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\title{\texorpdfstring{Hecke eigenvalues in level~$1$ modulo powers of $2$ and $3$:\\
Hatada's question (1981) and the Coleman--Stein conjecture}{Hecke eigenvalues in level 1
modulo powers of 2 and 3: Hatada's question (1981) and the Coleman-Stein conjecture}}
\author{David Niedbala Giraudin\\\small ORCID 0009-0009-1526-1178}
\date{}

\begin{document}
\maketitle

\begin{abstract}
Let $f$ be a normalized Hecke eigenform of level $1$ and weight $k$, with eigenvalues $a_n(f)$. In 1981 Hatada asked whether $a_p(f)\equiv 1+p \pmod{2^a}$ whenever $p\equiv\pm1\pmod{2^{a-1}}$ and $6\le a\le 13$. Rustom's congruence modulo $128$ answers this for $a\le 7$. We show that the answer is affirmative for $a\le 10$ and negative for $a=11$, $12$ and $13$, the counterexamples coming from weight $16$ and $p=12289$. The proof combines Rustom's theorem with a certificate for $T_5$ modulo $512$ on period polynomials, a $2$-adic Galois law (for instance $a_p\equiv 1+p^{k-1}\pmod{2^{10}}$ for every $p\equiv 1\pmod 8$), and a reduction to a computation in a finite group. As a consequence, $a_p(f)=0$ implies $p\equiv -1\pmod{2^{11}}$. At the prime $3$ we show that, on Hatada's lattice of period polynomials, $T_2^2\equiv 9(k-1)^2\pmod{27}$, with refinements modulo $81$. This gives an independent proof of Conjecture 9.1 of Rustom without any ramification hypothesis and of the level $1$ case of the Coleman--Stein conjecture, both of which also follow from Rustom's recent classification modulo $81$. We also prove that $a_n(f)\equiv\sigma_{k-1}(n)\pmod{81}$ for every $n\equiv 1\pmod 3$, and a $3$-adic Galois law modulo $243$ on the Galois group of $\mathbb{Q}(\sqrt{-3})$, which gives congruences modulo $243$ and $729$ for suitable split primes and a bilinear law for primes $\ell\equiv 2\pmod 3$. Beyond $729$, for every prime $\ell\equiv 2\pmod 3$ we prove $a_\ell^2\equiv\sigma_{k-1}(\ell)^2\pmod{3^{\min(2j+2,10)}}$, where $j=v_3(\ell+1)$; the exponent $2j+2$ is attained for $j\le 4$ and fails for $j=5$. Each statement is marked as proved, known, verified or open. The finite computations are certified by seven Python scripts, provided as ancillary files.
\end{abstract}

\section{The question}

For even $k$, let $S_k$ be the space of cusp forms of weight $k$ for
$\mathrm{SL}_2(\Z)$, and let $\lambda_p$ be an eigenvalue of the Hecke operator
$T_k(p)$, $p$ an odd prime. Hatada proved that
$\lambda_p\equiv 1+p \pmod 8$ for every odd $p$~\cite{H79}, and then that
\[
\lambda_p\equiv 1+p \pmod{16}\ \text{ if } p\equiv\pm1 \pmod 8,
\qquad
\lambda_p\equiv 1+p \pmod{32}\ \text{ if } p\equiv\pm1 \pmod{16}
\]
\cite[p.~305]{H81}. He then asks the following question \cite[p.~306]{H81}:
\begin{quote}
\emph{Is it true that $\lambda_p \equiv 1+p \pmod{2^a}$ if
$p\equiv\pm1\pmod{2^{a-1}}$ and $6\le a\le 13$\,?}
\end{quote}
together with a stronger version concerning the lattice $L^-_{w+2}$ of forms
whose odd periods are integers (stability of $L^-_{w+2}$ under
$(T_{w+2}(p)-1-p)/2^a$), which implies the former. In his Remark~1,
Hatada points out that for $a\ge 14$ the congruence may fail when
$p\equiv-1\pmod{2^a}$, after Swinnerton-Dyer~\cite{SD73}. According to
Rustom~\cite{R19}, the question was asked at Serre's suggestion.
Rustom proved $\lambda_p\equiv 1+p^{k-1}\pmod{128}$ for every odd $p$ and
every $k$ \cite[Thm~8.1]{R19}, which gives the affirmative answer for
$a\le 7$\footnote{Rustom states it for $a<7$; the case $a=7$ follows, since
$p\equiv\pm1\pmod{2^6}$ implies $p^2\equiv1\pmod{2^7}$, hence
$p^{k-1}\equiv p\pmod{2^7}$.}.

In residue characteristic~$3$, Hatada proved
$\lambda_p\equiv1+p\pmod3$ for $p\ne3$, $\lambda_3\equiv0\pmod9$, and
$\lambda_3\equiv0\pmod{27}$ if $3\nmid k$ \cite[congruences~(5) to~(7)]{H79}.
Coleman and Stein~\cite{CS04} conjectured
that there are only five residue classes modulo~$9$ of eigenforms
(levels $1$, $3$, $9$), and checked it up to weight~$74$ in level~$1$.
Rustom~\cite[Thm~9.2]{R19} states that finiteness modulo~$9$ in level~$1$
follows from the following statement.

\smallskip
\noindent\textbf{Conjecture 9.1 of \cite{R19}} (Conjecture~1.1 of \cite{Ch26};
Rustom writes the weight as $w+2$). \emph{For every eigenform $f$
of level~$1$ and weight $k$, one has $a_2(f)\equiv 3$ or $6\pmod{9\overline{\Z}_3}$
if $k\equiv 0,2\pmod 6$, and $a_2(f)\equiv 0\pmod{9\overline{\Z}_3}$ if
$k\equiv 4\pmod 6$.}

\smallskip
\noindent Chiriac~\cite{Ch26} proved unconditionally that $a_2\equiv0\pmod{9\overline{\Z}_3}$
implies $k\equiv4\pmod6$ \cite[Thm~1.2]{Ch26}, and the whole of Conjecture~9.1
when $3$ is unramified in the coefficient field
\cite[Cor.~1.3]{Ch26}; since Rustom checked this non-ramification for $k\le1000$
\cite{R19}, Conjecture~9.1 is proved for $k\le1000$.

\smallskip
\noindent\textbf{Related work.} The recent preprint \cite{R26} of Rustom classifies, in every weight, the prime-to-$p$ Hecke
eigensystems of level~$1$ modulo $2^8$, $3^4$, $5^3$ and $7^2$, by Hecke-equivariant
transfer maps between modular-symbol modules. At $p=3$, \cite[Thm~1.5]{R26} gives in
particular the literal congruence modulo~$27$ of every prime-to-$3$ eigensystem with an
Eisenstein twist; this implies Conjecture~9.1 of \cite{R19} without any ramification
hypothesis, and the level~$1$ case of the Coleman--Stein conjecture. Our Corollary~\ref{coL}
and Theorem~\ref{thM} were obtained independently, by a different route (explicit certificates
on Hatada's lattice), and \cite{R26} has priority for these two statements. At $p=2$,
\cite[Thm~1.4]{R26} is a classification modulo $2^8$ in the valuative sense
($v_2(a_\ell-\alpha)>7$); it does not address Hatada's question, and for
$p\equiv\pm1\pmod{2^7}$ it gives the valuative form of the case $a=8$. The congruences
modulo $2^9$ and $2^{10}$ of \S\ref{sec8}, the negative answer for $a=11,12,13$, and the
$3$-adic results beyond $81$ of \S\S\ref{sec7}--\ref{sec9} are not covered by \cite{R26}.
Earlier, Calegari and Emerton \cite[Lemma~3.3]{CE04} had proved
$a_\ell\equiv1+\ell\pmod9$ for $\ell\equiv\pm1\pmod9$.

\smallskip
The main results of this paper are the following. We first give again
(Theorem~\ref{thF}, \S\ref{sec3}), on Hatada's lattice, his congruences
$a_3(f)\equiv0\pmod9$, and $\pmod{27}$ if $3\nmid k$ \cite{H79}. Next
(\S\ref{sec5}), \emph{on Hatada's lattice,
$T_2^2\equiv9(k-1)^2\pmod{27}$}, and, more precisely, modulo~$81$,
$T_2^2\equiv\sigma_{k-1}(2)^2$ if $k\equiv0,2\pmod6$ and $T_2^2+9T_2\equiv0$ if
$k\equiv4\pmod6$ (Theorems~\ref{thI} and~\ref{thK}). It follows that
Conjecture~9.1 of \cite{R19} holds for every eigenform of level~$1$, without any
ramification hypothesis (this also follows from \cite[Thm~1.5]{R26}, see above), and even that $a_2(f)\equiv\pm\sigma_{k-1}(2)\pmod{27}$
when $k\equiv0,2\pmod6$ (Corollary~\ref{coL}). Together with \cite{R19}, we obtain the
level~$1$ case of the Coleman--Stein conjecture (Theorem~\ref{thM}). Next,
\S\ref{sec6} extends these congruences to the operators $T_\ell$, $\ell\le19$, and deduces,
with the structure theorem for the $3$-adic Hecke algebra of \cite{R19}, a
law valid in every weight: \emph{for every eigenform $f$ of level~$1$ and every
$n\equiv1\pmod3$, $a_n(f)\equiv\sigma_{k-1}(n)\pmod{81}$}, and, if $k\not\equiv4\pmod6$,
$a_n(f)\equiv\psi(n)\sigma_{k-1}(n)\pmod{27}$ for every $n$ prime to $3$, with
$\psi$ trivial or equal to the character of conductor~$3$ (Theorem~\ref{thO}).
Section~\ref{sec7} derives a Galois law: \emph{on the Galois group of
$\Q(\sqrt{-3})$, $\operatorname{tr}\rho_f\equiv1+\chi^{k-1}+(a_2^2-\sigma_{k-1}(2)^2)\,u^2
\pmod{243}$}, where $u$ measures the action on $\zeta_9$ (Theorem~\ref{thP}); hence
$a_\ell\equiv\sigma_{k-1}(\ell)\pmod{243}$ if $\ell\equiv1\pmod9$, a congruence
modulo~$729$ for the $\ell$ that split completely in $\Q(\zeta_9,\sqrt[3]3)$
(Theorem~\ref{thR}), a bilinear law for $\ell,\ell'\equiv2\pmod3$
(Theorem~\ref{thS}), and, if $k\not\equiv4\pmod6$, the determination of all $a_n$,
$3\nmid n$, modulo~$81$ by $a_2$ modulo~$81$ (Corollary~\ref{coT}). Then
\S\ref{sec8} establishes a $2$-adic Galois law (Theorem~\ref{thU}), from
the modulo~$128$ theorem of \cite{R19} and a certificate for $T_5$ modulo $512$:
\emph{$a_p\equiv1+p^{k-1}\pmod{2^{10}}$ for every $p\equiv1\pmod8$}, and explicit laws
on the other classes; with a reduction to a computation in a finite group (Theorem~\ref{thX}),
Hatada's question is completely answered: affirmatively for $a\le10$, negatively for
$a=11,12,13$ (Corollary~\ref{coY}), and $a_p(f)=0$ implies $p\equiv-1\pmod{2^{11}}$ for
every eigenform of level~$1$ (Corollary~\ref{coZ}). Finally, \S\ref{sec9} extends the
$3$-adic Galois law beyond $729$: \emph{for every prime $\ell\equiv2\pmod3$,
$a_\ell^2\equiv\sigma_{k-1}(\ell)^2\pmod{3^{\min(2j+2,10)}}$, where $j=v_3(\ell+1)$}
(Theorem~R$'$); the exponent $2j+2$ is attained for $j\le4$ and no longer holds for $j=5$.
It follows that $v_3(a_\ell)=v_3(\ell+1)$ if $v_3(\ell+1)\le4$ and $k\not\equiv4\pmod6$, and
that, for such $k$, $a_\ell(f)=0$ implies $\ell\equiv-1\pmod{2^{11}\cdot3^5}$ (Corollary~Y$'$). For
$\ell\equiv1\pmod{27}$ with $3$ a cube modulo $\ell$, $a_\ell\equiv\sigma_{k-1}(\ell)\pmod{3^7}$,
and $\pmod{3^8}$ if $k\equiv4\pmod6$ and $\ell=X^2+XY+547Y^2$ (Theorem~R$''$).

Each statement carries a status: \textbf{PROVED} (complete proof, possibly resting on
a finite exact computation whose certificate is checked by an ancillary script),
\textbf{KNOWN} (a result of the literature, given again or checked independently),
\textbf{VERIFIED} (checked by computation on the stated range only) or \textbf{OPEN}.

\section{\texorpdfstring{$2$}{2}-adic results}

For $k\in\{12,16,18,20,22,26\}$, let $\Delta_k=\sum_{n\ge1}a_k(n)q^n$ be the unique
normalised eigenform in $S_k$ (thus $\Delta_{12}=\Delta$ and
$a_{12}(n)=\tau(n)$). One has $\Delta_k=E_{k-12}\,\Delta$.

\begin{theoreme}[negative answer for $a=11,12,13$]\label{thA}
The answer to Hatada's question is negative for $a=11$, $a=12$ and $a=13$.
The prime $p=12289=3\cdot 2^{12}+1$ provides witnesses at the three levels:

\smallskip
\begin{center}
\small
\begin{tabular}{lcccccc}
\toprule
 & $\Delta_{12}$ & $\Delta_{16}$ & $\Delta_{18}$ & $\Delta_{20}$ & $\Delta_{22}$ & $\Delta_{26}$\\
\midrule
$a_k(p)\bmod 2^{16}$ & 18434 & 3074 & 11266 & 49154 & 40962 & 25602\\
$v_2\bigl(a_k(p)-1-p\bigr)$ & 11 & 10 & 10 & 12 & 12 & 10\\
failing levels $a$ & 12, 13 & 11, 12, 13 & 11, 12, 13 & 13 & 13 & 11, 12, 13\\
\bottomrule
\end{tabular}
\end{center}
\smallskip

\noindent In particular, for Ramanujan's function,
\[
\tau(12289)=13\,297\,575\,938\,818\,018\,265\,090,\qquad
v_2\bigl(\tau(12289)-12290\bigr)=11,
\]
whereas $12289\equiv1\pmod{2^{12}}$: the requested congruence fails modulo
$2^{12}$ and modulo $2^{13}$. A second witness at level $a=11$ is
$p=13313=13\cdot2^{10}+1$, for which $v_2(a_k(p)-1-p)=10$ if $k=16,18,26$.
\end{theoreme}
\statut{PROVED} Finite exact computation. All values were obtained by two
independent methods which agree: PARI/GP~\cite{PARI}
(\texttt{ramanujantau}; \texttt{mfinit}, \texttt{mfeigenbasis},
\texttt{mfcoef}) and a computation modulo $2^{23}$ by products of the
$\eta$ function and Eisenstein series. The ``lattice'' version of the question therefore
also fails for these three values of $a$.

\begin{proposition}[class congruences]\label{prC}
For $n$ odd and $n\equiv r\pmod 8$,
\[
a_k(n)\equiv c_{k,r}\,\sigma_{k-1}(n)\pmod{2^{A_{k,r}}},
\]
with the following pairs $(A_{k,r},\,c_{k,r})$ (for $r=3,5,7$, the value
given is a representative of the constant):

\smallskip
\begin{center}
\small
\begin{tabular}{ccccc}
\toprule
$k$ & $r=1$ & $r=3$ & $r=5$ & $r=7$\\
\midrule
12 & $(11,\,1)$ & $(13,\,1217)$ & $(12,\,1537)$ & $(14,\,705)$\\
16 & $(10,\,1)$ & $(12,\,609)$  & $(11,\,769)$  & $(13,\,353)$\\
18 & $(10,\,1)$ & $(12,\,97)$   & $(11,\,769)$  & $(13,\,865)$\\
20 & $(12,\,1)$ & $(14,\,3969)$ & $(13,\,3073)$ & $(15,\,2945)$\\
22 & $(12,\,1)$ & $(14,\,897)$  & $(13,\,3073)$ & $(15,\,3969)$\\
26 & $(10,\,1)$ & $(12,\,289)$  & $(11,\,257)$  & $(13,\,545)$\\
\bottomrule
\end{tabular}
\end{center}
\smallskip

\noindent For $k=12$, these are Kolberg's congruences~\cite{K62}.
The exponents are optimal: no constant works modulo
$2^{A_{k,r}+1}$.
\end{proposition}
\statut{PROVED} For odd $r$, the restriction
$f_r=\sum_{n\equiv r\,(8)}a(n)q^n$ of a form $f$ of level~$1$ is a
linear combination of the twists of $f$ by the Dirichlet characters
modulo~$8$, all of which lie in $M_k(\Gamma_0(64))$ with trivial character (these
characters being quadratic). With
$G_k=-\tfrac{B_k}{2k}+\sum_{n\ge1}\sigma_{k-1}(n)q^n$, the form
$g=(\Delta_k)_r-c_{k,r}(G_k)_r$ therefore belongs to
$M_k(\Gamma_0(64))\cap\Z[[q]]$. By Sturm's theorem~\cite{St87},
applied successively to $g,\,g/2,\,g/4,\dots$, it suffices to check
divisibility by $2^{A_{k,r}}$ of the coefficients of index
$n\le\tfrac{k}{12}\,[\mathrm{SL}_2(\Z):\Gamma_0(64)]=8k\le 208$. Checked for
all odd $n\le 2000$, for the six forms and the four classes
($24$ congruences); optimality is witnessed by some $n\le2000$ in each case.

\begin{theoreme}[exact threshold for the six rational forms]\label{thB}
Let $k\in\{12,16,18,20,22,26\}$ and $6\le a\le 13$. Hatada's question has
an affirmative answer for $\Delta_k$ if and only if $a\le a_k$, where
\[
a_{12}=11,\qquad a_{16}=a_{18}=a_{26}=10,\qquad a_{20}=a_{22}=12 .
\]
More precisely, the requested congruence holds for every $a\le 13$ when
$p\equiv-1\pmod{2^{a-1}}$; when $p\equiv+1\pmod{2^{a-1}}$, it holds
exactly for $a\le a_k=A_{k,1}$.
\end{theoreme}
\begin{proof}
Let $p\equiv\pm1\pmod{2^{a-1}}$ with $6\le a\le13$. Then $p\equiv\pm1\pmod8$
and $p^2\equiv1\pmod{2^a}$, whence $p^{k-1}\equiv p\pmod{2^a}$ since $k-2$ is
even.

If $p\equiv-1\pmod{2^{a-1}}$, Proposition~\ref{prC} (class $r=7$,
$A_{k,7}\ge13\ge a$) gives $a_k(p)\equiv c\,(1+p^{k-1})\pmod{2^a}$ with $c$
odd. As $1+p^{k-1}=(1+p)(1-p+\dots+p^{k-2})$, the second factor being
a sum of an odd number of odd terms, one has
$v_2(1+p^{k-1})=v_2(1+p)\ge a-1$, hence $(c-1)(1+p^{k-1})\equiv0\pmod{2^a}$, and
$a_k(p)\equiv1+p^{k-1}\equiv1+p\pmod{2^a}$.

If $p\equiv1\pmod{2^{a-1}}$ and $a\le A_{k,1}$, Proposition~\ref{prC}
(class $r=1$, $c_{k,1}=1$) gives directly
$a_k(p)\equiv1+p^{k-1}\equiv1+p\pmod{2^a}$.

