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\title{\Large\sc The First Newform on $\Gamma_0(N)$ such that
$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$}
\author{William A. Stein}
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\begin{document}
\maketitle
\begin{abstract}
We compute the minimum $N$ such that there
is a newform $f=\sum a_n q^n\in{}S_2(\Gamma_0(N))$ with
the odd property that
$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$.\footnote{Henri
Cohen told me that
he discoved this example many years ago, and Elkies ``explained'' it to him.
This example is not in the literature, and I found
it indepedently of Cohen and Elkies.}
\end{abstract}
\section{The Newform}
Using modular symbols, we computed the first few terms of each
newform $f=\sum a_n q^n$ on $S_2(\Gamma_0(N))$ for $N\leq 512$.
For each newform of level $<512$, we found that there
exists an integer~$n$ such that the field $K_f=\Q(\ldots, a_m, \ldots)$
generated by all Fourier coefficients of~$f$ equals the field
$\Q(a_n)$ generated by a single coefficient. This lead me
to suspect that we always have equality.
We do not always have equality.
The characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is
$
(x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2),
$
and if we let $f=\sum a_n q^n$
be one of the newforms in the four-dimensional kernel~$V$
of $T_3^2-6$, then
$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11}
-2\sqrt{3}q^{13}-6\sqrt{2}q^{15} +\cdots$$
provides an example in which $K_f\neq \Q(a_n)$ for some~$n$.
We will use the following theorem (see \cite[Prop.~3.64]{shimura}):
\begin{theorem}[Shimura]\label{thm:shimura}
Let $N, r, s, k>0$ be integers such that $s|N$ and let
$M$ be the least common multiple of $N$, $r^2$, and $rs$.
Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).
If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then
$$\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$
\end{theorem}
\begin{lemma}\label{preserves}
Suppose $n\geq 6$ and let~$\eps$ be a primitive
Dirichlet character of conductor dividing~$8$.
Then the map $f\mapsto f\tensor\eps$
preserves $S_k(\Gamma_0(2^n))$.
\end{lemma}
\begin{proof}
Set $r=8$ in Theorem~\ref{thm:shimura}, and note that $\eps^2=1$.
\end{proof}
It is clear from the $q$-expansion of~$f$ that
$\Q(\sqrt{2},\sqrt{3})\subset K_f$.
The newform~$f$ and its companions lie inside
of the kernel~$V$ of $T_3^2-6$. A modular symbols computation
shows that this kernel has dimension~$4$, which proves that
$K_f=\Q(\sqrt{2},\sqrt{3})$.
\begin{proposition}
There is no integer~$n$ such that $K_f\neq \Q(a_n)$.
\end{proposition}
\begin{proof}
Let~$\chi$ be a character of conductor dividing~$8$. By
Lemma~\ref{preserves}, $f\tensor\chi$ lies in $S_2(\Gamma_0(512))$.
We check computationally that $f\tensor\chi$ is one of the four
Galois-conjugates of~$f$, and that each conjugate is of the form
$f\tensor\chi$ for some~$\chi$. Let~$n>2$ be a positive integer and
let~$\sigma$ be an automorphism. Then we have just showed that
$\sigma(a_n) = \chi(n) a_n$ for some character~$\chi$ of order at
most~$2$, so $\chi(n)\in\{\pm 1\}$. Thus the action of
$\Gal(\Qbar/\Q)$ on $a_n$ factors through $\{ \pm 1\}$, so
$\Q(a_n)$ has degree at most~$2$.
\end{proof}
\vspace{2ex}
\noindent{\bf Acknowledgment:} It is a pleasure to thank Kevin Buzzard and
Ken Ribet for helpful comments.
\begin{thebibliography}{HHHHHHH}
%\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
\bibitem[1]{shimura} G. Shimura, {\em Introduction to the
Arithmetic Theory of Automorphic Functions}, Princeton
University Press, (1994).
