Last Theorem'', Ed.'s
322Cornell-Silverman-Stevens, (1997).
323\end{thebibliography} \normalsize\vspace*{1 cm}
324
325\end{document}
326
327
328\comment{%Letter to Luiz
329
330     THe endomorphism algebra of A_f (over Q) is a central simple algebra over
331     F_f which contains Q_f as a maximal conmutative subfield. Its degree over
332     Q is [Q_f : Q]*[Q_f : F_f] .
333
334Moreover, the central simple algebra has order either one or two
335in the Brauer group of F_f.  Thus
336
337                          /- matrix algebra over F_f
338     (End A_f) tensor Q =  or
339                          \- matrix algebra over quaternion algebra over F_f.
340
341Knowing which one gives the exact isogeny decomposition of A_f over Qbar.
342
343     This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms
344     of Abelian Varieties", Math. Ann. 253, 43-62 (1980)
345
346There is also some discussion in the (not so nicely typeset) paper of
347Ken's: Endomorphism algebras of abelian varieties attached
348         to newforms of weight 2''
349
350Have you tried to work out the isogeny structure (over Qbar) of the
351abelian variety A_f attached to your level 8192 form f?  I suspect
352that f has inner twists corresponding to all of the Dirichlet characters
353of conductor dividing 8, i.e., corresponding to the quadratic
354subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any
355inner twists by characters of degree > 2 or ramified outside of 2.
356I wonder what the order of (End A_f)tensor Q is in the Brauer group.
357Is it one or two?
358
359
360
361 Best,
362 William.
363
364
365Ken,
366
367Thanks for giving me a copy of your paper on extra twists.
368Let D be the endomorphism ring (tensor Q) of the abelian
369variety A_f corresponding to the newform in f in
370S_2(Gamma_0(512)) that we've been discussing (the one
371for which Q(a_n) never equals Q(f)).  In this case, the subfield
372of Q(f) fixed by the group which you call Gamma is the rational
373numbers Q. Thus D is a central simple Q-algebra.  I think I can use
374the theorem in your paper to show that D = M_2(K) where K is a division
375quaternion algebra with center Q, i.e., D has order two in the
376Brauer group of Q.  This example may be interesting in
377light of your remark on page 60 (of the 1980 Math. Annalen paper)
378that it does not seem easy to determine the order of D
379by pure thought''.  You showed it was 1 in many cases, but
380I don't think you gave conditions which imply that it must be 2.
381
382  Thanks again for the paper,
383  William
384}
385