Finally, if $A_{k,1}<a\le13$, the prime $p=12289$, which satisfies
$p\equiv1\pmod{2^{12}}$, is a counterexample (Theorem~\ref{thA}).
\end{proof}
\statut{PROVED}

\begin{corollaire}[non-vanishing]\label{coD}
For each of the six forms, if $p$ is an odd prime and
$v_2(p+1)<A_{k,r}$ with $r\equiv p\pmod8$, then $v_2\bigl(a_k(p)\bigr)=v_2(p+1)$;
in particular $a_k(p)\ne0$. Consequently $a_k(p)=0$ implies
\[
p\equiv-1\pmod{2^{A_{k,7}}},\qquad
2^{A_{k,7}}=2^{14},\,2^{13},\,2^{13},\,2^{15},\,2^{15},\,2^{13}
\quad\text{for } k=12,16,18,20,22,26.
\]
\end{corollaire}
\begin{proof}
By Proposition~\ref{prC}, $a_k(p)\equiv c\,(1+p^{k-1})\pmod{2^{A_{k,r}}}$
with $c$ odd, and $v_2(1+p^{k-1})=v_2(1+p)$. If $p\not\equiv7\pmod 8$, then
$v_2(p+1)\le2<A_{k,r}$.
\end{proof}
\statut{PROVED}

\begin{remarque}[Tian--Qin bounds]
The five bounds $B_k$ ($k=16,18,20,22,26$) of \cite[Prop.~5.2]{TQ}, taken
from the arXiv version 1504.07381v3, are primes satisfying the
congruence conditions of \cite{TQ}, but $v_2(B_k+1)=1,3,1,6,2$
respectively. By Corollary~\ref{coD}, $a_{B_k}(\Delta_k)\ne0$ and
$v_2\bigl(a_{B_k}(\Delta_k)\bigr)=v_2(B_k+1)$. Four of them were
already excluded by \cite{H79,H81}, which force $p\equiv-1\pmod{32}$ for a
zero, and all five by \cite{R19}, which forces $p\equiv-1\pmod{128}$. The values
of the published version of \cite{TQ} were not consulted.
\end{remarque}


\section{The threshold in all weights \texorpdfstring{$\le 100$}{<= 100}}

For even $k$, $12\le k\le 100$, $k\ne14$, and for each prime
$p\le 13313$ with $p\equiv\pm1\pmod{2^7}$ ($48$ primes), we compute the
minimal valuation $s(k,p)=\min_\lambda v_2(\lambda-1-p)$, where $\lambda$ runs over
the eigenvalues of $T_p$ on $S_k$ (all eigenforms, including those with
irrational coefficients; dimension up to~$8$). The computation is exact modulo
$2^{128}$ in a Miller basis; every valuation $\le 14$ obtained is
guaranteed.

\begin{theoreme}[failures in all weights $\le 100$]\label{thD}
For every $k\le 100$ with $k\equiv 0$ or $2\pmod 8$ and $k\ge 16$, there is an
eigenvalue $\lambda$ of $T_{12289}$ on $S_k$ with $v_2(\lambda-12290)=10$;
Hatada's question therefore fails at level $11$ in each of these weights. For
$k\equiv 4$ or $6\pmod 8$, $k\in\{12\}\cup[28,100]$, there is one with
$v_2(\lambda-12290)=11$, hence a failure at level~$12$.
\end{theoreme}
\statut{PROVED} Finite exact computation: for each such $k$, the Newton polygon at $2$ of the
characteristic polynomial of $T_{12289}-12290$ on $S_k$ (PARI/GP, see \S\ref{secverif}).

\begin{center}
\small
\begin{tabular}{lc}
\toprule
weight $k$ & first failing level $a$ observed\\
\midrule
$k\equiv 0,2\pmod 8$, $16\le k\le 98$ & $11$\\
$k\equiv 4,6\pmod 8$, $k=12$ and $28\le k\le 100$ & $12$\\
$k=20,\ 22$ & $13$\\
\bottomrule
\end{tabular}
\end{center}
\noindent{\small The failure at the stated level is proved (Theorems~\ref{thA} and~\ref{thD});
the absence of failure at lower levels is checked on the range of Statement~V.}

\smallskip
\noindent\textbf{Statement V.} For every eigenform of weight $12\le k\le100$, every
$6\le a\le10$ and every prime $p\le13313$ with $p\equiv\pm1\pmod{2^{a-1}}$, one has
$\lambda_p\equiv1+p\pmod{2^a}$. The same holds for $24\le k\le 54$ and all
primes $p\le 60000$ ($183$ primes $p\equiv\pm1\pmod{2^7}$, $48$ of which satisfy
$p\equiv\pm1\pmod{2^9}$). Level $10$ is attained with equality: for example
$v_2\bigl(\tau(7681)-7682\bigr)=10$, where $7681\equiv1\pmod{2^9}$.
\statut{VERIFIED} Exhaustive check on the stated range; the first failing
levels of the table above are unchanged when passing from $p\le13313$ to
$p\le60000$.

\smallskip
\noindent\textbf{Statement V$'$ (``lattice'' form).} For $12\le k\le 100$ and each
of the $194$ primes $p\le 13313$ such that $p\equiv\pm1\pmod{32}$, the lattice
$L^-_{w+2}$ of \cite{H81} is stable under $(T_{w+2}(p)-1-p)/2^a$ for every
$a\le 10$ such that $p\equiv\pm1\pmod{2^{a-1}}$ (ten of these primes reach
$a=10$). The first failing level of the lattice form is $11$, $12$ or $13$
according to $k$; it is strictly smaller than that of the eigenvalues ($11$
against $12$) for $k\in\{36,52,60,68,76,84,92,100\}$.
\statut{VERIFIED} Exact computation on a basis of $L^-_{w+2}$.

\begin{remarque}[conjecture]
The data suggest that Hatada's question has an affirmative answer
exactly for $a\le 10$, including in its ``lattice'' form: the congruence
$\lambda_p\equiv1+p\pmod{2^a}$ for $p\equiv\pm1\pmod{2^{a-1}}$ would hold for
every eigenform of level~$1$ as soon as $a\le10$, and it fails for $a=11$.
Section~\ref{sec8} proves the affirmative answer for $a\le10$ (Corollary~\ref{coY});
the ``lattice'' form remains only verified.
\end{remarque}


\section{\texorpdfstring{$3$}{3}-adic results}\label{sec3}

\paragraph{Notation.} Let $k\ge12$ be even and $w=k-2$. For
$\gamma=\mat abcd\in M_2(\Z)$ with non-zero determinant and $P\in\Q[X]$ of degree
$\le w$, put $(P|\gamma)(X)=(cX+d)^wP\bigl(\frac{aX+b}{cX+d}\bigr)$. With
$S=\mat0{-1}10$ and $U=\mat1{-1}10$, let $W^-$ be the space of odd polynomials
$P$ such that $P|(1+S)=P|(1+U+U^2)=0$. The map sending $f$ to the odd
part of $\int_0^{i\infty}f(z)(X-z)^w\,dz$ identifies $S_k(\mathbb R)$ with
$W^-\otimes\mathbb R$ and transforms $T_n$ into $P\mapsto P|\widetilde T_n$, where
$\widetilde T_n=\sum_{M\in\mathcal M_n}M$ and $\mathcal M_n$ is the set of
integer matrices $\mat abcd$ such that $ad-bc=n$, $a>c\ge0$ and $d>b\ge0$
\cite{Ma73,Z90}. For $f$ with real coefficients, this odd part is the
real part of the integral; since $\widetilde T_n$ is a sum of integer
matrices, it commutes with taking real parts, and $W^-$ is therefore stable
under $\widetilde T_n$.

The coefficient of $X^{w-j}$ in $\int_0^{i\infty}f(z)(X-z)^w\,dz$ is
$(-1)^j\binom wj r_j(f)$, with $r_j(f)=\int_0^{i\infty}f(z)z^j\,dz$. Hatada's lattice
$L^-_{w+2}$, formed by the forms whose odd periods $r_j(f)$
are integers \cite[p.~305]{H81}, therefore corresponds to
\[
L_k=\Bigl\{P=\sum_j\tbinom wj m_jX^j\in W^-\ :\ m_j\in\Z\Bigr\}.
\]
The polynomials $\sum_j\binom wj m_jX^j$, $m_j\in\Z$, form a module $B_w$
stable under $P\mapsto P|\gamma$ for every $\gamma\in M_2(\Z)$, by virtue of
the identity
$\binom wj\binom js\binom{w-j}{l-s}=\binom wl\binom ls\binom{w-l}{j-s}$.
Thus $L_k=W^-\cap B_w$ is a lattice of rank $\dim S_k$, saturated in $B_w$,
stable under each $\widetilde T_n$, whose eigenvalues are the $a_n(f)$.

For $P\in L_k$, we write $\mu_P$ for the $\Z$-linear form on the space
$\Z[x,y]_w$ of integral binary forms of degree $w$ defined by
$\mu_P(x^jy^{w-j})=m_j$. Then $P(X)=\mu_P\bigl((xX+y)^w\bigr)$ and
\[
\mu_{P|\gamma}(F)=\mu_P\bigl(F(ax+cy,\,bx+dy)\bigr).
\]

\begin{lemme}[period identities]\label{l1}
For every $P\in W^-$:
\begin{enumerate}
\item[(a)] $m_j=0$ for even $j$, $m_j=m_{w-j}$ for every $j$, and
$\mu_P\bigl(F(y,x)\bigr)=\mu_P(F)$;
\item[(b)] $P(1)=P'(1)=0$, $P(\tfrac12)=0$ and
$P(\tfrac13)+P(\tfrac23)=0$;
\item[(c)] $\mu_P\bigl((x+y)^w\bigr)=\mu_P\bigl(x(x+y)^{w-1}\bigr)=
\mu_P\bigl(y(x+y)^{w-1}\bigr)=0$ and
$\mu_P\bigl(x^2(x+y)^{w-2}\bigr)=\mu_P\bigl(y^2(x+y)^{w-2}\bigr)=-2m_1$,
$\mu_P\bigl(xy(x+y)^{w-2}\bigr)=2m_1$; the same identities hold
with $x+y$ replaced by $y-x$, with the values $0,0,0$ and $2m_1,2m_1,2m_1$.
\end{enumerate}
\end{lemme}
\begin{proof}
(a) $P$ is odd; the relation $X^wP(-1/X)=-P(X)$ gives $m_j=m_{w-j}$.
(b) The relations for $U$ (at the point $1+t$) and for $S$ (at the point $t$) give the
three-term relation $P(1+t)=P(t)-(1+t)^wP\bigl(\tfrac t{1+t}\bigr)$. At
$t=0$: $P(1)=0$; differentiating at $t=0$: $P'(1)=-wP(0)=0$; at $t=1$, with the
relation for $S$ at the point $2$ and oddness: $P(2)=-2^wP(\tfrac12)=2^wP(\tfrac12)$,
whence $P(\tfrac12)=0$; at $t=-\tfrac13$:
$P(\tfrac23)=P(-\tfrac13)-(\tfrac23)^wP(-\tfrac12)=-P(\tfrac13)$.
(c) One has $P(1)=\mu_P((x+y)^w)$ and $P'(1)=w\,\mu_P(x(x+y)^{w-1})$. The relation for
$U$ applied to $F=x^2y^{w-2}$ reads
$\mu_P(F)+\mu_P\bigl(x^{w-2}(x+y)^2\bigr)+\mu_P\bigl(y^2(x+y)^{w-2}\bigr)=0$,
that is $0+2m_1+\mu_P(y^2(x+y)^{w-2})=0$; the other values follow by
(a) and by writing $y=(x+y)-x$. The passage to $y-x$ is made by the substitution
$x\mapsto-x$, which changes $\mu_P$ into $-\mu_P$.
\end{proof}

\begin{lemme}[$T_3$ modulo $27$]\label{l2}
For every $P\in L_k$ and every form $F\in\Z[x,y]_w$,
\[
\mu_{P|\widetilde T_3}(F)\equiv m_1(P)\Bigl[\,3\sum_{t=0}^{2}\bigl(\partial_yF(1,t)+\partial_xF(t,1)\bigr)
+18\,\bigl(\tfrac12\partial_x^2+\partial_x\partial_y+\tfrac12\partial_y^2\bigr)F(1,-1)\Bigr]\pmod{27}.
\]
In particular $\mu_{P|\widetilde T_3}(F)\equiv0\pmod 9$ if $3\mid m_1(P)$, and
$\mu_{P|\widetilde T_3}(F)\equiv0\pmod{27}$ if $9\mid m_1(P)$.
\end{lemme}
\begin{proof}
The seven elements of $\mathcal M_3$ give the following substitutions
$(x,y)\mapsto(ax+cy,bx+dy)$, written in the form
$\lambda(x,y)\,u+3h(x,y)$:
\[
\begin{array}{lll}
(x,3y)=x\,(1,0)+3(0,y), & (x,x+3y)=x\,(1,1)+3(0,y), & (x,2x+3y)=x\,(1,2)+3(0,y),\\
(3x,y)=y\,(0,1)+3(x,0), & (3x+y,y)=y\,(1,1)+3(x,0), & (3x+2y,y)=y\,(2,1)+3(x,0),\\
\multicolumn{3}{l}{(2x+y,\,x+2y)=(y-x)\,(1,-1)+3(x,y).}
\end{array}
\]
Since the divided derivatives of an integral form are integral, Taylor's formula
gives, modulo $27$,
$F(\lambda u+3h)\equiv\lambda^wF(u)+3\lambda^{w-1}\sum_ih_i\,\partial_iF(u)
+9\lambda^{w-2}\sum_{i\le i'}h_ih_{i'}\,\partial^{[i,i']}F(u)$, with
$\partial^{[i,i]}=\frac12\partial_i^2$ and $\partial^{[1,2]}=\partial_1\partial_2$.
Apply $\mu_P$. By Lemma~\ref{l1}, $\mu_P(\lambda^w)=0$ for
$\lambda\in\{x,y,y-x\}$; $\mu_P(x^{w-1}y)=\mu_P(y^{w-1}x)=m_1$,
$\mu_P(x^w)=\mu_P(y^w)=0$, $\mu_P\bigl((y-x)^{w-1}g\bigr)=0$ for linear $g$;
at second order, $\mu_P(x^{w-2}y^2)=m_{w-2}=0$, $\mu_P(y^{w-2}x^2)=m_2=0$ and
$\mu_P\bigl((y-x)^{w-2}q\bigr)=2q(1,1)\,m_1$ for quadratic $q$. Collecting
the terms gives the stated formula.
\end{proof}

\begin{lemme}[the first period]\label{l3}
For every $P\in L_k$, $3\mid m_1(P)$; if moreover $6\nmid k$, then $9\mid m_1(P)$.
\end{lemme}
\begin{proof}
By Lemma~\ref{l1}(b), $3^w\bigl(P(\tfrac13)+P(\tfrac23)\bigr)=0$, that is
$\sum_j\binom wj3^{w-j}(1+2^j)m_j=0$. Putting $i=w-j$ and using
$m_{w-i}=m_i$, we get
\[
\sum_{i\ \text{odd}}c_i\,m_i=0,\qquad c_i=3^i\tbinom wi\bigl(1+2^{w-i}\bigr).
\]
The terms $i=1$ and $i=w-1$ both carry $m_1$. For odd $n$,
$v_3(1+2^n)=1+v_3(n)$ (lifting the exponent). Thus
$v_3(c_1)=2+v_3(w)+v_3(w-1)$ and, for odd $i\ge3$, with
$\binom wi=\frac wi\binom{w-1}{i-1}$,
\[
\delta_i:=v_3(c_i)-v_3(c_1)=(i-1)-v_3(i)+v_3\tbinom{w-1}{i-1}+v_3(w-i)-v_3(w-1).
\]
Let $e=v_3(w-1)$. If $v_3(i-1)<e$, the addition in base $3$ of $i-1$ and
$w-i$, with sum $w-1\equiv0\pmod{3^e}$, has a carry at each position from
$v_3(i-1)$ to $e-1$; by Kummer's theorem
$v_3\binom{w-1}{i-1}\ge e-v_3(i-1)$, whence
$\delta_i\ge(i-1)-v_3(i)-v_3(i-1)\ge1$, since at most one of the two
consecutive integers $i-1,i$ is divisible by $3$, with valuation at most
$\log_3 i$. If $v_3(i-1)\ge e$ and $e=0$, then $\delta_i\ge(i-1)-v_3(i)\ge1$. If
$v_3(i-1)\ge e\ge1$, then $3\nmid i$, $i-1\ge3^e$ and $\delta_i\ge(i-1)-e\ge2$.
Hence $\delta_i\ge1$ for every odd $i\ge3$; in particular
$v_3(c_1+c_{w-1})=v_3(c_1)$, and the relation gives
$v_3(c_1)+v_3(m_1)\ge v_3(c_1)+\min_{3\le i\le w-3}\delta_i$, that is $v_3(m_1)\ge1$.

If $6\nmid k$, then $w\not\equiv1\pmod3$, hence $e=0$. For $i\ge5$,
$\delta_i\ge(i-1)-v_3(i)\ge2$; for $i=3$,
$\delta_3=1+v_3(w-2)+v_3(w-3)\ge2$, since $3$ divides $w$ or $w-2$. Hence
$v_3(m_1)\ge2$.
\end{proof}

\begin{theoreme}[$a_3$ modulo $9$ and $27$; Hatada]\label{thF}
For every even $k\ge12$, $\widetilde T_3(L_k)\subset 9L_k$, and
$\widetilde T_3(L_k)\subset 27L_k$ if $6\nmid k$. Consequently, for every
eigenform $f$ of level~$1$ and weight $k$, $a_3(f)/9$ is an algebraic
integer, and so is $a_3(f)/27$ when $6\nmid k$. These bounds are
optimal: $\tau(3)=2^2\cdot3^2\cdot7$ $(k=12)$ and
$a_3(\Delta_{16})=-2^2\cdot3^3\cdot31$ $(k=16)$.
\end{theoreme}
\begin{proof}
Let $P\in L_k$. By Lemmas~\ref{l2} and~\ref{l3}, $\mu_{P|\widetilde T_3}(F)$ is
divisible by $9$ (by $27$ if $6\nmid k$) for every $F\in\Z[x,y]_w$, hence
$P|\widetilde T_3\in 9B_w$ (resp. $27B_w$). Since $P|\widetilde T_3\in W^-$ and
$L_k=W^-\cap B_w$ is saturated in $B_w$, we get $P|\widetilde T_3\in9L_k$
(resp. $27L_k$). The operator $\widetilde T_3/9$ (resp. $\widetilde T_3/27$) is
therefore an endomorphism of the $\Z$-module $L_k$; its characteristic polynomial is
monic with integer coefficients and its roots are the $a_3(f)/9$
(resp. $a_3(f)/27$).
\end{proof}
\statut{KNOWN} These are congruences~(6) and~(7) of \cite{H79}, proved by
Hatada on the same lattice, from the first period \cite[\S8]{H79}; the
proof above is a write-up of it. Checks (\texttt{VERIF\_A3.py}): the inequality
$\delta_i\ge1$ (and $\ge2$ if $w\not\equiv1\pmod3$) for all even $w\le3000$;
the identity of Lemma~\ref{l3}, the divisibility of $m_1$ and the inclusion
$\widetilde T_3(L_k)\subset9L_k$ (resp. $27L_k$) on a basis of
$L_k\otimes\Z_{(3)}$ for $12\le k\le100$; eigenvalues recovered for the six rational forms.
With PARI/GP, for every $k\le400$, the smallest $3$-adic slope of the
characteristic polynomial of $T_3$ on $S_k$ is $2$ if $6\mid k$; if $6\nmid k$,
it is $3$, except for $k\equiv10,14\pmod{18}$, where it lies between
$4$ and $6$ (\textbf{VERIFIED}).

\begin{theoreme}[Conjecture 9.1 up to weight $1000$; Chiriac]\label{thG}
For every even weight $12\le k\le1000$, $k\ne14$, every eigenform of level~$1$
and weight $k$ satisfies Conjecture~9.1 of \cite{R19}.
\end{theoreme}
\statut{KNOWN} \cite[Cor.~1.3]{Ch26} together with \cite{R19}. Independent check, which
does not use the non-ramification hypothesis ($494$ spaces, all eigenforms).
Certificate: let
$P$ be the characteristic polynomial of $T_2$ on $S_k$, of degree $d$, computed
exactly modulo $3^{2d+2}$. If $k\equiv4\pmod6$, the Newton polygon of $P$
at $3$ has all its $d$ slopes $\ge2$. If $k\equiv0,2\pmod6$, the number of slopes $\ge2$
of the Newton polygon of $P(x+3)$, plus that of $P(x+6)$, is $d$;
since $v_3\bigl((\lambda-3)-(\lambda-6)\bigr)=1$, no root can be counted
twice, so that every eigenvalue satisfies
$\lambda\equiv3$ or $6\pmod{9\overline{\Z}_3}$. At precision $3^{2d+2}$, the
slopes $<2$ are determined exactly.

\begin{theoreme}[level~$1$ case of the Coleman--Stein conjecture]\label{thH}
Let $f$ be an eigenform of level~$1$ and weight $k$. If $k\le1000$, or if
$3$ is unramified in the coefficient field of $f$, then
$a_3(f)\equiv0\pmod{9\overline{\Z}_3}$ and $f$ is congruent modulo
$9\overline{\Z}_3$, coefficient by coefficient, to one of the five eigenforms
of the table of \cite{CS04} (weights $12$, $16$, $20$, $24$, $32$). The five classes
are distinct and all attained.
\end{theoreme}
\begin{proof}
For $k\ge3$, one has $a_{3n}(f)=a_3(f)a_n(f)-3^{k-1}a_{n/3}(f)\equiv a_3(f)a_n(f)\pmod9$:
the reduction of $f$ modulo $9$ is an eigenvector of the operator $U$, with eigenvalue
$a_3(f)$, and of $t_\Lambda=[1+3]$, with eigenvalue $4^k$. By
\cite[Thm~7.2(ii)]{R19}, the Hecke algebra $\mathbb{T}(3,\Z_3)$ is generated by
$U$, $T_2$, $1+T_7$ and $t_\Lambda$; all the coefficients of $f$ modulo
$9\overline{\Z}_3$ are therefore determined by $(a_2,a_3,a_7,4^k)\bmod 9\overline{\Z}_3$.
One has $a_3\equiv0$ (\cite{H79}; Theorem~\ref{thF}); by \cite[Prop.~9.3]{R19},
$a_7\equiv5,8,2\pmod{9\overline{\Z}_3}$ according as $k\equiv0,2,4\pmod6$, and $4^k\bmod9$
depends only on $k\bmod3$; finally $a_2$ satisfies Conjecture~9.1
(Theorem~\ref{thG} if $k\le1000$, \cite[Cor.~1.3]{Ch26} under the
non-ramification hypothesis). At most five tuples remain: two for $k\equiv0\pmod6$
($a_2\equiv3$ or $6$), two for $k\equiv2\pmod6$, one for $k\equiv4\pmod6$
($a_2\equiv0$). They are realised by the weights $12$ and $24$, $32$
and $20$, and $16$ respectively.
\end{proof}
\statut{PROVED} A consequence of \cite[Thm~7.2, Prop.~9.3]{R19}, of \cite{Ch26}, of
Theorem~\ref{thG} and of \cite{H79} (Theorem~\ref{thF}). Coleman and Stein had checked their
conjecture up to weight $74$ in level~$1$.

\begin{remarque}
In the proof of \cite[Thm~9.2]{R19}, each $a_n(f)$ modulo~$9$ is a
polynomial with coefficients in $\Z/9\Z$ in $a_2(f)$, $a_3(f)$, $a_7(f)$ and the
eigenvalue of $t_\Lambda$; Conjecture~9.1 controls $a_2(f)$,
\cite[Prop.~9.3]{R19} controls $a_7(f)$, the eigenvalue of $t_\Lambda$ is
$4^k$, and $a_3(f)\equiv0\pmod{9\overline{\Z}_3}$ is taken there from \cite{H79}. Only
Conjecture~9.1 therefore remained; it is proved in~\S\ref{sec5}.
\end{remarque}


\section{The eigenvalue \texorpdfstring{$a_2$}{a2} modulo \texorpdfstring{$9$}{9} and \texorpdfstring{$27$}{27}}\label{sec5}

We keep the notation of \S\ref{sec3}. For $\gamma=\mat abcd$ with non-zero
determinant and $F\in\Z[x,y]_w$, put $\sigma_\gamma F(x,y)=F(ax+cy,\,bx+dy)$; thus
$\mu_{P|\gamma}=\mu_P\circ\sigma_\gamma$, $\sigma_\gamma\sigma_\delta=\sigma_{\gamma\delta}$,
and $\sigma_{-\gamma}=\sigma_\gamma$ since $w$ is even. We extend $\gamma\mapsto\sigma_\gamma$
by linearity to the ring $\Z[\mathcal M]$ of the monoid
$\mathcal M=M_2(\Z)\cap\mathrm{GL}_2(\Q)$. We write $T=\mat1101$, $T'=\mat1011$,
$\varepsilon=\mat{-1}001$, $F_x$, $F_y$ for the partial derivatives, and
$s=\sigma_{k-1}(2)=1+2^{k-1}$. Since $2^n=(3-1)^n\equiv-1+3n\pmod9$ for odd $n$,
\[
s\equiv3(k-1)\pmod 9 .
\]
Let $P\in W^-$. By definition, $P|(1+S)=P|(1+U+U^2)=P|(1+\varepsilon)=0$; since
$T=US^{-1}$ and $T'=-U^2S$, we deduce the three-term relation $P=P|T+P|T'$
(the one in the proof of Lemma~\ref{l1}(b)). Thus, for every form $H$,
\[
\mu_P(H+\sigma_\varepsilon H)=\mu_P(H+\sigma_SH)=\mu_P(H-\sigma_TH-\sigma_{T'}H)=0.
\tag{R}
\]
Applying the last one to $\sigma_\varepsilon H$ ($T\varepsilon=\varepsilon T^{-1}$,
$T'\varepsilon=\varepsilon T'^{-1}$), we also get
$\mu_P(H-\sigma_{T^{-1}}H-\sigma_{T'^{-1}}H)=0$. More generally, let $\mathcal I$ be
the right ideal of $\Z[\mathcal M]$ generated by $1+S$, $1+U+U^2$ and $1+\varepsilon$:
one has $\mu_P\circ\sigma_\zeta=0$ for every $\zeta\in\mathcal I$.