\end{thebibliography} \normalsize\vspace*{1 cm}
\end{document}
\section{Don't read this. It contains a mistake.}
We introduce some notation in order to recall Ribet's Theorem.
Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$. Let
$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f=
f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$
It is known that $\Gamma$ is an abelian subgroup
of $\Aut(E)$. Let $F$ be the subfield of $E$ fixed by $\Gamma$.
\begin{remark}
What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?
Embed $E$ in its normal closure $K$.
Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms
which preserve $E$. Since every automorphism extends,
$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup
of $G$ corresponding to $E$ via Galois theory.
Note that $J$ is normal in $H$ but need not be normal in $G$.
The inverse image $\Gamma'$ of
$\Gamma$ in $H$ is a subgroup of the same index as the index
of $\Gamma$ in $\Aut(E)$.
Unless I make a mistake computing indices, we get
$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]}
= \frac{[E:\Q]}{\#\Gamma}$$
which is just what we would expect.
\end{remark}
For a primitive character $\vphi$ of conductor $c$ define
$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$
For $\gamma,\delta\in\Gamma$, let
$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1})
g(\gamma(\chi_{\delta}^{-1}))}
{g(\chi_{\gamma\delta}^{-1})}.$$
In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$
in terms of generators and relations.
Let $\X$ be the $E$-vector space
$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$
where the $X_{\gamma}$ are formal symbols. By imposing
on the $X_{\gamma}$ the rules
\begin{eqnarray*}
X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},
\quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\
X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}
\end{eqnarray*}
we make $\X$ into an associative algebra.
\begin{theorem}[Ribet\cite{ribet}]\label{ribet}
There algebra $\X$ is a central simple
algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.
Furthermore, the $2$-cocycle $c$ corresponds to the class of
$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.
\end{theorem}
\section{Example in which the endomorphism ring has order $2$ in the Brauer group}
\comment{%PARI code
? nf=nfinit(subst(E[1],x,t));
? nffactor(nf,x^2-2)
? a=-1/12*t^2 + 3/2;
? nffactor(nf,x^2-3)
? b=-1/24*t^3 + 7/4*t;
? c=lift(Mod(a*b,t^4-36*t^2+36))
? d=1;
? v=subst(E[2],x,t);
? 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);
? w(f)=vector(4,i,polcoeff(f,i-1))~;;
? N*w(v[5])
? N*w(v[7])
}
\begin{theorem}
The endomorphism ring of $A_f$ is $M_2(\Q)$.
\end{theorem}
\begin{proof}
{\bf NOTE, added July 2002: } The proof below is wrong, as Pete
Clark has pointed out. I recommend ignoring it, and looking at
\cite{ghate:endo} instead (see their Example 4.3.1).
Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$
where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.
Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of
conductor dividing $8$, where $\chi_d$ corresponds to the field
$\Q(\sqrt{d})$.
We have
\begin{eqnarray*}
f & = & f\tensor\chi_1\\
\gamma_2 f = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\
\gamma_3 f = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
&=& f\tensor\chi_{-1}\\
\gamma_6 f = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
&=& f\tensor\chi_{-2}
\end{eqnarray*}
The sums are
\begin{eqnarray*}
g(\chi_1) &=& 1\\
g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\
g(\chi_{2}) &=&
e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}
= 2\sqrt{2}\\
g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}
- e^{2\pi i 5/8} - e^{2\pi i 7/8}
= 2i\sqrt{2}.
\end{eqnarray*}
Thus we can compute
$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})}
{g(\chi_{\gamma_i\gamma_j})}.$$
For example,
$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$
By Theorem~\ref{ribet} we obtain a presentation of
$(\End A)\tensor\Q$.
Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so
that the generators $X_{\gamma}$ commute. The endomorphism
ring is not commutative as $E$ does not commute with $X_{\gamma}$
for nontrivial $\gamma$.
The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$
corresponding to $(\End A)\tensor\Q$. We ask, does this element have
order $1$ or order $2$? Let $K=\Q_2(\sqrt{2},\sqrt{3})$.