\subsection{Modulo \texorpdfstring{$27$}{27}}

\begin{theoreme}[$T_2$ modulo $27$]\label{thI}
For every even $k\ge12$, $\widetilde T_2(L_k)\subset3L_k$ and, for every $P\in L_k$,
\[
P|\widetilde T_2^{\,2}\equiv9(k-1)^2\,P\pmod{27L_k}.
\]
\end{theoreme}

\begin{corollaire}\label{coJ}
For every eigenform $f$ of level~$1$ and weight $k$,
$a_2(f)\equiv0\pmod{3\overline{\Z}_3}$ and
$a_2(f)^2\equiv9(k-1)^2\pmod{27\overline{\Z}_3}$. If $k\equiv0,2\pmod6$, then
$a_2(f)\equiv3$ or $6\pmod{9\overline{\Z}_3}$, without any ramification hypothesis. If
$k\equiv4\pmod6$, then $v_3\bigl(a_2(f)\bigr)\ge3/2$.
\end{corollaire}
\begin{proof}
Let $\lambda=a_2(f)$. The operator $\widetilde T_2$ is an endomorphism of the
$\Z$-module $L_k$, with eigenvalues the $a_2(f)$; let
$v\in L_k\otimes\overline{\Z}_3$ be an eigenvector for $\lambda$ at least one of whose
coordinates is a unit ($\overline{\Z}_3$ is a valuation ring). By
Theorem~\ref{thI}, $\lambda v\in3(L_k\otimes\overline{\Z}_3)$ and
$\bigl(\lambda^2-9(k-1)^2\bigr)v\in27(L_k\otimes\overline{\Z}_3)$; reading a unit
coordinate, $\lambda=3\nu$ with $\nu\in\overline{\Z}_3$ and
$\nu^2\equiv(k-1)^2\pmod{3\overline{\Z}_3}$. If $k\not\equiv1\pmod3$, then
$(k-1)^2\equiv1$. The two factors of $(\nu-1)(\nu+1)\in3\overline{\Z}_3$ are not
both in the maximal ideal, since their difference $2$ is a unit; one of them
is therefore a unit, and the other lies in $3\overline{\Z}_3$. Hence $\nu\equiv\pm1$ and
$\lambda\equiv\pm3\pmod{9\overline{\Z}_3}$. If $k\equiv1\pmod3$, then
$\nu^2\in3\overline{\Z}_3$, that is $v_3(\lambda)\ge3/2$.
\end{proof}
\statut{PROVED} For $k\equiv0,2\pmod6$, one also has
$a_2(f)\equiv\pm s\pmod{9\overline{\Z}_3}$, since $s\equiv3(k-1)\pmod9$. The case
$k\equiv4\pmod6$ is treated in \S\ref{sec5b}.

\smallskip
The proof of Theorem~\ref{thI} occupies the rest of this subsection.

\begin{lemme}[expansion of $T_2$]\label{l4}
For $F\in\Z[x,y]_w$, put
\[
A^*F=y\,F_y(x,-y)+y\,F_y(x,x-y)+x\,F_x(-x,y)+x\,F_x(y-x,y).
\]
There is a $\Z$-linear endomorphism $B^*$ of $\Z[x,y]_w$ such that, for every
$P\in L_k$ and every $F$,
\[
\mu_{P|\widetilde T_2}(F)\equiv-3\mu_P(F)+3\mu_P(A^*F)+9\mu_P(B^*F)\pmod{27}.
\]
\end{lemme}
\begin{proof}
With $2=-1+3$, the four substitutions of $\mathcal M_2$ read
\[
\begin{array}{ll}
(x,2y)=(x,-y)+3(0,y), & (x,x+2y)=(x,x-y)+3(0,y),\\
(2x,y)=(-x,y)+3(x,0), & (2x+y,y)=(y-x,y)+3(x,0).
\end{array}
\]
In other words, each $M\in\mathcal M_2$ is written $M=N+3H$, where $N$ runs over
$N_1=\mat100{-1}$, $N_2=\mat110{-1}$, $N_3=\mat{-1}001$ and $N_4=\mat{-1}011$. Taylor's
formula, whose divided derivatives are integral, gives
$\sum_M\sigma_MF\equiv\sum_i\sigma_{N_i}F+3A^*F+9B^*F\pmod{27}$, the first-order
term being $A^*F$. Finally $\sigma_{N_1}=\sigma_{N_3}=\sigma_\varepsilon$,
$\sigma_{N_2}=\sigma_\varepsilon\sigma_T$ and $\sigma_{N_4}=\sigma_\varepsilon\sigma_{T'}$
(since $N_2=N_1T$ and $N_4=N_3T'$); by (R),
$\mu_P\bigl(\sum_i\sigma_{N_i}F\bigr)=-2\mu_P(F)-\mu_P(\sigma_TF+\sigma_{T'}F)=-3\mu_P(F)$.
\end{proof}

\begin{lemme}[an identity in characteristic~$3$]\label{l5}
Let $\mu$ be a linear form on $\Z[x,y]_w$, with values in $\mathbb F_3$, which
satisfies the relations (R) modulo~$3$. For every $F$,
\[
\mu\bigl(A^*A^*F+A^*F\bigr)=w(w+2)\,\mu(F)-w\,\mu\bigl(y\,\partial_x[(\sigma_T-1)^2F]\bigr).
\]
\end{lemme}
\begin{proof}
Write $E=\partial_x+\partial_y$; steps (i) to (iii) are exact identities.

(i) One has $A^*F=-\sigma_{N_1}(yF_y)+\sigma_{N_2}\bigl((x-y)F_y\bigr)-\sigma_{N_3}(xF_x)
+\sigma_{N_4}\bigl((y-x)F_x\bigr)$. With the equalities of the proof of
Lemma~\ref{l4} and $xF_x+yF_y=wF$, we get $A^*=\sigma_\varepsilon\circ(\Phi-w)$, where
$\Phi F=\sigma_T\bigl((x-y)F_y\bigr)-\sigma_{T'}\bigl((x-y)F_x\bigr)$.

(ii) Since $x-y=-\sigma_\varepsilon(x+y)$, $\partial_y\sigma_\varepsilon=\sigma_\varepsilon\partial_y$,
$\partial_x\sigma_\varepsilon=-\sigma_\varepsilon\partial_x$, $T\varepsilon=\varepsilon T^{-1}$ and
$T'\varepsilon=\varepsilon T'^{-1}$, one has, for every $H$,
$\Phi(\sigma_\varepsilon H)=-\sigma_\varepsilon X_H$ with
$X_H=\sigma_{T^{-1}}\bigl((x+y)H_y\bigr)+\sigma_{T'^{-1}}\bigl((x+y)H_x\bigr)$. With
$G=\Phi F$, we get $A^*A^*F+A^*F=-X_{G-wF}-w(G-wF)+\sigma_\varepsilon(G-wF)$, then,
by (R),
\[
\mu(A^*A^*F+A^*F)=-\mu(X_G+G)-w\,\mu(G)+w\,\mu(X_F)+w(w+1)\,\mu(F).
\]

(iii) Put $K_1=(x-y)F_y$ and $K_2=(x-y)F_x$, so that
$G=\sigma_TK_1-\sigma_{T'}K_2$. The rules $\partial_x\sigma_T=\sigma_TE$,
$\partial_y\sigma_T=\sigma_T\partial_y$, $\partial_x\sigma_{T'}=\sigma_{T'}\partial_x$,
$\partial_y\sigma_{T'}=\sigma_{T'}E$, $(x+y)\,\sigma_TY=\sigma_T(yY)$,
$(x+y)\,\sigma_{T'}Y=\sigma_{T'}(xY)$, $T^{-1}T'=ST$ and $T'^{-1}T=-ST'$ give
\[
X_G=y\,\partial_yK_1-x\,\partial_xK_2-\sigma_S\sigma_T(xEK_2)+\sigma_S\sigma_{T'}(yEK_1).
\]
Put $Z_1=xEK_2+K_1$ and $Z_2=yEK_1+K_2$. Euler's identities
$xF_{xx}+yF_{xy}=(w-1)F_x$ and $xF_{xy}+yF_{yy}=(w-1)F_y$ give
$Z_1+Z_2=w(x-y)EF$ and $y\,\partial_yK_1-x\,\partial_xK_2-Z_2=-wF-w(x-y)F_x$. By (R),
$\mu\circ\sigma_S=-\mu$ and $\mu\circ\sigma_{T'}=\mu-\mu\circ\sigma_T$; hence
\[
\mu(X_G+G)=\mu\bigl(y\,\partial_yK_1-x\,\partial_xK_2-Z_2\bigr)+\mu\bigl(\sigma_T(Z_1+Z_2)\bigr)
=w\,\mu\bigl(-F-(x-y)F_x+\sigma_T((x-y)EF)\bigr).
\]

(iv) Substituting into (ii), using again $\mu\circ\sigma_{T'}=\mu-\mu\circ\sigma_T$ and
$\mu\circ\sigma_{T'^{-1}}=\mu-\mu\circ\sigma_{T^{-1}}$, together with
$\sigma_T\bigl((x-y)EF\bigr)=-y\,\partial_x(\sigma_TF)$ and
$\sigma_{T^{-1}}\bigl((x+y)(F_x-F_y)\bigr)=y\,\partial_x(\sigma_{T^{-1}}F)$, we obtain
the exact identity
\[
\mu(A^*A^*F+A^*F)=w(w+2)\,\mu(F)+3w\,\mu(xF_x)-w\,\mu\bigl(y\,\partial_x[(1-2\sigma_T+\sigma_{T^{-1}})F]\bigr).
\]
Modulo~$3$, the term $3w\,\mu(xF_x)$ vanishes, $\sigma_{T^{-1}}=\sigma_{T^2}$ (since
$T^3\equiv I$) and $1-2\sigma_T+\sigma_{T^2}\equiv1+\sigma_T+\sigma_{T^2}\equiv(\sigma_T-1)^2$.
\end{proof}

\begin{lemme}[information modulo~$9$]\label{l6}
For every $P\in L_k$ and every $F\in\Z[x,y]_w$,
$\mu_P\bigl(y\,\partial_x[(\sigma_T-1)^2F]\bigr)\equiv0\pmod3$.
\end{lemme}
\begin{proof}
Let $\mathcal G$ be the group of matrices in $\mathrm{GL}_2(\Z/9\Z)$ of determinant
$\pm1$, modulo $\pm I$; it has $648$ elements, and the action $F\mapsto\sigma_\gamma F$
on $\Z[x,y]_w/9$ factors through $\mathcal G$, whence an action of the ring
$\mathcal R=(\Z/9\Z)[\mathcal G]$. Let $\mathcal I_9$ be the image of $\mathcal I$ in
$\mathcal R$, and $\mathcal C\subset\mathcal R$ the two-sided ideal generated by $3$ and
the $\kappa-1$, $\kappa\equiv\pm I\pmod3$. A finite exact computation in $\mathcal R$
(an explicit combination of $290$ terms, redone by \texttt{VERIF\_A2.py}) shows
that
\[
\xi:=T\,(1+T'+T'^2)(1+T+T^2)\in\mathcal I_9+\mathcal C^2.
\]
The elements of $\mathcal I_9$ annihilate $\mu_P$ modulo~$9$. Each $\sigma_\kappa$,
$\kappa\equiv\pm I\pmod3$, is congruent to the identity modulo~$3$; every element of
$\mathcal C$ therefore acts by an operator vanishing modulo~$3$, and $\mathcal C^2$ acts by
$0$ modulo~$9$. Thus $\mu_P(\sigma_\xi F)\equiv0\pmod9$. On the other hand,
multiplying on the right the relation $P=P|T+P|T'$ by $T'^j$ ($j=0,1,2$) and summing,
we get $P|T(1+T'+T'^2)=P-P|T'^3$; with $H=(1+\sigma_T+\sigma_{T^2})F$ and
$\sigma_{T'^3}H(x,y)=H(x+3y,y)$, Taylor's formula gives
\[
\mu_P(\sigma_\xi F)=\mu_P(H)-\mu_P(\sigma_{T'^3}H)\equiv-3\,\mu_P(y\,\partial_xH)\pmod9.
\]
Hence $\mu_P(y\,\partial_xH)\equiv0\pmod3$, and one concludes with
$1+\sigma_T+\sigma_{T^2}=(\sigma_T-1)^2+3\sigma_T$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thI}]
Let $P\in L_k$. By Lemma~\ref{l4}, $\mu_{P|\widetilde T_2}(F)\equiv0\pmod3$ for
every $F$, hence $P|\widetilde T_2\in3B_w\cap W^-=3L_k$. Put $Q=P|\widetilde T_2$.
Lemma~\ref{l4} gives $\mu_Q(G)\equiv3\mu_P(A^*G-G)\pmod9$ for every $G$, and,
applied to $Q$, $\mu_{Q|\widetilde T_2}(F)\equiv3\mu_Q(A^*F-F)+9\mu_Q(B^*F)
\equiv9\mu_P\bigl((A^*-1)^2F\bigr)\pmod{27}$, since $\mu_Q\equiv0\pmod3$. Modulo~$3$,
$(A^*-1)^2\equiv A^*A^*+A^*+1$; Lemmas~\ref{l5} and~\ref{l6}, applied to the
reduction of $\mu_P$, give
$\mu_P\bigl((A^*-1)^2F\bigr)\equiv\bigl(1+w(w+2)\bigr)\mu_P(F)=(k-1)^2\mu_P(F)\pmod3$.
Thus $P|\widetilde T_2^{\,2}-9(k-1)^2P\in27B_w\cap W^-=27L_k$.
\end{proof}
\statut{PROVED} Lemma~\ref{l6} rests on a finite exact computation, whose
certificate is checked by \texttt{VERIF\_A2.py}. Checks: the congruence of
Theorem~\ref{thI} checked directly on $L_k$ for $12\le k\le290$ (PARI/GP);
the identity of Lemma~\ref{l5} on all solutions modulo~$3$ of the relations, and the
condition of Lemma~\ref{l6} on $L_k$, for $k\le40$ (\texttt{VERIF\_A2.py}).

\subsection{Modulo \texorpdfstring{$81$}{81}}\label{sec5b}

\begin{theoreme}[$T_2$ modulo $81$]\label{thK}
Let $k\ge12$ be even and $P\in L_k$. Then
\[
\begin{array}{ll}
P|\widetilde T_2^{\,2}\equiv s^2P\pmod{81L_k} & \text{if }k\equiv0,2\pmod6,\\[2pt]
P|\widetilde T_2^{\,2}+9\,P|\widetilde T_2\equiv0\pmod{81L_k} & \text{if }k\equiv4\pmod6.
\end{array}
\]
\end{theoreme}

\begin{corollaire}\label{coL}
Let $f$ be an eigenform of level~$1$ and weight $k$. If $k\equiv0,2\pmod6$,
then $a_2(f)\equiv\pm\sigma_{k-1}(2)\pmod{27\overline{\Z}_3}$. If
$k\equiv4\pmod6$, then $a_2(f)\equiv0\pmod{9\overline{\Z}_3}$. In particular,
Conjecture~9.1 of \cite{R19} holds for every eigenform of level~$1$.
\end{corollaire}
\begin{proof}
Let $\lambda=a_2(f)$. As in the proof of Corollary~\ref{coJ},
Theorem~\ref{thK} gives $\lambda^2-s^2\in81\overline{\Z}_3$ if $k\equiv0,2\pmod6$, and
$\lambda^2+9\lambda\in81\overline{\Z}_3$ if $k\equiv4\pmod6$.

If $k\equiv4\pmod6$ and $v_3(\lambda)<2$, then $v_3(\lambda+9)=v_3(\lambda)$ and
$v_3(\lambda^2+9\lambda)=2v_3(\lambda)<4$, which is excluded; hence
$\lambda\in9\overline{\Z}_3$.

If $k\equiv0,2\pmod6$, then $3\nmid k-1$, hence $s\equiv3(k-1)\equiv\pm3\pmod9$ and
$v_3(s)=1$. From $\lambda^2\equiv s^2\pmod{81}$ we get $v_3(\lambda)=1$. The elements
$\lambda-s$ and $\lambda+s$ lie in $3\overline{\Z}_3$, and their sum $2\lambda$ has
valuation $1$: one of the two has valuation exactly $1$, and the other, whose
product with the first lies in $81\overline{\Z}_3$, has valuation $\ge3$.
Hence $\lambda\equiv\pm s\pmod{27\overline{\Z}_3}$; reducing modulo~$9$,
$\lambda\equiv3$ or $6\pmod{9\overline{\Z}_3}$.
\end{proof}
\statut{PROVED} From Theorem~\ref{thK}, whose proof is completed below (Lemma~\ref{l8}).
The same consequences follow from \cite[Thm~1.5]{R26}, obtained independently by another method.

\begin{lemme}[a vanishing criterion]\label{l7}
Let $n\ge1$ and $r\in\{0,1,2\}$. Let $\mathcal H$ be a system of integral
representatives of $\mathrm{GL}_2(\mathbb F_3)/\{\pm1\}$, and let $X=\sum_gc_g\,g$ be a
finite combination, with integer coefficients, of integer matrices $g$ with determinant
prime to $3$. Write $g\equiv\pm h_g\kappa_g\pmod{3^n}$ with $h_g\in\mathcal H$ and
$\kappa_g=\mat pquv\equiv I\pmod3$. For $\delta\in\Z$ and $A,B\ge0$, put
\[
c_{\delta,A,B}(\kappa_g)=\sum u^a\,q^b\,(p-1)^m\,(v-1)^{m'}\tbinom Aa\tbinom Bb,
\]
the sum running over the $a,b,m,m'\ge0$ such that $b-a=\delta$, $a+m=A$ and $b+m'=B$;
and, for $h\in\mathcal H$,
\[
f_{h,\delta}(i,t)=\sum_{h_g=h}c_g\sum_{A+B<n}c_{\delta,A,B}(\kappa_g)\tbinom iA\tbinom{3t+r-i}B,
\]
where $\binom zB=z(z-1)\cdots(z-B+1)/B!$. If, for all $h$, $\delta$ and $A+B<n$, the
finite difference $\Delta_i^A\Delta_t^Bf_{h,\delta}(0,0)$ is divisible by $3^n$,
then $\sigma_X(\Z[x,y]_w)\subset3^n\Z[x,y]_w$ for every even $w$ such that
$w\equiv r\pmod3$.
\end{lemme}
\begin{proof}
Let $w$ be even, $w=3t+r$. The operator $\sigma_g$ modulo $3^n$ depends only on $g$
modulo $3^n$, and $\sigma_{-g}=\sigma_g$; hence $\sigma_g\equiv\sigma_{h_g}\sigma_{\kappa_g}$.
For $\kappa=\mat pquv$ and $i+j=w$,
\[
\sigma_\kappa(x^iy^j)=(px+uy)^i(qx+vy)^j
=\sum_{a,b}\tbinom ia\tbinom jb\,p^{i-a}u^a\,q^bv^{j-b}\,x^{i-a+b}y^{j+a-b}.
\]
Expand $p^{i-a}=\sum_m\binom{i-a}m(p-1)^m$ and
$v^{j-b}=\sum_{m'}\binom{j-b}{m'}(v-1)^{m'}$, and use
$\binom ia\binom{i-a}m=\binom i{a+m}\binom{a+m}a$. The terms with $a+b+m+m'\ge n$ are
divisible by $3^n$, since $u$, $q$, $p-1$ and $v-1$ are divisible by~$3$. We get
\[
\sigma_X(x^iy^{w-i})\equiv\sum_{h\in\mathcal H}\sigma_h\Bigl(\sum_\delta f_{h,\delta}(i,t)\,x^{i+\delta}y^{w-i-\delta}\Bigr)\pmod{3^n}.
\]
The polynomial $f_{h,\delta}$ has total degree $<n$ and takes integer values on
$\Z^2$; by Newton's formula,
$f_{h,\delta}(i,t)=\sum_{A+B<n}\Delta_i^A\Delta_t^Bf_{h,\delta}(0,0)\binom iA\binom tB$.
Under the hypothesis, all its values are divisible by $3^n$, and
$\sigma_X(x^iy^{w-i})\in3^n\Z[x,y]_w$ for every $i$.
\end{proof}

\begin{lemme}[three certificates]\label{l8}
Let $\mathcal T=\sum_{M\in\mathcal M_2}M$ and $\Lambda_2=\mat2002$, and put
$\Xi_\sigma=\mathcal T^2-(1+2\Lambda_2)^2$ and $\Xi_4=\mathcal T^2+9\mathcal T$. There
are explicit elements $\zeta_0,\zeta_1,\zeta_2\in\mathcal I$, sums of
$977$, $1005$ and $1079$ terms, such that $\Xi_\sigma-\zeta_0$, $\Xi_\sigma-\zeta_1$ and
$\Xi_4-\zeta_2$ satisfy the hypothesis of Lemma~\ref{l7} with $n=4$ and $r=0$, $1$,
$2$ respectively.
\end{lemme}
\begin{proof}
Each term has the form
$c\,\gamma\,h\,(\kappa_1-1)^{\alpha_1}(\kappa_2-1)^{\alpha_2}(\kappa_3-1)^{\alpha_3}(\kappa_4-1)^{\alpha_4}$,
with $c$ an integer, $\gamma\in\{1+S,\,1+U+U^2,\,1+\varepsilon\}$, $h\in\mathcal H$,
$\kappa_1=\mat4001$, $\kappa_2=\mat1301$, $\kappa_3=\mat1031$, $\kappa_4=\mat1004$ and
$\alpha_1+\alpha_2+\alpha_3+\alpha_4\le3$; it belongs to $\mathcal I$, which is a
right ideal. Checking the hypothesis of Lemma~\ref{l7} is a finite exact
computation: one expands $\Xi-\zeta_r$ as a combination of matrices modulo~$81$, then
computes the finite differences. This computation, and the three certificates, are in
\texttt{VERIF\_A2.py}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thK}]
Let $P\in L_k$, $w=k-2$ and $r\in\{0,1,2\}$ such that $w\equiv r\pmod3$: $r=0$ if
$k\equiv2\pmod6$, $r=1$ if $k\equiv0\pmod6$ and $r=2$ if $k\equiv4\pmod6$. Let
$\Xi=\Xi_\sigma$ if $r\le1$, $\Xi=\Xi_4$ if $r=2$, and $\zeta=\zeta_r$. Since
$\mu_{P|\gamma}=\mu_P\circ\sigma_\gamma$ and $\Lambda_2$ acts on $\Z[x,y]_w$ by
$2^w$, the form $\mu_P\circ\sigma_\Xi$ is that of $P|\widetilde T_2^{\,2}-s^2P$ if
$r\le1$, of $P|\widetilde T_2^{\,2}+9\,P|\widetilde T_2$ if $r=2$. One has
$\mu_P\circ\sigma_\zeta=0$ since $\zeta\in\mathcal I$, and, by
Lemmas~\ref{l7} and~\ref{l8}, $\sigma_{\Xi-\zeta}(F)\in81\Z[x,y]_w$ for every $F$.
Since $\mu_P$ takes integer values on $\Z[x,y]_w$, we get
$\mu_P(\sigma_\Xi F)\equiv0\pmod{81}$ for every $F$: the polynomial under consideration
lies in $81B_w\cap W^-=81L_k$.
\end{proof}
\statut{PROVED} Finite exact computation (Lemma~\ref{l8}), checked by \texttt{VERIF\_A2.py}
(part E). Checks: direct action of the remainders $\Xi-\zeta_r$ on $\Z[x,y]_w$
modulo~$81$ for all even $w\le60$ in the relevant class (part G);
congruences of Theorem~\ref{thK} checked directly on
$L_k$ for $12\le k\le290$ and, if $k\equiv4\pmod6$, up to $k=334$ (PARI/GP).
For $k\equiv0,2\pmod6$, an operator congruence of the same kind for $T_2^2$ modulo $81$, on
a lattice of even modular symbols, appears independently in \cite[\S5.2.2]{R26}.

\begin{remarque}
Theorem~\ref{thK} implies Theorem~\ref{thI}, given
$\widetilde T_2(L_k)\subset3L_k$ (Lemma~\ref{l4}). For $k\equiv4\pmod6$, one does not have
$\widetilde T_2(L_k)\subset9L_k$ in general (for example for $k=40$): the
quadratic congruence of Theorem~\ref{thK} is the natural form of the statement. Corollaries~\ref{coJ}
and~\ref{coL} contain \cite[Thm~1.2 and Cor.~1.3]{Ch26}; the
difference is that the congruences of Theorems~\ref{thI} and~\ref{thK} concern
the operator $\widetilde T_2$ acting on the whole lattice $L_k$, and therefore pass
to all eigenvalues, whatever the coefficient field. For
the rational forms of weights $12$, $18$, $20$ and $26$, one has
$a_2=-24$, $-528$, $456$, $-48$, and even $a_2\equiv-s\pmod{27}$; for example
$\tau(2)=-24\equiv-2049\pmod{27}$. For $k=16$ and $22$, $a_2=216$ and $-288$ are
divisible by~$9$.
\end{remarque}

\begin{theoreme}[level~$1$ case of the Coleman--Stein conjecture]\label{thM}
Let $f$ be an eigenform of level~$1$ and weight $k$. Then
$a_3(f)\equiv0\pmod{9\overline{\Z}_3}$, and $f$ is congruent modulo
$9\overline{\Z}_3$, coefficient by coefficient, to one of the five eigenforms of
the table of \cite{CS04}: the one of weight $16$ if $k\equiv4\pmod6$; the one of weight
$12$ or the one of weight $24$ if $k\equiv0\pmod6$, and the one of weight $32$ or the one of
weight $20$ if $k\equiv2\pmod6$, according as $a_2(f)\equiv3$ or $6\pmod{9\overline{\Z}_3}$.
\end{theoreme}
\begin{proof}
This is the proof of Theorem~\ref{thH}, with Corollaries~\ref{coJ}
and~\ref{coL} in place of Conjecture~9.1.
\end{proof}
\statut{PROVED} A consequence of \cite[Thm~7.2, Prop.~9.3]{R19}, of \cite{H79}
(Theorem~\ref{thF}) and of Corollaries~\ref{coJ} and~\ref{coL}, without any hypothesis on
$k$ or on the coefficient field. This statement also follows from \cite[Thm~1.5]{R26},
obtained independently by another method.


\section{An Eisenstein law modulo \texorpdfstring{$81$}{81}}\label{sec6}

We write $\chi$ for the Dirichlet character of conductor~$3$ ($\chi(n)=1$ if
$n\equiv1$, $-1$ if $n\equiv2\pmod3$), $\mathbf 1_3$ for the principal character
modulo~$3$, $\sigma(n)=\sigma_{k-1}(n)$ and, for a prime $\ell\ne3$,
$s_\ell=\sigma_{k-1}(\ell)=1+\ell^{k-1}$. If $\ell\equiv2\pmod3$,
$v_3(s_\ell)=v_3(\ell+1)+v_3(k-1)$; for $\ell\not\equiv-1\pmod9$ (for example
$\ell=2,5,11$), this equals $1$ if $k\not\equiv4\pmod6$ and is at least $2$ otherwise.

\subsection{Other operators \texorpdfstring{$T_\ell$}{T\_l}}

Theorem~\ref{thK} extends to other primes by the same method:
$\widetilde T_\ell=\sum_{M\in\mathcal M_\ell}M$ replaces $\widetilde T_2$
($|\mathcal M_\ell|=15$, $25$, $54$, $77$, $93$, $109$ for $\ell=5,7,11,13,17,19$),
$\Lambda_\ell=\mat\ell00\ell$, which acts on $\Z[x,y]_w$ by $\ell^w$, replaces
$\Lambda_2$, and Lemma~\ref{l7} applies with any modulus $3^n$.

\begin{theoreme}[$T_\ell$ modulo $81$ and $243$, $\ell\le19$]\label{thN}
Let $k\ge12$ be even and $P\in L_k$.
\begin{enumerate}
\item[(a)] For $\ell\in\{7,13\}$: $P|\widetilde T_\ell\equiv s_\ell P\pmod{81L_k}$ if
$k\not\equiv4\pmod6$, and $\pmod{27L_k}$ if $k\equiv4\pmod6$. For $\ell=19$: the
same congruences modulo $243L_k$ and $81L_k$ respectively.
\item[(a$'$)] If $k\equiv4\pmod6$:
$P|\widetilde T_7-s_7P-18\,P|\widetilde T_2\equiv0\pmod{81L_k}$.
\item[(a$''$)] If $k\equiv4\pmod6$:
$P|\widetilde T_{19}-s_{19}P-54\,P|\widetilde T_2\equiv0\pmod{243L_k}$.
\item[(b)] For $\ell\in\{5,11\}$ and $k\not\equiv4\pmod6$:
$P|\widetilde T_\ell^{\,2}\equiv s_\ell^2P\pmod{81L_k}$; for $\ell=17$, modulo $243L_k$.
\item[(c)] For $\ell\in\{5,11\}$ and $k\equiv4\pmod6$: there are explicit
polynomials $a_\ell,b_\ell\in(\Z/81\Z)[X]$ of degree $5$, with $b_\ell(\ell^w)\equiv0$
and $a_\ell(\ell^w)\equiv0\pmod9$ for every even $w\equiv2\pmod3$, such that
$P|\widetilde T_\ell^{\,2}+a_\ell(\ell^w)\,P|\widetilde T_\ell+b_\ell(\ell^w)P\equiv0\pmod{81L_k}$.
\end{enumerate}
In particular, for every eigenform $f$ of level~$1$ and weight $k$,
$a_7(f)\equiv s_7\pmod{81\overline{\Z}_3}$ and $a_{19}(f)\equiv s_{19}\pmod{243\overline{\Z}_3}$ whatever $k$.
\end{theoreme}
\begin{proof}
Each congruence on $L_k$ is established by a certificate $\zeta\in\mathcal I$ and the
criterion of Lemma~\ref{l7}, exactly as in Lemma~\ref{l8}: nineteen certificates
(from $229$ to $1600$ terms), all checked by \texttt{VERIF\_EISENSTEIN.py}, which
also checks, for (c), the two conditions on $a_\ell(\ell^w)$ and $b_\ell(\ell^w)$
(they depend only on $\ell^w$ modulo~$81$, which takes $27$ values). For the
last assertion, if $k\equiv4\pmod6$, (a$'$) gives
$a_7-s_7\equiv18\,a_2\pmod{81}$ and (a$''$) gives $a_{19}-s_{19}\equiv54\,a_2\pmod{243}$, and $9\mid a_2$ (Corollary~\ref{coL}).
\end{proof}
\statut{PROVED} Finite exact computation. For $\ell=7$, the congruence modulo~$9$ is
\cite[Prop.~9.3]{R19}; the congruence $a_7\equiv s_7$ modulo $81$ in the valuative sense
is also in \cite[(5.5)]{R26}.

\subsection{The law}

We use the full Hecke algebra $\mathbb T=\mathbb T(3,\Z_3)$ of \cite{R19}:
it acts on the space $D(3,\cdot)$ of divided congruences (Katz) of cusp forms
of level~$3$, is generated by the $T_n$, $3\nmid n$, and by $U$, and
contains the operator $t=[1+3]$ of the action of $\Z_3^\times$, which acts by $x^k$ on
forms of weight~$k$, so that $t$ acts by $t_k=4^k$. Let
$V^{\mathrm{par}}$ be the space of cuspidal $3$-adic modular functions of tame
level~$1$ (in Katz's sense: vanishing at all cusps of the Igusa tower), and
$V\supset V^{\mathrm{par}}$ that of all $3$-adic modular functions of tame
level~$1$. For every finite quotient $R$ of the ring of integers $O$ of a
finite extension of $\Q_3$, one has $D(3,R)=D(3,O)\otimes R=V^{\mathrm{par}}(O)\otimes R$
\cite[Prop.~2.1 and Lemma~3.1]{R19}. We use three of its properties.
\begin{enumerate}
\item[(F1)] \cite[Thm~7.2(ii)]{R19}: $\mathbb T$ is local, with residue field
$\mathbb F_3$, and the set $\mathcal S=\{U,T_2,1+T_7,t-1\}$ generates it as a
$\Z_3$-algebra (by Nakayama's lemma: $\mathfrak m=(3,\mathcal S)+\mathfrak m^2$).
By induction, $\mathbb T=\Z_3[\mathcal S]+\mathfrak m^N$ for every $N$; since every
finite quotient of $\mathbb T$ is local Artinian, hence killed by a power of
$\mathfrak m$, the image of $\mathbb T$ in every finite quotient is that of
$\Z_3[\mathcal S]$.
\item[(F2)] \cite[Thm~2.2, Prop.~3.3]{R19} (Katz): the $q$-expansion is
injective on $V^{\mathrm{par}}(O)$. If $F\in V^{\mathrm{par}}(O)$ satisfies $a_1(F)=1$ and
is an eigenvector of the $T_n$ ($3\nmid n$) and of $U$, its reduction modulo~$81$ is an
element of $D(3,O/81)=V^{\mathrm{par}}(O)\otimes O/81$ which is an eigenvector of $\mathbb T$, and
$\tau\mapsto a_1(\tau F)$ defines a character $\varphi_F:\mathbb T\to O/81O$, with
$\varphi_F(T_n)=a_n(F)$ for $3\nmid n$ (since $a_1(T_nG)=a_n(G)$). Moreover
$V^{\mathrm{par}}(O)$ is saturated in $O[[q]]$: if the expansion of
$F\in V^{\mathrm{par}}(O)$ lies in $3\,O[[q]]$, then $F\in3V^{\mathrm{par}}(O)$, since
$D(3,O/3)=V^{\mathrm{par}}(O)\otimes O/3$ injects into $(O/3)[[q]]$.
\item[(F3)] If $G\in V(O)$ is an eigenvector of weight $k$ of the $T_\ell$
($\ell\ne3$), with eigenvalues $\lambda_\ell$, the series
$\theta^jG=\sum n^ja_n(G)q^n$, $\theta=q\,d/dq$, is a $3$-adic modular form
in Serre's sense \cite{Se73}, a limit of forms of level~$1$ with coefficients in $O$,
hence an element of $V(O)$ \cite{Ka75}; it has weight $k+2j$ (the action of $x\in\Z_3^\times$ is multiplied by $x^{2j}$), and is an eigenvector
with eigenvalues $\ell^j\lambda_\ell$. Since $x$ acts by a scalar, the
constant term of $\theta^jG$, which vanishes at the cusp~$\infty$, vanishes at all cusps:
$\theta^jG\in V^{\mathrm{par}}(O)$. For $j=2\cdot3^m$ (resp.
$j=3^m$), $n^j$ tends, as $m\to\infty$, to $\mathbf 1_3(n)$ (resp. $\chi(n)$),
uniformly in~$n$. The limits $G\otimes\mathbf 1_3=\sum_{3\nmid n}a_n(G)q^n$ and
$G\otimes\chi$ are therefore in $V^{\mathrm{par}}(O)$, eigenvectors of the
$T_\ell$ with eigenvalues $\lambda_\ell$ and $\chi(\ell)\lambda_\ell$, killed
by $U$, and $t$ acts on them by the same scalar as on $G$ (since $4^{2j}\to1$).
\end{enumerate}

\begin{theoreme}[Eisenstein law modulo $81$]\label{thO}
Let $f$ be a normalised cuspidal eigenform of level~$1$ and weight $k$, and
fix an embedding of its coefficient field into $\overline\Q_3$.
\begin{enumerate}
\item[(i)] For every $n\equiv1\pmod3$: $a_n(f)\equiv\sigma_{k-1}(n)\pmod{81\overline{\Z}_3}$.
\item[(ii)] If $k\not\equiv4\pmod6$, let $\psi=\mathbf 1_3$ if
$a_2(f)\equiv s_2\pmod{27}$ and $\psi=\chi$ if $a_2(f)\equiv-s_2\pmod{27}$
(Corollary~\ref{coL}); then $a_n(f)\equiv\psi(n)\,\sigma_{k-1}(n)\pmod{27\overline{\Z}_3}$
for every $n$ prime to~$3$.
\item[(iii)] If $k\equiv4\pmod6$: $a_n(f)\equiv\sigma_{k-1}(n)\pmod{9\overline{\Z}_3}$
for every $n$ prime to~$3$.
\end{enumerate}
\end{theoreme}
\begin{proof}
Let $O$ be the ring of integers of a finite extension of $\Q_3$ containing the
coefficients of $f$, and $R=O/81O$. The weight $k$ being fixed, let $E=-\frac{2k}{B_k}G_k$,
where $G_k=-\frac{B_k}{2k}+\sum\sigma(n)q^n$; by von Staudt--Clausen, $v_3(B_k)=-1$,
hence $c=-2k/B_k$ satisfies $v_3(c)=1+v_3(k)\ge1$, and $E=1+c\sum\sigma(n)q^n$ is a
form of level~$1$ with coefficients in $\Z_3$, hence an element of $V(\Z_3)$. By (F3),
applied to $f$ and to $E$, then the saturation (F2) (to divide by $c$), the four series
\[
f\otimes\mathbf 1_3,\qquad f\otimes\chi,\qquad E^{\mathbf 1}=\sum_{3\nmid n}\sigma(n)q^n,
\qquad E^{\chi}=\sum_{3\nmid n}\chi(n)\sigma(n)q^n
\]
lie in $V^{\mathrm{par}}(O)$, with $a_1=1$, are eigenvectors of the $T_\ell$, are killed
by $U$, and $t$ acts on them by $t_k$. By (F2), they define four characters
$\varphi,\varphi^\chi,\varphi_E,\varphi_E^\chi:\mathbb T\to R$. On the generators of
(F1), they take the values
\[
U\mapsto0,\qquad t\mapsto t_k,\qquad T_2\mapsto a_2,\ -a_2,\ s_2,\ -s_2,\qquad
1+T_7\mapsto1+a_7,\ 1+a_7,\ 1+s_7,\ 1+s_7,
\]
and $a_7\equiv s_7\pmod{81}$ (Theorem~\ref{thN}). Let $n$ be prime to~$3$. The image of
$\mathbb T$ in $R^4$ is finite; (F1) therefore provides a polynomial $P$ with
coefficients in $\Z_3$ such that $T_n$ and $P(U,T_2,1+T_7,t)$ have the same image under
the four characters. Put $Q(x)=P(0,x,1+s_7,t_k)\in R[x]$. Then
\[
a_n(f)\equiv Q(a_2),\quad \chi(n)a_n(f)\equiv Q(-a_2),\quad
\sigma(n)\equiv Q(s_2),\quad \chi(n)\sigma(n)\equiv Q(-s_2)\pmod{81}.
\]
(i) If $n\equiv1\pmod3$, write $Q(x)=Q_0(x^2)+xQ_1(x^2)$. From $Q(a_2)\equiv Q(-a_2)$
we get $2a_2Q_1(a_2^2)\equiv0$, and likewise for $s_2$; hence
$a_n(f)\equiv Q_0(a_2^2)$ and $\sigma(n)\equiv Q_0(s_2^2)$. Now $a_2^2\equiv s_2^2\pmod{81}$:
this is Theorem~\ref{thK} if $k\not\equiv4\pmod6$; if $k\equiv4\pmod6$, one has
$a_2^2\equiv0$ and $s_2^2\equiv0\pmod{81}$, since $9\mid a_2$ (Corollary~\ref{coL}) and
$9\mid s_2$. Hence $a_n(f)\equiv\sigma(n)\pmod{81}$.