Then, by inf-res, $c$ arises from an
element of
$$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$
Note that inf {\em is} injective because of Hilbert's
Theorem 90; see Proposition 6 on page 156 of \cite{serre}.
I think that $c$ must have order dividing $2$ and
$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$
is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.
Using the Local symbol in Chapter XIV of we can write down the
nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.
Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$. What is
$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?
We have:
$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad
c(\tau,\tau) = -4.$$
This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;
see Washington's article \cite{washington}. Thus this cocycle is
trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.
If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.
But this is impossible as the ramification degree of
$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$
is $4$. Thus we finally conclude that $c$ is nontrivial and hence
obtain the theorem. {\bf No, it is trivial.}
\end{proof}
\begin{thebibliography}{HHHHHHH}
\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
\bibitem[1]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators
for $\Gamma_0(N)$ and class fields over quadratic number fields},
J. Math. Soc. Japan, {\bf 25} (1973), 629--643.
\bibitem[2]{koike} M. Koike, {\em On certain abelian varieties
obtained from new forms of weight $2$ on
$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,
{\bf 62} (1976), 29--39.
\bibitem[3]{ribet} K. Ribet, {\em Twists of Modular Forms and
Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},
(1980), 43--62.
\bibitem[4]{serre} J-P. Serre, {\em Local Fields}, Springer-Verlag,
(1979).
\bibitem[5]{shimura} G. Shimura, {\em Introduction to the
Arithmetic Theory of Automorphic Functions}, Princeton
University Press, (1994).
\bibitem[6]{washington} L.C. Washington, {\em Galois Cohomology},
In ``Modular Forms and Fermat's Last Theorem'', Ed.'s
Cornell-Silverman-Stevens, (1997).
\end{thebibliography} \normalsize\vspace*{1 cm}
\end{document}
\comment{%Letter to Luiz
THe endomorphism algebra of A_f (over Q) is a central simple algebra over
F_f which contains Q_f as a maximal conmutative subfield. Its degree over
Q is [Q_f : Q]*[Q_f : F_f] .
Moreover, the central simple algebra has order either one or two
in the Brauer group of F_f. Thus
/- matrix algebra over F_f
(End A_f) tensor Q = or
\- matrix algebra over quaternion algebra over F_f.
Knowing which one gives the exact isogeny decomposition of A_f over Qbar.
This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms
of Abelian Varieties", Math. Ann. 253, 43-62 (1980)
There is also some discussion in the (not so nicely typeset) paper of
Ken's: ``Endomorphism algebras of abelian varieties attached
to newforms of weight 2''
Have you tried to work out the isogeny structure (over Qbar) of the
abelian variety A_f attached to your level 8192 form f? I suspect
that f has inner twists corresponding to all of the Dirichlet characters
of conductor dividing 8, i.e., corresponding to the quadratic
subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any
inner twists by characters of degree > 2 or ramified outside of 2.
I wonder what the order of (End A_f)tensor Q is in the Brauer group.
Is it one or two?
Best,
William.
Ken,
Thanks for giving me a copy of your paper on extra twists.
Let D be the endomorphism ring (tensor Q) of the abelian
variety A_f corresponding to the newform in f in
S_2(Gamma_0(512)) that we've been discussing (the one
for which Q(a_n) never equals Q(f)). In this case, the subfield
of Q(f) fixed by the group which you call Gamma is the rational
numbers Q. Thus D is a central simple Q-algebra. I think I can use
the theorem in your paper to show that D = M_2(K) where K is a division
quaternion algebra with center Q, i.e., D has order two in the
Brauer group of Q. This example may be interesting in
light of your remark on page 60 (of the 1980 Math. Annalen paper)
that it does not seem easy to determine the order of D
by ``pure thought''. You showed it was 1 in many cases, but
I don't think you gave conditions which imply that it must be 2.
Thanks again for the paper,
William
}