(ii) If $\psi(2)s_2\equiv a_2\pmod{27}$, then $Q(a_2)\equiv Q(\psi(2)s_2)\pmod{27}$,
that is, $a_n(f)\equiv\psi(n)\sigma(n)\pmod{27}$.

(iii) If $k\equiv4\pmod6$, $a_2\equiv s_2\equiv0\pmod9$, hence
$a_n(f)\equiv Q(a_2)\equiv Q(s_2)\equiv\sigma(n)\pmod9$.
\end{proof}
\statut{PROVED} From \cite[Prop.~2.1, Thm~2.2, Lemma~3.1, Prop.~3.3, Thm~7.2(ii)]{R19},
from the theory of $3$-adic modular forms \cite{Se73,Ka75} and from
Theorems~\ref{thK} and~\ref{thN}. The principle --- $a_n(f)$ is a polynomial in the
eigenvalues of the generators of $\mathbb T$ --- is that of the proof of
Theorem~8.1 of \cite{R19} ($a_\ell\equiv1+\ell^{k-1}\pmod{128}$ for odd $\ell$, at
$p=2$). Statements (ii) and (iii) can also be read off from the literal classification
modulo $27$ of \cite[Thm~1.5]{R26}.
Independent check, by Sturm's theorem: the series
$\sum_{n\equiv1(3)}(a_n(f)-\sigma(n))q^n=\tfrac12\bigl((f-E')\otimes\mathbf 1_3+(f-E')\otimes\chi\bigr)$,
where $E'=\sum_{n\ge1}\sigma(n)q^n$, lies in $M_k(\Gamma_0(9))$ (the constant term of
$G_k$ disappears under twisting), and the index of $\Gamma_0(9)$ is $12$; it
therefore suffices to check (i) for $n\le k$. This check was made for every $k\le212$
(Newton polygons of $\mathrm{charpoly}(T_n-\sigma(n))$ on $S_k$, PARI/GP), and
(ii) for $k\le140$ on an eigenbasis embedded in $\overline\Q_3$; part~C
of \texttt{VERIF\_EISENSTEIN.py} redoes (i) for $k\le44$ on the lattice $L_k$.

\begin{remarque}
(1) Statement (i) is equivalent, by multiplicativity, by Chebotarev's theorem and by the
identity $\operatorname{tr}g\,\operatorname{tr}h=\operatorname{tr}gh+\det g\,\operatorname{tr}g^{-1}h$
(which reduces $a_{\ell}a_{\ell'}$, $\ell\equiv\ell'\equiv2\pmod3$, to traces on the subgroup of index~$2$),
to: $\operatorname{tr}\rho_f(g)\equiv1+\chi_{\mathrm{cyc}}^{\,k-1}(g)\pmod{81}$ for
every $g$ in the absolute Galois group of $\Q(\sqrt{-3})$, where $\rho_f$ is the
$3$-adic representation of $f$. For $\ell\equiv2\pmod3$, it gives
$a_\ell^2\equiv s_\ell^2\pmod{81}$ (via $a_{\ell^2}=a_\ell^2-\ell^{k-1}$), hence
$a_\ell\equiv\pm s_\ell\pmod{27}$ if $v_3(s_\ell)=1$ (that is,
$\ell\not\equiv-1\pmod9$ and $k\not\equiv4\pmod6$), and $a_\ell\equiv0\pmod9$ otherwise:
Theorem~\ref{thO} contains Conjecture~9.1 of \cite{R19} and, modulo~$9$, its
analogue for every prime $\ell\equiv2\pmod3$.

(2) For $\Delta$, (i) and (ii) are contained in Swinnerton-Dyer's theorem
$\tau(n)\equiv n^{-610}\sigma_{1231}(n)\pmod{3^6}$ for $3\nmid n$ \cite{SD73}: since
$(\Z/81\Z)^\times$ is cyclic of order $54$, one has $n^{27}\equiv\chi(n)\pmod{81}$;
for $de=n$, $n^{-610}d^{1231}\equiv d^{27}e^{38}\equiv\chi(n)\,e^{11}\pmod{81}$, whence
$\tau(n)\equiv\chi(n)\,\sigma_{11}(n)\pmod{81}$ for $3\nmid n$. Modulo~$3$,
(i) to (iii) give back Hatada's theorem $a_\ell\equiv1+\ell\pmod3$ \cite{H79}. Modulo~$9$, (ii),
(iii) and $a_3\equiv0\pmod9$ give exactly the five classes of \cite{CS04}: two
for $k\equiv0$ and two for $k\equiv2\pmod6$ (according to $\psi$), one for
$k\equiv4\pmod6$.

(3) The statements are optimal: (ii) is false modulo~$81$ (for $k=18$,
$a_2+s_2=-528+131073=27\cdot4835$, $3\nmid4835$), and for $k\equiv4\pmod6$ nothing
survives beyond $a_\ell\equiv0\pmod9$, $\ell\equiv2\pmod3$ (for $k=40$, the three
values of $a_2/9$ modulo~$3$ are $0$, $1$ and $2$).

(4) The depth of (i) increases by one when $\ell\equiv\pm1\pmod9$: this is
Corollary~\ref{coQ} of \S\ref{sec7}, which also describes the other classes modulo~$243$, and
Theorem~\ref{thR} gives depth~$6$ when $\ell$ splits completely in
$\Q(\zeta_9,\sqrt[3]3)$ or $\ell\equiv-1\pmod9$.
\end{remarque}

\section{A Galois law modulo \texorpdfstring{$243$}{243} and \texorpdfstring{$729$}{729}}\label{sec7}

Let $f$ be a normalised cuspidal eigenform of level~$1$ and weight $k$, $O$
the ring of integers of a finite extension of $\Q_3$ containing its coefficients, and
$\rho_f:G_\Q\to\mathrm{GL}_2(O)$ its $3$-adic representation \cite{De71}: it is
unramified outside $3$, $\operatorname{tr}\rho_f(\mathrm{Frob}_\ell)=a_\ell$ and
$\det\rho_f=\chi^{k-1}$, where $\chi$ is the $3$-adic cyclotomic character. We put
$T=\operatorname{tr}\rho_f$, $D=\chi^{k-1}$, $T_E=1+\chi^{k-1}$ (trace of
$\mathbf 1\oplus\chi^{k-1}$), $K=\Q(\sqrt{-3})=\Q(\zeta_3)$ and $G_K=\operatorname{Gal}(\overline\Q/K)$.
On $G_K$, $\chi\equiv1\pmod3$; we define two homomorphisms $u,v:G_K\to\Z/3\Z$ by
\[
\chi(g)\equiv1+3u(g)\pmod9,\qquad g(\sqrt[3]3)=\zeta_3^{\,v(g)}\sqrt[3]3 .
\]
Thus $u(\mathrm{Frob}_{\mathfrak l})=0$ if and only if $\ell\equiv1\pmod9$, and
$v(\mathrm{Frob}_{\mathfrak l})=0$ if and only if $3$ is a cube modulo $\ell$
($\mathfrak l\mid\ell$, $\ell\equiv1\pmod3$); $\ker u\cap\ker v=G_{\Q(\zeta_9,\sqrt[3]3)}$.

\begin{lemme}[Fricke reduction]\label{l9}
Let $\Gamma$ be a group, $A$ a commutative ring, $D:\Gamma\to A^\times$ a
homomorphism and $T:\Gamma\to A$ a function such that $T(1)=2$, $T(gh)=T(hg)$ and
\begin{equation}\label{eqps}
T(gh)+D(g)\,T(g^{-1}h)=T(g)\,T(h)\qquad(g,h\in\Gamma).
\end{equation}
For every word $W$ in the free group on two letters, there is a polynomial
$P_W\in\Z[t_1,t_2,t_3,d_1^{\pm1},d_2^{\pm1}]$, depending only on $W$, such that
$T(W(x,y))=P_W(T(x),T(y),T(xy),D(x),D(y))$ for all $\Gamma,A,D,T$ as above and
all $x,y\in\Gamma$.
\end{lemme}
\begin{proof}
For $a,b,z\in\Gamma$, \eqref{eqps} with $(g,h)=(z,zba)$ and invariance under conjugation
give
\[
\text{(R1)}\ \ T(az^2b)=T(z)T(azb)-D(z)T(ab),\qquad
\text{(R2)}\ \ T(az^{-1}b)=D(z)^{-1}\bigl(T(z)T(ab)-T(azb)\bigr),
\]
(R2) being obtained by applying (R1) to $(az^{-1},z,b)$. We argue by induction on
the pair (length of $W$, number of negative exponents), ordered
lexicographically, $W$ being taken cyclically reduced (which does not change $T$).
If a syllable of $W$ has an exponent $\ge2$, (R1) reduces to shorter words; if
a syllable has a negative exponent, (R2) reduces to a shorter word and to a word of the same
length with one negative exponent fewer; otherwise $W$ is $1$, $x$, $y$, or conjugate
to $(xy)^m$, $m\ge1$, and (R1) with $z=xy$ gives
$T((xy)^m)=T(xy)T((xy)^{m-1})-D(x)D(y)T((xy)^{m-2})$. The algorithm depends only on
$W$.
\end{proof}
\noindent For traces of $2\times2$ matrices, this is the classical theorem of Fricke
and Vogt \cite{Go09}; \eqref{eqps} holds for $T=\operatorname{tr}\rho$,
$D=\det\rho$, by Cayley--Hamilton ($g+\det(g)g^{-1}=\operatorname{tr}(g)$).

\begin{lemme}[the group]\label{l10}
Let $E$ be the maximal pro-$3$-extension of $K$ unramified outside $3$ and
$\widehat G=\operatorname{Gal}(E/K)$.
\begin{enumerate}
\item[(a)] The cyclic extensions of degree $3$ of $K$ unramified outside $3$ are
the subfields of degree $3$ of $K(\zeta_9,\sqrt[3]3)$; $u$ and $v$ factor through
$\widehat G$ and $(u,v):\widehat G\to(\Z/3\Z)^2$ is surjective.
\item[(b)] Two elements $x,y\in\widehat G$ such that $(u,v)(x)=(1,0)$ and
$(u,v)(y)=(0,1)$ topologically generate $\widehat G$.
\item[(c)] $T$, $D$ and $T_E$, restricted to $G_K$, factor through $\widehat G$.
\end{enumerate}
\end{lemme}
\begin{proof}
(a) $K$ contains $\mu_3$ and has class number one; by Kummer theory, such an
extension is $K(\sqrt[3]a)$ with $a\in O_K[1/3]^\times$, and
$O_K[1/3]^\times/(O_K[1/3]^\times)^3$ has dimension $2$, generated by $\zeta_3$ and
$1-\zeta_3$, or equally by $\zeta_3$ and $3=-\zeta_3^2(1-\zeta_3)^2$; and
$K(\sqrt[3]{\zeta_3})=K(\zeta_9)$. (b) $\widehat G$ is a pro-$3$-group and
$\operatorname{Hom}(\widehat G,\Z/3\Z)$ is generated by $u,v$; this is Burnside's basis
theorem \cite{Ko02}. (c) By Chebotarev and Hatada's congruence
$a_\ell\equiv1+\ell^{k-1}\pmod3$ \cite{H79}, $T\equiv2$ and $D\equiv1\pmod3$ on $G_K$:
every $\rho_f(g)$, $g\in G_K$, has characteristic polynomial $(X-1)^2$ modulo
the maximal ideal, so the image of $\rho_f(G_K)$ in $\mathrm{GL}_2$ of the residue field
consists of unipotents and is a $3$-group; the kernel of the reduction being a
pro-$3$-group, $\rho_f(G_K)$ is a pro-$3$-group, and $\rho_f$ being unramified outside
$3$, $\rho_f|_{G_K}$ factors through $\widehat G$. Likewise for $\chi|_{G_K}$, with
values in $1+3\Z_3$.
\end{proof}
\noindent By Shafarevich \cite{Sh63}, $\widehat G$ is even free on two
generators; this fact is used only in \S\ref{sec9}. This group, and the two conditions
``$\ell$ modulo $9$'' and ``$3$ a cube modulo $\ell$'', also govern the structure of
the Hecke algebra of level~$1$ \emph{modulo~$3$} \cite{BK15,Me16}; the
statements below concern the eigenvalues in characteristic~$0$.

\begin{lemme}[a quadratic form]\label{l11}
Let $O'$ be the ring of integers of a finite extension of $\Q_3$, and $T_1,T_2:\widehat
G\to O'$ two continuous functions satisfying the hypotheses of Lemma~\ref{l9} with the
same $D$, where $D\equiv1$ and $T_2\equiv2\pmod{3}$. If $T_1-T_2\in3^nO'$ with $n\ge1$,
there are $\alpha,\beta,\gamma\in O'/3O'$ such that
\[
T_1(g)-T_2(g)\equiv3^n\bigl(\alpha\,u(g)^2+\beta\,u(g)v(g)+\gamma\,v(g)^2\bigr)\pmod{3^{n+1}O'}
\qquad(g\in\widehat G).
\]
If moreover $T_1$ and $T_2$ are invariant under an automorphism $\sigma$ of
$\widehat G$ such that $u\circ\sigma=u$ and $v\circ\sigma=-v$, then $\beta=0$.
\end{lemme}
\begin{proof}
Let $x,y$ be as in Lemma~\ref{l10}(b), $x_1=x$, $x_2=y$, $x_3=xy$, and
$\varepsilon=(T_1-T_2)/3^n$ modulo~$3$. For a word $W$, Lemma~\ref{l9} and
Taylor's formula (the two points differ only in the first three
coordinates, by at most $3^n$, and $2n\ge n+1$) give
$\varepsilon(W)=\sum_jc_j(W)\varepsilon(x_j)$ with
$c_j(W)=\partial_jP_W(2,2,2,1,1)\bmod3$, since $T_2\equiv2$ and $D\equiv1\pmod3$. For a
quadratic form $Q=\alpha u^2+\beta uv+\gamma v^2$ with coefficients in $O'/3$, the
function $T_Q=2+\delta Q$, with values in $(O'/3)[\delta]/(\delta^2)$, with $D=1$,
satisfies the hypotheses of Lemma~\ref{l9}: this is the parallelogram identity
$Q(a+b)+Q(b-a)=2Q(a)+2Q(b)$. Since $P_W(2,2,2,1,1)=2$ (constant function $T=2$, $D=1$),
we also have $Q(W)=\sum_jc_j(W)Q(x_j)$. Choose $\alpha=\varepsilon(x)$,
$\gamma=\varepsilon(y)$, $\beta=\varepsilon(xy)-\alpha-\gamma$: then $\varepsilon=Q$
on the subgroup generated by $x,y$, which is dense by Lemma~\ref{l10}(b), hence everywhere by
continuity. Finally, if $\sigma$ exists, $\varepsilon\circ\sigma=\varepsilon$ gives
$Q(a,-b)=Q(a,b)$, whence $2\beta=0$ and $\beta=0$.
\end{proof}

Conjugation by a complex conjugation $c$ provides such a $\sigma$: $E$ is
Galois over $\Q$, $\chi(cgc^{-1})=\chi(g)$, and, $c$ fixing the real $\sqrt[3]3$ and
sending $\zeta_3$ to $\zeta_3^{-1}$, $v(cgc^{-1})=-v(g)$; $T$, $D$, $T_E$ are
class functions on $G_\Q$.

\begin{theoreme}[Galois law modulo $243$]\label{thP}
For every $g\in G_K$,
\[
\operatorname{tr}\rho_f(g)\equiv1+\chi(g)^{k-1}+\bigl(a_2^2-s_2^2\bigr)\,u(g)^2\pmod{243\,O}.
\]
\end{theoreme}
\begin{proof}
By Theorem~\ref{thO}(i) and Chebotarev (the $\mathrm{Frob}_{\mathfrak l}$, $\mathfrak
l$ of degree~$1$, are dense in $G_K$), $T-T_E\in81O$ on $G_K$. Lemma~\ref{l11}
($n=4$, $T_1=T$, $T_2=T_E$, $\sigma$ conjugation by $c$) gives
$T-T_E\equiv81(\alpha u^2+\gamma v^2)\pmod{243}$. For the Frobenius element $g_1$ of
the prime ideal $2O_K$ (inert, residue field $\mathbb F_4$), $g_1=\mathrm{Frob}_2^2$,
$\chi(g_1)=4$, hence $u(g_1)=1$, and $v(g_1)=0$ since $3=1$ is a cube in $\mathbb F_4$;
$T(g_1)=a_2^2-2\cdot2^{k-1}$, $T_E(g_1)=1+4^{k-1}$, whence $81\alpha\equiv a_2^2-s_2^2$.
For $g_2=\mathrm{Frob}_{\mathfrak l}$, $\mathfrak l\mid19$: $u(g_2)=0$ since
$19\equiv1\pmod9$, $v(g_2)\ne0$ since $3^6\equiv7\pmod{19}$, whence
$81\gamma\equiv a_{19}-s_{19}\equiv0\pmod{243}$ by Theorem~\ref{thN}: (a) if
$k\not\equiv4\pmod6$, (a$''$) if $k\equiv4\pmod6$.
\end{proof}
\statut{PROVED} From Theorem~\ref{thO}, hence from \cite{R19}, and from Theorem~\ref{thN}.

\begin{corollaire}[primes]\label{coQ}
Let $\ell\ne3$ be prime.
\begin{enumerate}
\item[(i)] If $\ell\equiv1\pmod3$: $a_\ell\equiv s_\ell\pmod{243}$ if $\ell\equiv1\pmod9$,
and $a_\ell\equiv s_\ell+a_2^2-s_2^2\pmod{243}$ if $\ell\equiv4,7\pmod9$.
\item[(ii)] If $\ell\equiv2\pmod3$: $a_\ell^2\equiv s_\ell^2\pmod{243}$ if
$\ell\equiv-1\pmod9$, and $a_\ell^2\equiv s_\ell^2+a_2^2-s_2^2\pmod{243}$ if
$\ell\equiv2,5\pmod9$.
\end{enumerate}
\end{corollaire}
\begin{proof}
(i) $g=\mathrm{Frob}_{\mathfrak l}$. (ii) $g=\mathrm{Frob}_\ell^2$, the Frobenius of
the inert ideal $\ell O_K$: $T(g)=a_\ell^2-2\ell^{k-1}$, $T_E(g)=1+\ell^{2k-2}$,
$\chi(g)=\ell^2$, and $u(g)=0$ if and only if $\ell\equiv\pm1\pmod9$.
\end{proof}
\noindent In particular, the depth of the Eisenstein law increases by one for
every $\ell\equiv\pm1\pmod9$, in every weight. The correction $a_2^2-s_2^2$ is necessary:
without it, the congruence fails modulo $243$ for $\ell=7$, $31$ and $\ell^2=25$ in all
the weights $12\le k\le70$ examined except $k=12$ and $k=22$, where $a_2^2\equiv s_2^2\pmod{243}$.

\begin{theoreme}[modulo $729$]\label{thR}
For every $g\in G_{\Q(\zeta_9,\sqrt[3]3)}$,
$\operatorname{tr}\rho_f(g)\equiv1+\chi(g)^{k-1}\pmod{729\,O}$. In particular
$a_\ell\equiv s_\ell\pmod{729}$ if $\ell\equiv1\pmod9$ and $3$ is a cube modulo $\ell$, and
$a_\ell^2\equiv s_\ell^2\pmod{729}$ if $\ell\equiv-1\pmod9$.
\end{theoreme}
\begin{proof}
Put $\kappa=k-1$ and, for $g\in G_K$, $\lambda(g)=\frac13\log_3\chi(g)\in\Z_3$;
$\lambda$ is a continuous homomorphism and $\lambda\equiv u\pmod3$. For $s$ in
the ring of integers $O'$ of a finite extension of $\Q_3$ containing $O$, let
$\psi_s=\exp(3s\lambda)$, a continuous character of $G_K$ with values in $O'^\times$, and
$M_s=\psi_s+\chi^\kappa\psi_s^{-1}$, the trace of $\psi_s\oplus\chi^\kappa\psi_s^{-1}$;
$M_s$ satisfies \eqref{eqps} with the same $D$, is invariant under $\sigma$, and
\[
M_s-T_E=-(e^{A}-1)(e^{B}-1),\qquad A=3s\lambda,\quad B=3(\kappa-s)\lambda .
\]
Let $\alpha\in O/3$ be defined by $81\alpha\equiv a_2^2-s_2^2\pmod{243}$. If $3\nmid\kappa$,
take $s=9t$ with $t\equiv-\alpha\kappa^{-1}\pmod3$: $v_3(A)\ge3$, $v_3(B)\ge1$, and
$M_s-T_E\equiv-AB\equiv-81t\kappa\lambda^2\equiv81\alpha u^2\pmod{243}$ (the other terms
of the expansion have valuation $\ge5$). If $3\mid\kappa$ ($k\equiv4\pmod6$), take
$s=3t$, where $t\in O'$ is a root of $X^2-\frac\kappa3X-\tilde\alpha$ ($\tilde\alpha\in O$
lifting $\alpha$): $v_3(A),v_3(B)\ge2$ and $M_s-T_E\equiv-81t(\tfrac\kappa3-t)\lambda^2
\equiv81\alpha u^2\pmod{243}$. In both cases, Theorem~\ref{thP} gives
$T\equiv M_s\pmod{243O'}$ on $G_K$, and Lemma~\ref{l11} ($n=5$, $T_1=T$, $T_2=M_s$)
gives $T-M_s\equiv243\,Q'(u,v)\pmod{729}$ for a quadratic form $Q'$, which vanishes if
$u=v=0$. Finally, if $u(g)=0$, then $3\mid\lambda(g)$, hence
$v_3(A)+v_3(B)\ge6$ in both cases, and $v_3(M_s(g)-T_E(g))=v_3(A)+v_3(B)\ge6$. For
the Frobenius of $\ell O_K$, $\ell\equiv-1\pmod9$, one has $u=0$, and $v=0$ since $3$ is a cube
in $\mathbb F_\ell^\times$, hence in $\mathbb F_{\ell^2}$.
\end{proof}
\statut{PROVED} From Theorem~\ref{thP}.

\begin{theoreme}[bilinear law]\label{thS}
For a prime $\ell\equiv2\pmod3$, let $\tilde u(\ell)=-1$, $1$, $0$ according as
$\ell\equiv2$, $5$, $8\pmod9$. For all primes $\ell,\ell'\equiv2\pmod3$,
\[
a_\ell\,a_{\ell'}\equiv s_\ell\,s_{\ell'}+\tilde u(\ell)\tilde u(\ell')\bigl(a_2^2-s_2^2\bigr)\pmod{243}.
\]
\end{theoreme}
\begin{proof}
Let $g,g'$ be Frobenius elements at $\ell,\ell'$; $gg'$ and $g^{-1}g'$ lie in $G_K$, and
$T(g)T(g')=T(gg')+D(g)T(g^{-1}g')$, likewise for $T_E$. One has $-\ell\equiv1+3\tilde
u(\ell)\pmod9$, hence $u(gg')=\tilde u(\ell)+\tilde u(\ell')$ and $u(g^{-1}g')=\tilde
u(\ell')-\tilde u(\ell)$. By Theorem~\ref{thP}, and $D(g)\equiv-1\pmod3$,
$a_\ell a_{\ell'}-s_\ell s_{\ell'}\equiv(a_2^2-s_2^2)\bigl((\tilde u+\tilde u')^2-(\tilde
u'-\tilde u)^2\bigr)=4\tilde u\tilde u'(a_2^2-s_2^2)\pmod{243}$, and $4\equiv1\pmod3$.
\end{proof}
\statut{PROVED}

\begin{corollaire}[law modulo $81$]\label{coT}
If $k\not\equiv4\pmod6$, for every prime $\ell\equiv2\pmod3$,
\[
a_\ell\equiv\frac{s_2\,s_\ell-\tilde u(\ell)\,(a_2^2-s_2^2)}{a_2}\pmod{81\overline\Z_3},
\]
and, for every $n$ prime to $3$, $a_n(f)$ modulo $81$ depends only on $k$ and on
$a_2(f)$ modulo $81$.
\end{corollaire}
\begin{proof}
Take $\ell'=2$ in Theorem~\ref{thS} ($\tilde u(2)=-1$) and divide by
$a_2$, which has valuation $1$ (Corollary~\ref{coL}). The numerator modulo $243$ and the quotient
modulo $81$ do not change if $a_2$ is replaced by $a_2+81e$. For
$\ell\equiv1\pmod3$, use Corollary~\ref{coQ}(i), then multiplicativity and
$a_{\ell^{r+1}}=a_\ell a_{\ell^r}-\ell^{k-1}a_{\ell^{r-1}}$.
\end{proof}
\noindent Modulo~$27$, this formula gives back Theorem~\ref{thO}(ii) for $\ell\equiv2\pmod3$; the corollary is
the analogue modulo~$81$ of the classification of \cite{CS04}. If $k\equiv4\pmod6$ and
$9\parallel a_2$, the same division determines $a_\ell$ modulo $27$ for $\ell\equiv2\pmod3$:
$a_2$ obeys no uniform law modulo $27$ (Remark~(3) of \S\ref{sec6}), but it
governs the others.

\begin{remarque}
\textbf{VERIFIED} (PARI/GP, exact Newton polygons on $S_k$). Theorem~\ref{thP}
for all primes $\ell<200$ and $12\le k\le70$; Theorem~\ref{thS} for
$1836$ pairs ($\ell,\ell'<110$, $12\le k\le80$); $a_{19}\equiv s_{19}\pmod{243}$ for
all $k\equiv4\pmod6$, $k\le298$. Minimal depth $\min v_3$ observed for $12\le k\le100$
(for $\ell\equiv-1\pmod9$, on $a_\ell^2-s_\ell^2$): $5$ for $\ell=19,37,109,127,163,181,199$
($3$ not a cube); $6$ for $\ell=73$ and for $\ell=17,71,89,179,197,233,251$; $8$ for
$\ell=53,107,269$ ($\ell\equiv-1\pmod{27}$) and for $\ell=271$ ($\equiv1\pmod{27}$, $3$ a
cube) if $k\not\equiv4\pmod6$, $7$ for $\ell=271$ if $k\equiv4\pmod6$. Theorems~\ref{thP}
and~\ref{thR} explain exactly the depths $5$ and $6$; the
depths $7$ and $8$ are treated in \S\ref{sec9} (Theorems~R$'$ and~R$''$).
\end{remarque}


\section{A \texorpdfstring{$2$}{2}-adic Galois law and the answer to Hatada's question}\label{sec8}

\subsection{A \texorpdfstring{$2$}{2}-adic certificate}

Lemma~\ref{l7} holds, with the same proof, with $3$ replaced by $2$:

\begin{lemme}[vanishing criterion, $p=2$]\label{l12}
Let $n\ge1$, $\mathcal H_2$ a system of integral representatives of
$\mathrm{GL}_2(\mathbb F_2)$, and $X=\sum_gc_g\,g$ a finite combination, with integer
coefficients, of integer matrices with odd determinant. Write $g\equiv h_g\kappa_g\pmod{2^n}$
with $h_g\in\mathcal H_2$ and $\kappa_g\equiv I\pmod2$, define $c_{\delta,A,B}$ as
in Lemma~\ref{l7}, and then
$f_{h,\delta}(i,t)=\sum_{h_g=h}c_g\sum_{A+B<n}c_{\delta,A,B}(\kappa_g)\binom iA\binom{2t-i}B$.
If all the finite differences $\Delta_i^A\Delta_t^Bf_{h,\delta}(0,0)$, $A+B<n$, are
divisible by $2^n$, then $\sigma_X(\Z[x,y]_w)\subset2^n\Z[x,y]_w$ for every even $w$.
\end{lemme}
\begin{proof}
That of Lemma~\ref{l7}: $u$, $q$, $p-1$, $v-1$ are divisible by $2$, so the terms
with $a+b+m+m'\ge n$ are divisible by $2^n$, and $w=2t$.
\end{proof}

\begin{lemme}[$a_5$ modulo $512$]\label{l13}
There is an explicit element $\zeta\in\mathcal I$, a sum of $190$ terms
$c\,\gamma\,h\,(\kappa'_1-1)^{\alpha_1}\cdots(\kappa'_4-1)^{\alpha_4}$, with
$\gamma\in\{1+S,1+U+U^2,1+\varepsilon\}$, $h\in\mathcal H_2$, $\kappa'_1=\mat3001$,
$\kappa'_2=\mat1201$, $\kappa'_3=\mat1021$, $\kappa'_4=\mat1003$ and $\sum\alpha_i\le8$,
such that $\widetilde T_5-(1+5\Lambda_5)-\zeta$ satisfies the hypothesis of Lemma~\ref{l12} with
$n=9$. Consequently $P|\widetilde T_5\equiv s_5P\pmod{512L_k}$ for every $k$ and every
$P\in L_k$, and $a_5(f)\equiv1+5^{k-1}\pmod{512\overline\Z_2}$ for every eigenform $f$
of level~$1$.
\end{lemme}
\begin{proof}
Finite exact computation (\texttt{VERIF\_HATADA2.py}, parts A and B); the conclusion is drawn
as in the proof of Theorem~\ref{thK}.
\end{proof}
\statut{PROVED}

\subsection{The law}

Let $f$ be an eigenform of level~$1$ and weight $k$, $O$ the ring of integers of a finite
extension of $\Q_2$ containing its coefficients, $\rho_f:G_\Q\to\mathrm{GL}_2(O)$ its
$2$-adic representation, $T=\operatorname{tr}\rho_f$, $D=\chi^{k-1}$ ($\chi$ the $2$-adic
cyclotomic character), $T_E=1+\chi^{k-1}$, $\Delta=T-T_E$, and $s_p=1+p^{k-1}$, $d_3=a_3-s_3$. Let $L$
be the maximal pro-$2$-extension of $\Q$ unramified outside $\{2,\infty\}$ and
$G=\operatorname{Gal}(L/\Q)$. We call the \emph{class} of $g\in G$ the class of $\chi(g)$
modulo~$8$.
\begin{enumerate}
\item[(G1)] Since $a_\ell\equiv1+\ell\pmod8$ \cite{H79}, $T\equiv0$ and $D\equiv1\pmod2$;
the residual image is unipotent, $\rho_f(G_\Q)$ is a pro-$2$-group, and $T$, $D$, $T_E$
factor through $G$. By \cite[Thm~8.1]{R19} and Chebotarev, $\Delta(G)\subset2^7O$.
\item[(G2)] The quadratic extensions of $\Q$ unramified outside $2$ are $\Q(i)$,
$\Q(\sqrt2)$, $\Q(\sqrt{-2})$: $\Phi(G)$ is the set of elements of class~$1$, and
$G/\Phi(G)\simeq(\Z/8\Z)^\times$ via the class.
\item[(G3)] For a complex conjugation $c$, $T(c)=T_E(c)=0$ ($\rho_f$ is odd).
\item[(G4)] If $\chi(g)\equiv1\pmod4$, $v_2(T_E(g))=1$; if $\chi(g)\equiv3\pmod4$,
$v_2(T_E(g))=v_2(1+\chi(g))$, since $k-1$ is odd. As $T\equiv T_E\pmod{2^7}$, $T$ has the
same valuation as long as it is $<7$.
\item[(G5)] $H=\{g:\chi(g)\equiv\pm1\pmod8\}=\operatorname{Gal}(L/\Q(\sqrt2))$ and
$N=\Phi(G)=\operatorname{Gal}(L/\Q(\zeta_8))$; $L$ is also the maximal pro-$2$-extension
of $\Q(\sqrt2)$ and of $\Q(\zeta_8)$ unramified outside $2$ and $\infty$ (that extension is
Galois over $\Q$, with group an extension of a finite $2$-group by a pro-$2$-group, hence a
pro-$2$-extension of $\Q$ unramified outside $\{2,\infty\}$, contained in $L$). These fields have
class number one, and Kummer theory gives isomorphisms
$H/\Phi(H)\simeq(\Z/2)^3$ via $(\theta_{-1},\theta_{1+\sqrt2},\theta_{\sqrt2})$ and
$N/\Phi(N)\simeq(\Z/2)^3$ via $(\theta_{\zeta_8},\theta_{1+\sqrt2},\theta_{1-\zeta_8})$,
where $\theta_x(g)\in\Z/2$ is defined by $g(\sqrt x)=(-1)^{\theta_x(g)}\sqrt x$; and
$\theta_x(s^{-1}as)=\theta_{s(x)}(a)$.
\end{enumerate}

\begin{lemme}[parity]\label{l14}
Let $n\ge1$ and $g,h\in G$ be such that $\Delta(gh),\Delta(g^{-1}h)\in2^nO$ and
$\Delta(g)T(h)+T_E(g)\Delta(h)\in2^{n+1}O$. Then
$\Delta(gh)\equiv\Delta(g^{-1}h)\pmod{2^{n+1}}$. In particular, if $\Delta(K)\subset2^nO$
for a closed subgroup $K$ of $G$, the function $\Delta/2^n\bmod2$ is constant on the
classes of $K$ modulo $\Phi(K)$, and vanishes on $\Phi(K)$.
\end{lemme}
\begin{proof}
Subtracting \eqref{eqps} for $T_E$ from \eqref{eqps} for $T$:
$\Delta(gh)+D(g)\Delta(g^{-1}h)=\Delta(g)T(h)+T_E(g)\Delta(h)$, and $D(g)$ is odd. If
$g,h\in K$, the hypothesis is satisfied since $T$ and $T_E$ are even; with $(g,gh)$ instead
of $(g,h)$, we get invariance under left multiplication by squares, which
topologically generate $\Phi(K)$.
\end{proof}

\begin{theoreme}[$2$-adic law]\label{thU}
For every odd prime $p$:
\[
a_p\equiv s_p\ (2^{10})\ \text{if } p\equiv1\ (8);\quad
a_p\equiv s_p\ (2^8)\ \text{if } p\equiv5\ (8);\quad
a_p-s_p\equiv d_3\ (2^8)\ \text{if } p\equiv3\ (8);
\]
\[
a_p-s_p\equiv2d_3\ (2^9)\ \text{if } p\equiv7\ (16);\qquad
a_p\equiv s_p\ (2^9)\ \text{if } p\equiv15\ (16).
\]
\end{theoreme}
\begin{proof}
\emph{Step 1 (modulo $2^8$).} By (G1) and Lemma~\ref{l14} ($K=G$, $n=7$),
$\varepsilon=\Delta/2^7\bmod2$ depends only on the class, and $\varepsilon=0$ on
class~$1$. On class~$7$, $\varepsilon(c)=0$ by (G3). On class~$5$,
$\varepsilon(\mathrm{Frob}_5)=0$ by Lemma~\ref{l13}. On class~$3$,
$\Delta\equiv\Delta(\mathrm{Frob}_3)=d_3\pmod{2^8}$.

\emph{Step 2 (modulo $2^9$ on $H$).} By step~1, $\Delta(H)\subset2^8O$; by
Lemma~\ref{l14}, $\varepsilon_2=\Delta/2^8\bmod2$ is a function on
$V=H/\Phi(H)\simeq(\Z/2)^3$, vanishing at $0$ and at $\bar c$. Let $s$ be of class $3$ or $5$ and
$a,b\in N$; for $g=sa$, $h=sb$, of the same class, the hypothesis of Lemma~\ref{l14} ($n=8$)
is satisfied by (G4) and step~1, and $gh=s^2(s^{-1}as)b$, $g^{-1}h=a^{-1}b$. Since
$s(1+\sqrt2)=-(1+\sqrt2)^{-1}$, $s(\sqrt2)=-\sqrt2$ and $\theta_{-1}(a)=0$, one has
$\overline{s^{-1}as}=\bar a$, whence $\varepsilon_2(\overline{s^2}+w)=\varepsilon_2(w)$ for
every $w\in V_N=\{\theta_{-1}=0\}$, the image of $N$. By Euler's criterion in $\mathbb F_9$ and
$\mathbb F_{25}$ (\texttt{VERIF\_HATADA2.py}, part C),
$\overline{\mathrm{Frob}_3^{\,2}}=(0,1,0)$ and $\overline{\mathrm{Frob}_5^{\,2}}=(0,0,1)$, which
generate $V_N$: hence $\varepsilon_2=0$ on $V_N$, that is, $a_p\equiv s_p\pmod{2^9}$
if $p\equiv1\pmod8$. For $g$ of class $3$ and $h$ of class $5$ (or the reverse), $gh$ and
$g^{-1}h$ are of class $7$, and (G4) and step~1 give
$\Delta(g)T(h)+T_E(g)\Delta(h)\equiv2^8\varepsilon(3)\pmod{2^9}$, whence
$\varepsilon_2(gh)+\varepsilon_2(g^{-1}h)=\varepsilon(3)$; with $g=\mathrm{Frob}_3a$,
$h=\mathrm{Frob}_5b$, we get $\varepsilon_2(z+\overline{\mathrm{Frob}_3^{\,2}})=
\varepsilon_2(z)+\varepsilon(3)$ for every $z$ in the image $V_7$ of class $7$, and
likewise with $\mathrm{Frob}_5^2$. Since $2+\sqrt2=\sqrt2\,(1+\sqrt2)$, one has $\theta_{2+\sqrt2}=\theta_{1+\sqrt2}+\theta_{\sqrt2}$,
which takes the value $1$ at $\mathrm{Frob}_3^{\,2}$ and at $\mathrm{Frob}_5^{\,2}$ (part C).
Since $\varepsilon_2(\bar c)=0$,
$\varepsilon_2(g)=\varepsilon(3)\,\theta_{2+\sqrt2}(g)$ on class $7$ ($2+\sqrt2$ is
totally positive, so $\theta_{2+\sqrt2}(c)=0$). Now $\theta_{2+\sqrt2}(g)=0$ if and
only if $g$ fixes $\Q(\zeta_{16})^+=\Q(\sqrt{2+\sqrt2})$, that is,
$\chi(g)\equiv15\pmod{16}$ for $g$ of class $7$; and
$2^8\varepsilon(3)\equiv2d_3\pmod{2^9}$.

\emph{Step 3 (modulo $2^{10}$ on $N$).} By step~2, $\Delta(N)\subset2^9O$, and
$\varepsilon_3=\Delta/2^9\bmod2$ is a function on $V'=N/\Phi(N)\simeq(\Z/2)^3$, vanishing
at~$0$. (i) For $g=ca$, $h=cb$ ($a,b\in N$), of class~$7$, $\Delta(g),\Delta(h)\in2^8O$
and $v_2(T(h)),v_2(T_E(g))\ge3$: Lemma~\ref{l14} ($n=9$) gives
$\varepsilon_3(\overline{cac^{-1}}+\bar b)=\varepsilon_3(\bar a+\bar b)$. Since
$\zeta_8\,c(\zeta_8)=1$, $(1+\sqrt2)\,c(1+\sqrt2)=(1+\sqrt2)^2$ and
$(1-\zeta_8)\,c(1-\zeta_8)=-\zeta_8^{-1}(1-\zeta_8)^2$, the element
$\overline{cac^{-1}}+\bar a$ equals $(0,0,\theta_{\zeta_8}(a))$: $\varepsilon_3$ is invariant
under $e_3=(0,0,1)$. (ii) For $g=\mathrm{Frob}_3a$, $h=\mathrm{Frob}_3b$:
$\Delta(g)\equiv\Delta(h)\equiv d_3\pmod{2^8}$, $T(h)=4u$, $T_E(g)=4u'$; here $u'=T_E(g)/4$
lies in $\Z_2^\times$, and $u\equiv T_E(h)/4\pmod{2^5}$ with $T_E(h)/4\in\Z_2^\times$, so that
$u,u'\in\Z_2^\times+2^5O$ and $u+u'\in2O$; hence $\Delta(g)T(h)+T_E(g)\Delta(h)\equiv4(u+u')d_3\equiv0\pmod{2^{10}}$. Here
$\zeta_8\,\zeta_8^3$ and $(1+\sqrt2)(1-\sqrt2)$ are squares, so
$\overline{s^{-1}as}+\bar a\in\langle e_3\rangle$, and $\varepsilon_3$ is invariant under
$\overline{\mathrm{Frob}_3^{\,2}}$, whose first two coordinates are $(1,1)$
(part C). Thus $\varepsilon_3=\beta\,(\theta_{\zeta_8}+\theta_{1+\sqrt2})$ for some
$\beta\in O/2$. (iii) For $g$ of class $5$, $g^2\in N$ has first two coordinates
$(1,0)$ (part C, and the same computation as in (ii)), and
$\Delta(g^2)=\Delta(g)\bigl(T(g)+T_E(g)\bigr)$, since $T(g^2)=T(g)^2-2D(g)$; now
$\Delta(g)\in2^8O$ (step 1) and $v_2(T(g)+T_E(g))=v_2(2T_E(g)+\Delta(g))=2$, hence
$\Delta(g^2)\in2^{10}O$ and $\beta=\varepsilon_3(g^2)=0$.
\end{proof}
\statut{PROVED} From \cite[Thm~8.1]{R19} and Lemma~\ref{l13}.

\begin{corollaire}[Hatada's question, $a=8$ and $9$]\label{coV}
For every eigenform of level~$1$ and every weight, Hatada's question has an
affirmative answer for $a=8$ and for $a=9$; for $a=10$, the requested congruence holds for
every $p\equiv1\pmod{2^9}$.
\end{corollaire}
\begin{proof}
If $p\equiv\pm1\pmod{2^{a-1}}$ and $a\ge4$, then $p^2\equiv1\pmod{2^a}$ and
$p^{k-1}\equiv p\pmod{2^a}$. If $p\equiv1\pmod{2^{a-1}}$, then $p\equiv1\pmod8$ and
Theorem~\ref{thU} gives the congruence modulo $2^{10}$. If $p\equiv-1\pmod{2^{a-1}}$ and
$a\in\{8,9\}$, then $p\equiv15\pmod{16}$, and Theorem~\ref{thU} gives the congruence
modulo $2^9$.
\end{proof}
\statut{PROVED}

\subsection{The case \texorpdfstring{$p\equiv-1$}{p = -1} and the complete answer}

Let $t\in G$ be a Frobenius element at $5$: $\chi(t)=5$, and the images of $t$ and of
$c$ generate $G/\Phi(G)\simeq(\Z/8\Z)^\times$, so $c$ and $t$ topologically
generate $G$. Put $x=\Delta(t)=a_5-s_5$ and $y=\Delta(ct)$: by
Lemma~\ref{l13}, $x\in2^9O$; $ct$ is of class $3$, so $y\in2^7O$. Let
$D=5^{k-1}$ and, in $\mathrm{GL}_2(\Z_2[\eta]/(\eta^2))$,
\[
\mathcal C=\mat100{-1},\qquad
\mathcal T_X=\mat110D+\eta\mat10{D+1}1,\qquad
\mathcal T_Y=\mat110D+\eta\mat10{D-1}{-1}.
\]
For a word $W$ in two letters, let $\tau_X(W)$ (resp. $\tau_Y(W)$) be the coefficient of
$\eta$ in the trace of $W(\mathcal C,\mathcal T_X)$ (resp. $W(\mathcal C,\mathcal T_Y)$), and
$\chi(W)=(-1)^{a}5^{b}$ if $W$ contains $a$ letters $c$ and $b$ letters $t$ (with sign).

\begin{lemme}[universal family]\label{l15}
For every word $W$,
\[
\Delta(W(c,t))\equiv2^7\frac x{2^8}\,\tau_X(W)+2^6\frac y{2^7}\,\tau_Y(W)\pmod{2^{14}O}.
\]
\end{lemme}
\begin{proof}
For indeterminates $u,v$, let $\rho_{u,v}$ be the representation of the free group on $c,t$
given by $c\mapsto\mat100{-1}$ and $t\mapsto\mat\alpha1\gamma\delta$ with
$\alpha=1+\frac{u+v}2$, $\delta=D+\frac{u-v}2$, $\gamma=\alpha\delta-D$: its trace at $t$
is $1+D+u=T_E(t)+u$, at $ct$ it is $1-D+v=T_E(ct)+v$, its determinant at $c$ is
$-1$ and at $t$ is $D$. By Lemma~\ref{l9}, applied on the one hand to $T$ on $G$ (with
$T(c)=0$, $D(c)=-1$) and on the other hand to $\rho_{u,v}$,
$T(W(c,t))=P_W(0,T_E(t)+x,T_E(ct)+y,-1,D)=\operatorname{tr}\rho_{x,y}(W)$. The polynomial
$P_W$ has integer coefficients; the Taylor expansion at $(u,v)=(0,0)$ therefore has
terms of order $\ge2$ in $(x,y)^2O\subset2^{14}O$, the constant term $T_E(W)$, and
first-order terms equal to the coefficients of $\eta$ computed with the derivatives
$\partial_u\rho(t)=\frac12\mat10{D+1}1$ and $\partial_v\rho(t)=\frac12\mat10{D-1}{-1}$, whence
the formula with $x=2^8\cdot\frac x{2^8}$ and $y=2^7\cdot\frac y{2^7}$.
\end{proof}

\begin{theoreme}[depth on the class of $-1$]\label{thX}
Let $g\in G$. If $\chi(g)\equiv1\pmod8$, then $\Delta(g)\in2^{10}O$; if
$\chi(g)\equiv5\pmod8$, then $\Delta(g)\in2^9O$. If
$v_2(1+\chi(g))=m\ge2$, then $\Delta(g)\in2^{\min(m+5,\,11)}O$. In particular, for every
prime $p\equiv-1\pmod{32}$, $a_p\equiv1+p^{k-1}\pmod{2^{10}}$.
\end{theoreme}
\begin{proof}
For powers of two $N$ and $N'$, the map $W\mapsto\bigl(W(\mathcal C,\mathcal T)\bmod N,\ \chi(W)\bmod N'\bigr)$
is a morphism to a finite group, which depends on $k$ only through $D\bmod N$, hence through
$k-1$ modulo $N/4$ (the order of $5$ modulo $N$). For $(N,N')=(16,64)$ and then $(32,512)$, and
for all these classes of $k-1$, \texttt{VERIF\_HATADA2.py} (part E) enumerates these images
($512$ to $32\,768$ elements) and checks that $\tau_X\equiv0\pmod8$ and
$\tau_Y\equiv0\pmod{16}$ when $\chi\equiv1\pmod8$ or $\chi\equiv-1\pmod{32}$, and that
$2^7\tau_X$ and $2^6\tau_Y$ are divisible by $2^{\min(m+5,11)}$ when
$v_2(1+\chi)=m\ge2$ (each of these assertions is tested by the script); finally that $\tau_X\equiv0\pmod2$ and $\tau_Y\equiv0\pmod8$ when
$\chi\equiv5\pmod8$, which, with $x\in2^9O$, gives $2^9\mid\Delta$ on class~$5$.
By Lemma~\ref{l15}, the conclusion holds for every word $W(c,t)$. For
arbitrary $g\in G$, $T$ modulo $2^{11}$ and $\chi$ modulo $2^{12}$ factor through a
finite quotient of $G$, in which the image of the dense subgroup generated by $c$ and $t$ is
everything: there is a word $W$ such that $T(W)\equiv T(g)\pmod{2^{11}}$ and
$\chi(W)\equiv\chi(g)\pmod{2^{12}}$, hence $T_E(W)\equiv T_E(g)\pmod{2^{11}}$.
\end{proof}
\statut{PROVED} Finite exact computation, from \cite[Thm~8.1]{R19} and
Lemma~\ref{l13}. This computation also gives back, by a route independent of
Kummer theory, the congruence modulo $2^{10}$ of Theorem~\ref{thU} for $p\equiv1\pmod8$.

\begin{corollaire}[complete answer to Hatada's question]\label{coY}
Hatada's question has an affirmative answer for $6\le a\le10$, for every eigenform
of level~$1$ and every weight, and a negative answer for $a=11$, $12$ and $13$.
\end{corollaire}
\begin{proof}
For $a=6,7$, this is \cite[Thm~8.1]{R19} (Section~1 and its footnote); for $a=8,9$, it is
Corollary~\ref{coV}; for $a=10$, $p\equiv1\pmod{2^9}$ falls under
Corollary~\ref{coV} and $p\equiv-1\pmod{2^9}$ under Theorem~\ref{thX}. For
$a\in\{11,12,13\}$, Theorem~\ref{thA} gives counterexamples ($\Delta_{16}$ and
$p=12289$).
\end{proof}
\statut{PROVED}

\begin{corollaire}[non-vanishing]\label{coZ}
For every eigenform $f$ of level~$1$ and every odd prime $p$ such that
$v_2(p+1)\le10$, one has $v_2(a_p(f))=v_2(p+1)$. In particular, $a_p(f)=0$ implies
$p\equiv-1\pmod{2^{11}}$, and even $p\equiv-1\pmod{2^{11}\cdot27}$ if
$k\not\equiv4\pmod6$.
\end{corollaire}
\begin{proof}
One has $v_2(s_p)=v_2(1+p)$: if $p\equiv1\pmod4$, both equal $1$; otherwise, $k-1$ being
odd, $1+p^{k-1}=(1+p)(1-p+\dots+p^{k-2})$ with an odd second factor. If
$p\equiv1\pmod4$, $a_p\equiv s_p\pmod{2^8}$ (Theorem~\ref{thU}), hence $v_2(a_p)=1$. If
$v_2(p+1)=m\in[2,10]$, Theorem~\ref{thX} gives $a_p-s_p\in2^{\min(m+5,11)}O$ and
$\min(m+5,11)>m$, hence $v_2(a_p)=m$. If $k\not\equiv4\pmod6$, $v_3(k-1)=0$, and
Theorem~\ref{thO} gives $a_p\equiv s_p\not\equiv0\pmod3$ if $p\equiv1\pmod3$, and
$a_p\equiv\pm s_p\pmod{27}$ with $v_3(s_p)=v_3(p+1)$ if $p\equiv2\pmod3$; hence $a_p=0$
implies $p\equiv-1\pmod{27}$.
\end{proof}
\statut{PROVED}
\statut{PROVED by the optional part F of the script, disabled by default} With
$(N,N')=(64,2048)$ (about $500\,000$ elements), the same computation gives $\Delta(g)\in2^{12}O$ as soon as $v_2(1+\chi(g))\ge7$, whence
$p\equiv-1\pmod{2^{12}}$. This sieve is effective but concerns only $p$; for fixed $p$,
the (non-effective) finiteness of the set of forms of level~$1$ such that $a_p(f)=0$
is due to Calegari and Sardari \cite{CSa21}.

\begin{remarque}
\textbf{VERIFIED} (PARI/GP, $12\le k\le80$, $p<300$): the minimal depths observed
are $10$ for $p\equiv1\pmod8$ (optimal, Theorem~\ref{thD}), $7$ for $p\equiv3\pmod8$,
$9$ for $p\equiv5\pmod8$, and $8$, $9$, $10$ for $p\equiv7\pmod{16}$, $15\pmod{32}$,
$31\pmod{32}$, in accordance with Theorems~\ref{thU} and~\ref{thX}. The laws of
Theorem~\ref{thU} were checked in $1247$ cases ($k\le70$, $p<200$) without failure; the
correction $2d_3$ is necessary in $84$ of them. The depth $9$ on class $5$ (Theorem~\ref{thX}) is therefore optimal.
\end{remarque}


\section{Beyond \texorpdfstring{$729$}{729}: the law of inert primes}\label{sec9}

Notation of \S\ref{sec7}; $\kappa=k-1$ and $s_\ell=1+\ell^{\kappa}$. We still write $\sigma$
for the automorphism of $\widehat G$ induced by conjugation by a complex conjugation $c$.
Let $x\in\widehat G$ be the Frobenius element of a prime ideal of $O_K$ above $7$, and
$y=\sigma(x)$. Since $7\equiv7\pmod9$ and $3$ is not a cube modulo $7$, one has
$u(y)=u(x)\ne0$ and $v(y)=-v(x)\ne0$; the images of $x$ and $y$ in $(\Z/3\Z)^2$ form a
basis, so $x$ and $y$ topologically generate $\widehat G$ (Burnside's basis
theorem).

\smallskip
\noindent\textbf{The anticyclotomic character.} The group $\widehat G$ is free on two
generators \cite{Sh63}, so $\operatorname{Hom}(\widehat G,\Z_3)$ is free of rank $2$;
$\sigma$ acts on it as an involution, and its $+1$ and $-1$ eigenparts, of rank $1$,
reduce modulo $3$ onto $\langle u\rangle$ and $\langle v\rangle$. The $+1$ part is
generated by $\lambda=\frac13\log_3\chi$ (proof of Theorem~\ref{thR}); we fix a
generator $\mu$ of the $-1$ part. Then $(\lambda,\mu)$ is a basis of
$\operatorname{Hom}(\widehat G,\Z_3)$, and $\operatorname{Hom}(\widehat G,A)=A\lambda\oplus
A\mu$ for every finite $\Z_3$-module $A$. The fixed field of $\ker(\mu\bmod9)$ is therefore the unique
cyclic extension of degree $9$ of $K$, unramified outside $3$, on whose Galois group
$c$ acts by $-1$. The ring class field $H_{27}$ of the order $\Z+27O_K$, of
discriminant $-2187$, has these properties: it is unramified outside $3$, $c$ acts by $-1$
on $\operatorname{Gal}(H_{27}/K)$ \cite[\S9]{Co13}, and this group, isomorphic to the
Picard group of the order, is cyclic of order $9$. For a prime $\ell\equiv1\pmod3$ and
$\mathfrak l\mid\ell$, one therefore has $\mu(\mathrm{Frob}_{\mathfrak l})\in9\Z_3$ if and only if
$\ell$ splits completely in $H_{27}$, that is \cite[\S9]{Co13}, if and only
if $\ell=X^2+XY+547Y^2$ with $X,Y\in\Z$. Likewise, $v(\mathrm{Frob}_{\mathfrak l})=0$ if and
only if $\ell=X^2+XY+61Y^2$.

\smallskip
\noindent\textbf{The base point.} Let $s\in O'$ be as in the proof of
Theorem~\ref{thR} ($s=9t$ if $3\nmid\kappa$, $s=3t$ if $3\mid\kappa$; in the latter case,
changing the lift $\tilde\alpha$ if necessary, we assume $2s\ne\kappa$, which ensures $w<\infty$ and $\Delta\ne0$ in Lemma~\ref{l17}). Put
$\psi_1=\psi_s$, $\psi_2=\chi^\kappa\psi_s^{-1}$,
$\theta=\psi_1/\psi_2=\exp\bigl(3(2s-\kappa)\lambda\bigr)$ and $w=v_3(2s-\kappa)$: $w=0$ if
$3\nmid\kappa$, $w\ge1$ if $3\mid\kappa$. Thus $M_s=\psi_1+\psi_2$, and, by the
proof of Theorem~\ref{thR}, $T\equiv M_s\pmod{243\,O'}$ on $G_K$. The functions
$\psi_1,\psi_2,\theta,\lambda$ are invariant under $\sigma$, and $\theta\equiv1\pmod{3}$ on
$\widehat G$.

\begin{lemme}[an irreducible tangent]\label{l16}
There are continuous functions $\beta,\gamma,z:\widehat G\to O'$ such that, for
$h,h'\in\widehat G$,
\[
\beta(hh')=\beta(h)+\theta(h)\beta(h'),\quad
\gamma(hh')=\gamma(h)+\theta(h)^{-1}\gamma(h'),\quad
z(hh')=z(h)+z(h')+\theta(h)\gamma(h)\beta(h'),
\]
$\beta(x)=\gamma(x)=1$, $z(x)=0$, $\beta\circ\sigma=-\beta$, $\gamma\circ\sigma=-\gamma$,
$z\circ\sigma=z$, and such that the function $e=\psi_2\bigl((1-\theta)z+\theta\beta\gamma\bigr)$
satisfies $e(1)=0$, $e(hh')=e(h'h)$, $e\circ\sigma=e$ and, with $D=\chi^\kappa$,
\begin{equation}\label{eqlin}
e(hh')+D(h)\,e(h^{-1}h')=e(h)M_s(h')+M_s(h)e(h')\qquad(h,h'\in\widehat G).
\end{equation}
Moreover:
\begin{enumerate}
\item[(a)] for every $h\in\widehat G$,
$e(\sigma(h)h)=2\psi_2(h)^2\bigl(1-\theta(h)\bigr)F(h)$, where $F=(1+\theta)z-\theta\beta\gamma$;
\item[(b)] $\beta\equiv\gamma\equiv v/v(x)\pmod{3O'}$, and there is $a\in O'$ such that
$F\equiv a\,u\pmod{3O'}$;
\item[(c)] if $w\ge1$, there are $b,b'\in O'$ such that $\beta\equiv b\mu$ and
$\gamma\equiv b'\mu\pmod{9O'}$.
\end{enumerate}
\end{lemme}
\begin{proof}
Let $\varepsilon$ be such that $\varepsilon^2=0$, $C=\mathrm{diag}(1,-1)$, and $\Gamma_1$ the
subgroup of $\mathrm{GL}_2(O'[\varepsilon])$ formed by the matrices congruent, modulo the ideal
generated by the maximal ideal of $O'$ and $\varepsilon$, to an upper unipotent
matrix; it is a pro-$3$-group. The matrices
\[
R=\mat{\psi_1(x)}{\psi_2(x)}{0}{\psi_2(x)}\Bigl(1+\varepsilon\mat{0}{0}{\theta(x)}{0}\Bigr)
\quad\text{and}\quad CRC
\]
lie in $\Gamma_1$. Since $\widehat G$ is a free pro-$3$-group of rank $2$ \cite{Sh63} generated by $x$ and $y$, the pair $(x,y)$ is a basis of it (a free pro-$3$-group of finite rank is Hopfian \cite{Ko02}), and there is
a continuous homomorphism $\rho:\widehat G\to\Gamma_1$ such that $\rho(x)=R$ and $\rho(y)=CRC$;
one has $\rho\circ\sigma=C\rho\, C$, the two sides agreeing at $x$ and $y$. Write
$\rho=\rho_0(1+\varepsilon Z)$, where $\rho_0$ is the reduction of $\rho$ modulo $\varepsilon$.
Then $\rho_0$ is upper triangular, with diagonal $(\psi_1,\psi_2)$ (two characters
which agree with $\psi_1,\psi_2$ at $x$ and at $y$), $Z(hh')=\rho_0(h')^{-1}Z(h)\rho_0(h')+Z(h')$,
and $\operatorname{tr}Z$ is a homomorphism vanishing at $x$ and $y$, hence zero; thus
$\det\rho=\det\rho_0=D$. Put $\beta=(\rho_0)_{12}/\psi_2$, $\gamma=Z_{21}/\theta$ and
$z=Z_{22}$: the three relations are obtained by reading these identities coefficient by
coefficient, and $\operatorname{tr}\rho=M_s+\varepsilon e$. The function $\operatorname{tr}\rho$
satisfies \eqref{eqps} over $O'[\varepsilon]$ with the same $D$, so $e$ satisfies
\eqref{eqlin}; the symmetries come from $\rho\circ\sigma=C\rho\, C$, which also gives
$\beta(y)=\gamma(y)=-1$ and $z(y)=0$.

(a) The relations give $\beta(\sigma(h)h)=(\theta(h)-1)\beta(h)$,
$\gamma(\sigma(h)h)=(\theta(h)^{-1}-1)\gamma(h)$, $z(\sigma(h)h)=2z(h)-\theta(h)\gamma(h)\beta(h)$,
and $\theta(\sigma(h)h)=\theta(h)^2$, $\psi_2(\sigma(h)h)=\psi_2(h)^2$. With $t=\theta(h)$ and
$P=\beta(h)\gamma(h)$,
\[
\frac{e(\sigma(h)h)}{\psi_2(h)^2}=(1-t^2)\bigl(2z(h)-tP\bigr)-t(1-t)^2P
=2(1-t)\bigl((1+t)z(h)-tP\bigr).
\]
(b) Since $\theta\equiv1\pmod3$, $\beta$ and $\gamma$ are, modulo $3$, homomorphisms
$\widehat G\to O'/3$ negated by $\sigma$ and equal to $1$ at $x$: they are
multiples of $v$, equal to $v/v(x)$. Then $z-\frac12\beta\gamma$ is, modulo $3$, a
homomorphism (the term $\gamma(h)\beta(h')-\frac12(\beta(h)\gamma(h')+\beta(h')\gamma(h))$
vanishes modulo~$3$ since $\beta\equiv\gamma$) invariant under $\sigma$, hence a multiple
$a'u$ of $u$, and $F\equiv2z-\beta\gamma\equiv2a'u\pmod3$.
(c) If $w\ge1$, then $\theta\equiv1\pmod9$ on $\widehat G$; $\beta$ and $\gamma$ are,
modulo $9$, homomorphisms negated by $\sigma$, hence lie in $(O'/9)\mu$.
\end{proof}

\begin{lemme}[expansion at the point $M_s$]\label{l17}
There are $A',B'$ in the fraction field of $O'$, with $v_3(A')\ge4-w$ and
$v_3(B')\ge5$, such that, for every $h\in\widehat G$,
\[
T(h)-M_s(h)\equiv A'\,\lambda(h)\bigl(\psi_1(h)-\psi_2(h)\bigr)+B'\,e(h)\pmod{3^{10}O'} .
\]
\end{lemme}
\begin{proof}
Let $e_r=\lambda(\psi_1-\psi_2)$; it is the $\varepsilon$-part of the trace of
$\psi_1e^{\varepsilon\lambda}\oplus\psi_2e^{-\varepsilon\lambda}$, which has determinant $D$, so $e_r$ satisfies \eqref{eqlin}; it is
central and invariant under $\sigma$. One has $e(x)=\psi_1(x)$, a unit; $e_r(x)=\lambda(x)
\psi_2(x)(\theta(x)-1)$, of valuation $1+w$ ($\lambda(x)$ is a unit since $u(x)\ne0$);
$xy=\sigma(y)y$ and $F(y)=-\theta(y)$, hence, by Lemma~\ref{l16}(a),
$e(xy)=-2\psi_2(x)^2\theta(x)(1-\theta(x))$, of valuation $1+w$; finally
$e_r(xy)=2\lambda(x)\psi_2(x)^2(\theta(x)^2-1)$, of valuation $1+w$. The determinant
$\Delta=e_r(x)e(xy)-e(x)e_r(xy)$ therefore has valuation exactly $1+w$. Let
$\delta=T-M_s$, with values in $3^5O'$ on $\widehat G$, and $A',B'$ defined by
$A'e_r+B'e=\delta$ at $x$ and at $xy$; Cramer's formulas give $v_3(A')\ge
5-(1+w)$ and $v_3(B')\ge(5+1+w)-(1+w)$. The function $L=A'e_r+B'e$ satisfies \eqref{eqlin} and
$L(y)=L(x)=\delta(x)=\delta(y)$ ($\sigma$-invariance). Apply Lemma~\ref{l9} to the
pseudo-representation $M_s+\varepsilon L$ (over $\mathrm{Frac}(O')[\varepsilon]$) and to $T$,
$M_s$, with $x_1=x$, $x_2=y$, $x_3=xy$ and $m=(M_s(x),M_s(y),M_s(xy),D(x),D(y))$: for every
word $W$, $L(W)=\sum_j\partial_jP_W(m)\,\delta(x_j)$ and
\begin{align*}
T(W)-M_s(W)&=P_W\bigl(m+(\delta(x_1),\delta(x_2),\delta(x_3),0,0)\bigr)-P_W(m)\\
&\equiv\sum_j\partial_jP_W(m)\,\delta(x_j)\pmod{3^{10}O'},
\end{align*}
since $P_W$ has integer coefficients, $D(x)$, $D(y)$ are units, and the Taylor coefficients
of order $\ge2$ of $P_W$ at $m$ therefore lie in $O'$, while
$\delta(x_j)\in3^5O'$. One concludes by density of the subgroup generated by $x$ and $y$.
\end{proof}

\begin{thRp}[law of inert primes]
Let $\ell\equiv2\pmod3$ be prime and $j=v_3(\ell+1)$. For every eigenform $f$ of level~$1$
and weight $k$,
\[
a_\ell^2\equiv s_\ell^2\pmod{3^{\min(2j+2,\,10)}O}.
\]
In particular $a_\ell^2\equiv s_\ell^2\pmod{3^8}$ if $\ell\equiv-1\pmod{27}$, and
$\pmod{3^{10}}$ if $\ell\equiv-1\pmod{81}$.
\end{thRp}
\begin{proof}
Let $f_\ell$ be a Frobenius element at $\ell$ and $h=cf_\ell\in G_K$; then
$f_\ell^2=chch=\sigma(h)h$, $\chi(h)=-\ell$, and $T(f_\ell^2)-T_E(f_\ell^2)=a_\ell^2-2\ell^\kappa-
(1+\ell^{2\kappa})=a_\ell^2-s_\ell^2$. One has $v_3(\lambda(h))=v_3(\log_3(-\ell))-1=j-1$, hence
$v_3(\theta(h)-1)=j+w$, $v_3(\lambda(f_\ell^2))=j-1$ and $v_3(\theta(f_\ell^2)-1)=j+w$;
finally $u(h)=0$ if and only if $j\ge2$. Write
$T-T_E=(M_s-T_E)+(T-M_s)$ at $f_\ell^2$ and estimate the terms.
(i) $M_s-T_E=-(\psi_s-1)(\chi^\kappa\psi_s^{-1}-1)$ has valuation at least
$(j+v_3(s))+(j+v_3(\kappa-s))\ge2j+2$, since $v_3(s)\ge2$ if $3\nmid\kappa$, and
$v_3(s),v_3(\kappa-s)\ge1$ if $3\mid\kappa$.
(ii) $A'e_r(f_\ell^2)$ has valuation at least $(4-w)+(j-1)+(j+w)=2j+3$.
(iii) By Lemma~\ref{l16}(a), $B'e(f_\ell^2)=2B'\psi_2(h)^2(1-\theta(h))F(h)$ has
valuation at least $5+j+w+v_3(F(h))$, and $v_3(F(h))\ge1$ if $j\ge2$ by
Lemma~\ref{l16}(b): hence at least $j+6\ge\min(2j+2,10)$ if $j\ge2$, and at least $6$ if $j=1$.
Lemma~\ref{l17} concludes.
\end{proof}
\statut{PROVED} From Theorem~\ref{thR}. For $j=1$ and $j=2$, one recovers
$a_\ell^2\equiv s_\ell^2$ modulo $81$ and modulo $729$ (Theorems~\ref{thO} and~\ref{thR}).

\begin{coYp}[non-vanishing, $3$-adic]
If $k\not\equiv4\pmod6$ and if $\ell\equiv2\pmod3$ is a prime with $v_3(\ell+1)\le4$, then
$v_3(a_\ell(f))=v_3(\ell+1)$. Consequently, if $k\not\equiv4\pmod6$, $a_\ell(f)=0$ implies
$\ell\equiv-1\pmod{2^{11}\cdot3^5}$.
\end{coYp}
\begin{proof}
If $3\nmid\kappa$, $v_3(s_\ell)=v_3(\ell+1)+v_3(\kappa)=j$, and Theorem~R$'$ gives
$a_\ell^2\equiv s_\ell^2\pmod{3^{2j+2}}$ for $j\le4$, whence $v_3(a_\ell^2)=2j$. If $a_\ell=0$,
then $\ell\ne2$ (Corollary~\ref{coL}), $\ell\ne3$ (Corollary~\ref{coZ}), $\ell\not\equiv1\pmod3$ ($a_\ell\equiv s_\ell\equiv2
\pmod3$), and, by what precedes, $j\ge5$; one concludes with Corollary~\ref{coZ}.
\end{proof}
\statut{PROVED}

\begin{thRpp}[split primes]
Let $\ell\equiv1\pmod{27}$ be a prime such that $3$ is a cube modulo $\ell$. For
every eigenform $f$ of level~$1$ and weight $k$, $a_\ell\equiv s_\ell\pmod{3^7O}$. If
moreover $k\equiv4\pmod6$ and $\ell=X^2+XY+547Y^2$ ($X,Y\in\Z$), then $a_\ell\equiv
s_\ell\pmod{3^8O}$, and $\pmod{3^9O}$ if $\ell\equiv1\pmod{81}$.
\end{thRpp}
\begin{proof}
Let $h=\mathrm{Frob}_{\mathfrak l}$, $\mathfrak l\mid\ell$, and $n=v_3(\lambda(h))=
v_3(\ell-1)-1\ge2$; $v(h)=0$, so $\beta(h),\gamma(h)\in3O'$ (Lemma~\ref{l16}(b)).
(i) $M_s(h)-T_E(h)$ has valuation at least $(1+v_3(s)+n)+(1+v_3(\kappa-s)+n)\ge2n+4$.
(ii) $A'e_r(h)$ has valuation at least $(4-w)+n+(1+w+n)=2n+5$.
(iii) $e(h)=\psi_2(h)\bigl((1-\theta(h))z(h)+\theta(h)\beta(h)\gamma(h)\bigr)$ has
valuation at least $\min(1+w+n,2)=2$, so $B'e(h)$ has valuation at least $7$.
Lemma~\ref{l17} gives the first assertion. If $k\equiv4\pmod6$, then $w\ge1$; if moreover
$\ell=X^2+XY+547Y^2$, then $\mu(h)\in9\Z_3$, so $\beta(h),\gamma(h)\in9O'$ by
Lemma~\ref{l16}(c), and $e(h)$ has valuation at least $\min(2+n,4)$; the minimum of
$2n+4$, $2n+5$, $5+\min(2+n,4)$ and $10$ is $8$ if $n=2$ and $9$ if $n\ge3$.
\end{proof}
\statut{PROVED} From Theorem~\ref{thR}.

\begin{remarque}
\textbf{VERIFIED} (PARI/GP, exact Newton polygons on $S_k$). \emph{Inert primes},
$12\le k\le94$: the minimal depth of $a_\ell^2-s_\ell^2$ is exactly $8$ for the
eight primes $\ell\equiv-1\pmod{27}$, $\ell\not\equiv-1\pmod{81}$, $\ell<1200$
($53,107,269,431,593,701,863,1187$), and $10$ for $\ell=647$ and $809$ ($j=4$), except for
$k=12$ and $k=22$, where it is $9$ and $11$: the exponent $2j+2$ is attained for $j\le4$. For
$\ell=971$ ($j=5$), it is $11$ if $k\not\equiv4\pmod6$ ($12$ for $k=12$) and $12$ if
$k\equiv4\pmod6$ ($13$ for $k=22$): the exponent $2j+2$ does not hold for $j=5$, which
corresponds to term (iii) of the proof. \emph{Split primes}, $12\le k\le78$, for the seven
primes $\ell\equiv1\pmod{27}$, $\ell<3000$, such that $3$ is a cube modulo $\ell$
($271,757,919,1621,1783,2269,2971$): if $k\equiv4\pmod6$, the minimal depth of
$a_\ell-s_\ell$ is $7$ for $271,919,1621,1783$, which are not of the form
$X^2+XY+547Y^2$, $8$ for $757$ and $2971$, and $9$ for $2269$ (except $k=22$, where it is
larger); if $k\not\equiv4\pmod6$, it is $7$ for $757$ and $2971$, and at least $8$ for
the other five. The depth $8$ for $\ell\equiv1\pmod{27}$ when
$k\not\equiv4\pmod6$ remains \textbf{OPEN}: modulo $3^8$, only the term
$B'\psi_2\theta\beta\gamma$ of the proof can obstruct it, and
$\beta/3$, $\gamma/3$ are then characters of $\operatorname{Gal}(E/K(\sqrt[3]3))$ which,
unlike the case $3\mid\kappa$ (Lemma~\ref{l16}(c)), do not factor through
$\operatorname{Gal}(H_{27}/K)$ (script \texttt{VERIF\_INERTES.py}, part~A).
\end{remarque}

\section{What remains open}

\begin{itemize}
\item Hatada's question: \textbf{SOLVED} (Corollary~\ref{coY}): affirmative answer
for $6\le a\le10$, negative for $a=11,12,13$. ``Lattice'' form for $a\le10$:
verified (Statement~V$'$), not proved.
\item Gain in depth for $\ell\equiv\pm1\pmod9$: \textbf{PROVED} in every weight
(Corollary~\ref{coQ}, Theorem~\ref{thR}). Beyond: \textbf{PROVED} for inert primes
up to $3^{10}$, with the optimal exponent $2v_3(\ell+1)+2$ for $v_3(\ell+1)\le4$
(Theorem~R$'$), and, for $\ell\equiv1\pmod{27}$ with $3$ a cube, up to $3^7$, or $3^8$ and
$3^9$ if $k\equiv4\pmod6$ and $\ell=X^2+XY+547Y^2$ (Theorem~R$''$). \textbf{OPEN}: the
depth $8$ for $\ell\equiv1\pmod{27}$ when $k\not\equiv4\pmod6$ (Remark of
\S\ref{sec9}), and the depths beyond $3^{10}$.
\item Value of $a_2(f)$ modulo $27$ when $k\equiv4\pmod6$: there is
no uniform law (Remark~(3) of \S\ref{sec6}), but $a_2$ determines the other
$a_\ell$, $\ell\equiv2\pmod3$, by Theorem~\ref{thS}; the possible values of $a_2(f)$
modulo $81$ (valuative sense) are listed, by class of $k$ modulo $54$, in
\cite[Thm~1.5 and Table~3]{R26}.
\item Coleman--Stein conjecture in levels $3$ and $9$: \textbf{OPEN}
(checked by Coleman and Stein up to weight $40$).
\end{itemize}


\section{Verification}\label{secverif}

\begin{itemize}
\item PARI/GP:\\
\texttt{ramanujantau(12289)}\\
\texttt{F = mfeigenbasis(mfinit([1,16],1))[1]; mfcoef(F,13313)}\\
\texttt{charpoly(mfheckemat(mfinit([1,k],1),3))} ($3$-adic slopes of $T_3$)\\
In Sage: \texttt{pari('ramanujantau(12289)')}.
\item Script \texttt{VERIF\_HATADA.py} (Python, numpy only, a few seconds):
self-test on known values and on the exact value of $\tau(12289)$;
the $24$ class congruences for all odd $n\le2000$; the two
witnesses; a scan of all primes $p\le13313$, which shows that the
smallest witness at each failing level is $12289$, and that no
$a\le10$ fails for the six forms.
\item Script \texttt{VERIF\_MOD9.py} (Python only): the criteria of
Theorems~\ref{thG} and~\ref{thH} ($T_2$ and $T_3$), for increasing weights until
a time budget set in the file is exhausted, with self-test. With the values set in
the file it covers $12\le k\le400$ in a few minutes; the range $k\le1000$ of
Theorem~\ref{thG} ($494$ spaces) was covered by the same certificate computed independently with
FLINT and PARI/GP (Miller basis, characteristic polynomials of $T_2$ and $T_3$, Newton
polygons at $3$), in about forty minutes.
\item Script \texttt{VERIF\_A3.py} (Python only, a few minutes at most):
the checks of Theorem~\ref{thF} described above, with self-test on the period
of $\Delta$, proportional to $(48,-25,20,-25,48)$, and on $\tau(3)=252$.
\item Script \texttt{VERIF\_A2.py} (Python only, under a minute):
the certificate of Lemma~\ref{l6} in $(\Z/9\Z)[\mathcal G]$ ($290$ terms); the three
certificates of Lemma~\ref{l8} and the criterion of Lemma~\ref{l7}, which prove
Theorem~\ref{thK}; direct check of Theorems~\ref{thI} and~\ref{thK} on $L_k$
for $12\le k\le100$; the identity of Lemma~\ref{l5} on all solutions
modulo~$3$ of the relations, and the condition of Lemma~\ref{l6} on $L_k$, for $k\le40$;
direct action of the remainders of Lemma~\ref{l8} on $\Z[x,y]_w$ for $w\le60$, with
out-of-class witnesses; self-test on the period of $\Delta$ and on $\tau(2)=-24$.
\item Script \texttt{VERIF\_EISENSTEIN.py} (Python only, about two minutes): the
nineteen certificates of Theorem~\ref{thN} (criterion of Lemma~\ref{l7}, moduli $27$,
$81$, $243$), their check by direct action on $\Z[x,y]_w$, statement (i) of
Theorem~\ref{thO} for $k\le44$ (characteristic polynomials of $T_n$ on $L_k$,
$n\equiv1\pmod3$, $n\le k$), and Theorems~\ref{thP}, \ref{thR}, \ref{thS} for
$k\le40$ and $\ell\le43$, $\ell=73$, on $L_k$, with witnesses where the correction term
is necessary; self-test on $\tau(2)$ and $\tau(7)$.
\item Script \texttt{VERIF\_HATADA2.py} (Python only, under a minute): the
certificate of Lemma~\ref{l13} ($190$ terms, criterion of Lemma~\ref{l12} modulo $512$),
the direct action of the remainder on $\Z[x,y]_w$ for $w\le40$ (witnesses of odd degree), the
Kummer symbols in the proof of Theorem~\ref{thU}, the five laws of that
theorem on $L_k$ for $k\le40$, and the enumeration of the finite groups of
Theorem~\ref{thX} (part E; part F, heavier, is optional); self-test on
$\tau(3)$ and $\tau(5)$.
\item Script \texttt{VERIF\_INERTES.py} (Python only, under a minute): the
identities of Lemma~\ref{l16} and the valuations of the system of Lemma~\ref{l17} in the
dual-number representation, on random words modulo $3^{14}$, with witnesses;
Theorem~R$'$ and Corollary~Y$'$ on $L_k$ for $k\le40$ and
$\ell\in\{5,11,17,53,71,107\}$, the exponent being attained for each $\ell$;
Theorems~R$'$, R$''$ and Corollary~Y$'$ for the six weights of dimension~$1$ and all
primes $\ell<3000$ ($1332$ inert cases, $42$ split cases), with the witnesses
$v_3(a_{971}^2-s_{971}^2)=11$ for $k=18$ and $v_3(a_{271}-s_{271})=7$ for $k=16$; finally the
cyclicity of the Picard groups of the orders of conductor $9$ and $27$ in $\Z[\zeta_3]$ and
their consistency with the forms $X^2+XY+61Y^2$, $X^2+XY+547Y^2$ and the cube criterion for
$\ell<20\,000$; self-test on $\tau(n)$, $n\le11$.
\item Theorem~\ref{thD} (PARI/GP): for $12\le k\le100$, $k\ne14$,\\
\texttt{P = charpoly(mfheckemat(mfinit([1,k],1),12289));}\\
\texttt{newtonpoly(subst(P,x,x+12290),2)}\\
contains the slope $10$ if $k\equiv0,2\pmod8$ ($k\ge16$) and the slope $11$ if $k\equiv4,6\pmod8$ and $k\in\{12\}\cup[28,100]$.
\item Check of Theorem~\ref{thO} by Sturm's theorem (PARI/GP): Newton polygons
of $\mathrm{charpoly}(T_n-\sigma_{k-1}(n))$ on $S_k$ for $n\equiv1\pmod3$,
$n\le k$, $12\le k\le212$; statement (ii) on an eigenbasis embedded in
$\overline{\Q}_3$ for $k\le140$.
\end{itemize}

\section*{Acknowledgements}
Technical assistance: Claude (Anthropic).

\begin{thebibliography}{99}
\small
\bibitem{BK15} J.~Bella\"iche, C.~Khare, \emph{Level 1 Hecke algebras of modular forms
modulo $p$}, Compos.\ Math.\ \textbf{151} (2015), 397--415,
doi:10.1112/S0010437X1400774X.
\bibitem{CSa21} F.~Calegari, N.~T.~Sardari, \emph{Vanishing Fourier coefficients of
Hecke eigenforms}, Math.\ Ann.\ \textbf{381} (2021), 1197--1215, doi:10.1007/s00208-021-02178-7
(arXiv:2003.07570).
\bibitem{CE04} F.~Calegari, M.~Emerton, \emph{The Hecke algebra $T_k$ has large index},
Math.\ Res.\ Lett.\ \textbf{11} (2004), 125--137, doi:10.4310/MRL.2004.v11.n1.a12.
\bibitem{Ch26} L.~Chiriac, \emph{Congruences for Hecke eigenvalues via period
polynomials}, Bull.\ Aust.\ Math.\ Soc., online first, published online 21 September 2026,
doi:10.1017/S0004972726101889.
\bibitem{Co13} D.~A.~Cox, \emph{Primes of the form $x^2+ny^2$: Fermat, class field
theory, and complex multiplication}, 2nd ed., Wiley, Hoboken (2013).
\bibitem{CS04} R.~F.~Coleman, W.~A.~Stein, \emph{Approximation of eigenforms of
infinite slope by eigenforms of finite slope}, Geometric aspects of Dwork theory,
vol.~I, de Gruyter, Berlin (2004), 437--449.
\bibitem{De71} P.~Deligne, \emph{Formes modulaires et repr\'esentations $\ell$-adiques},
S\'eminaire Bourbaki 1968/69, exp.~355, Lecture Notes in Math.\ \textbf{179}, Springer
(1971), 139--172.
\bibitem{Go09} W.~M.~Goldman, \emph{Trace coordinates on Fricke spaces of some simple
hyperbolic surfaces}, Handbook of Teichm\"uller theory, vol.~II, IRMA Lect.\ Math.\
Theor.\ Phys.\ \textbf{13}, EMS (2009), 611--684.
\bibitem{H79} K.~Hatada, \emph{Eigenvalues of Hecke operators on
$\mathrm{SL}(2,\mathbb Z)$}, Math.\ Ann.\ \textbf{239} (1979), 75--96.
\bibitem{H81} K.~Hatada, \emph{Congruences for eigenvalues of Hecke operators
on $\mathrm{SL}_2(\mathbb Z)$}, Manuscripta Math.\ \textbf{34} (1981),
305--326.
\bibitem{Ko02} H.~Koch, \emph{Galois theory of $p$-extensions}, Springer Monographs in
Mathematics, Springer, 2002.
\bibitem{K62} O.~Kolberg, \emph{Congruences for Ramanujan's function
$\tau(n)$}, \AA rbok Univ.\ Bergen Mat.-Natur.\ Ser.\ 1962, no.~11.
\bibitem{Ka75} N.~M.~Katz, \emph{Higher congruences between modular forms}, Ann.\ of
Math.\ (2) \textbf{101} (1975), 332--367.
\bibitem{Ma73} Yu.~I.~Manin, \emph{Periods of parabolic forms and $p$-adic Hecke
series}, Mat.\ Sb.\ (N.S.) \textbf{92(134)} (1973), 378--401; Math.\ USSR-Sb.\
\textbf{21} (1973), 371--393.
\bibitem{Me16} A.~Medvedovsky, \emph{An explicit Galois representation on the mod-$3$
Hecke algebra}, manuscript, 23 December 2016,
\url{https://people.mpim-bonn.mpg.de/medved/Mathwriting/Mod3galoisrep.pdf}.
\bibitem{PARI} The PARI Group, \emph{PARI/GP}, version 2.17.2, Univ.\ Bordeaux,
\url{https://pari.math.u-bordeaux.fr}.
\bibitem{R19} N.~Rustom, \emph{Congruences modulo prime powers of Hecke
eigenvalues in level~1}, Res.\ Number Theory \textbf{5} (2019), Paper~10,
doi:10.1007/s40993-018-0147-5 (arXiv:1710.00928).
\bibitem{R26} N.~Rustom, \emph{Prime-power congruences for level-one eigenforms},
preprint (2026), arXiv:2609.16750.
\bibitem{Se73} J.-P.~Serre, \emph{Formes modulaires et fonctions z\^eta $p$-adiques},
in \emph{Modular Functions of One Variable III}, Lecture Notes in Math.\ \textbf{350},
Springer, 1973, 191--268.
\bibitem{Sh63} I.~R.~Shafarevich, \emph{Extensions with given points of ramification},
Publ.\ Math.\ IH\'ES \textbf{18} (1963), 71--92 (in Russian).
\bibitem{St87} J.~Sturm, \emph{On the congruence of modular forms}, Lecture
Notes in Math.\ \textbf{1240}, Springer (1987), 275--280.
\bibitem{SD73} H.~P.~F.~Swinnerton-Dyer, \emph{On $\ell$-adic representations
and congruences for coefficients of modular forms}, Lecture Notes in Math.\
\textbf{350}, Springer (1973), 1--55.
\bibitem{TQ} P.~Tian, H.~Qin, \emph{Non-vanishing Fourier coefficients of
$\Delta_k$}, Appl.\ Math.\ Comput.\ \textbf{339} (2018), 507--515,
doi:10.1016/j.amc.2018.07.022 (arXiv:1504.07381, under the title \emph{Non-vanishing
Fourier coefficients of modular forms}).
\bibitem{Z90} D.~Zagier, \emph{Hecke operators and periods of modular forms},
Israel Math.\ Conf.\ Proc.\ \textbf{3} (1990), 321--336.
\end{thebibliography}


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