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\myhead{Congruences between modular forms}{0.1}
\section*{Introduction}
This is an account of the paper \cite{sturm}
which is of importance to the explicit computation of Hecke algebras. 
Suppose $f=\sum a_n q^n$ and $g=\sum b_n q^n$ are modular
forms in $S_k(\Gamma)$.   Suppose that the field $K$ generated
by all of the $a_n$ and $b_n$ is a finite extension of $\Q$.
Let $\lambda$ be a prime of $K$.  Sturm deduces
an integer $B$, depending in a simple way on $k$ and the
index of $\Gamma$ in $\sltwoz$, such that if $a_n-b_n\in\lambda$ for $n\leq B$
then $a_n-b_n\in\lambda$ for all $n$.

\section{Notation and terminology}
Fix a positive integer $N$ and let
$$\Gamma(N) = \{\abcd{a}{b}{c}{d} \in \sltwoz\, :\, 
       \abcd{a}{b}{c}{d} \con \abcd{1}{0}{0}{1}\!\!\!\!\!\mod{N}\}.$$
A subgroup $\Gamma$ of $\sltwoz$ is a called a {\em congruence subgroup} of level $N$ if it
contains $\Gamma(N)$ for some $N$.  Fix such a subgroup $\Gamma$.
Let $$\h=\{z\in\C : \Im(z)>0\}$$ be the complex upper half-plane.
For any function $f:\h\ra\C$, matrix $\gamma\in\sltwoz$, and 
integer $k$, define $f|[\gamma]_k:\h\ra\C$ by
$$(f|[\gamma]_k)(z) = (cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right).$$
Let $M_k(\Gamma)$ be the space of modular forms for $\Gamma$ of weight
$k$, that is, the complex vector space of function $f:\h\ra\C$ such that
(1) $f|[\gamma]_k=f$ for all $\gamma\in\Gamma$, 
(2) $f$ is holomorphic, and
(3) $f|[\gamma]_k$ is bounded on $\{z:\Im(z)>1\}$ for all $\gamma\in\sltwoz$. 
If $f\in M_k(\Gamma)$ then $f(z)$ has a Fourier expansion
$$f(z) = \sum_{\substack{n\geq 0\\n\in \frac{1}{N}\Z}} a_n q^n$$
where $q=e^{2\pi i z}$ and $a_n=a_n(f)\in\C$. 
Let $S_k(\Gamma)$ be the subspace of cusp forms, that is, 
those $f\in M_k(\Gamma)$ such that
$a_0(f|[\gamma]_k)=0$ for all $\gamma\in\sltwoz$.
If $R$ is a subring of $\C$, denote by 
$M_k(\Gamma,R)$ the subset of $M_k(\Gamma)$ consisting of modular
forms whose $q$-expansion coefficients $a_n(f)$ lie in $R$.

\section{Congruences}
Let $k\geq 1$ be an integer, $\Gamma$ a congruence subgroup,
 and let $\mu=[\sltwoz:\Gamma]$.
Let $K$ be a number field (considered as a subfield of $\C$)
with ring of integers $\O$. Fix a prime ideal $\p$ of $\O$ 
and let $\F=\O/\p$ denote the residue class field. Let
$\O_{\p}\subset K$ denote the localization of $\O$ at $\p$,
i.e., the domain obtained by inverting all elements not in $\p$. 
The valuation on $\F[[q]]$ induces a valuation $v_\p$ on $M_k(\Gamma,\O_\p)$ via
the natural map $M_k(\Gamma,\O_\p)\ra \F[[q]]$.  
Thus $$v_\p(f) = \min\{ n \,:\, a_n(f) \not\con 0 \!\pmod{\p}\}.$$

\begin{theorem}\label{theorem}
If $f\in M_k(\Gamma,\O_{\p})$ and
$\displaystyle v_\p(f) > \frac{\mu{}k}{12},$
then
$f\con 0\mod{\p}$. 
\end{theorem}
We recall some notation and a lemma before giving the proof.
Let 
\begin{eqnarray*}
\Delta(q) &=& q - 24q^2 + 252q^3 - 1472q^4 +  4830q^5 + \cdots \in S_{12}(\sltwoz)\\
      j(q)&=& \frac{1}{q} + 744 + 196884q + 21493760q^2+864299970q^3 + \cdots \in K(X(1))
\end{eqnarray*}
Note that $j(q)$ is an automorphic function for $\sltwoz$ of weight $0$
whose only pole is at infinity (i.e., $j|[\gamma]_0=j$ for all $\gamma\in\sltwoz$).  
The Fourier coefficients of both $\Delta$
and $j$ lie in $\Z$.

For any positive integer $n$, let $\zeta_n=e^{2\pi i/n}$ be a primitive
$n$th root of unity.

\begin{lemma}\label{lemma}
Let $N$ be a positive integer.
Then any $\gamma\in\sltwoz$ induces a linear isomorphism
$$[\gamma]_k: M_k(\Gamma(N))\ra M_k(\Gamma(N))$$
which preserves $M_k(\Gamma(N),\Q(\zeta))$.
\end{lemma}
\begin{proof}
That $[\gamma]_k$ leaves $M_k(\Gamma(N))$ invariant
follows because $\Gamma(N)$ is a normal subgroup of $\sltwoz$:
if $\alp\in\Gamma(N)$ then there is $\alp'\in\Gamma(N)$
such that $\gamma\alp=\alp'\gamma$, so for any
$f\in M_k(\Gamma(N))$, 
$$(f|[\gamma]_k)|[\alp]_k = f|[\gamma\alp]_k
        = f|[\alp'\gamma]_k
        = (f|[\alp']_k)|[\gamma]_k
        = f|[\gamma]_k.$$
Furthermore, $[\gamma]_k$ is an isomorphism because 
$[\gamma^{-1}]_k$ is its inverse. 
The second part follows because $X(N)$ has
a model over $\Q(\zeta_N)$ and $\sltwo(\Z/N\Z)$
acts via automorphisms of the function field of $X(N)$. 
The action of $[\gamma]_k$ is induced by this action.
This is discussed in Chapter 6 of \cite{shimura}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{theorem}]
We assume that $f\neq 0$, otherwise we are done.
In each case we will give the argument for $\O$ and then 
note that it works for $\O_{\p}$ as well.

{\em Case one: $\Gamma=\sltwoz$.}
In the ring $\O\{\{q\}\}$ of Laurant series,
$$\frac{f^{12}}{\Delta^k} = 
  f^{12} \cdot \left(\frac{1}{q}+24+324q+\cdots\right)^k = \sum_{n\geq -k} b_n q^n\in\O\{\{q\}\}.$$
Since
$$v_\p\left(\frac{f^{12}}{\Delta^k}\right) 
   = 12v_\p(f) - kv_\p(\Delta) = 12 v_\p(f) - k > 0$$
we see that $b_n\in\p$ for $n\leq 0$. 
By inductively subtracting off multiples
of powers of $j(q)=\frac{1}{q}+744+\cdots$ from $\frac{f^{12}}{\Delta^k}$
(reducing the order of the pole by at least one at each step) we
find a polynomial $R(j)$ in $\O[j]$ such that 
$$\frac{f^{12}}{\Delta^k} - R(j) \in q\O[[q]].$$
But this difference is a holomorphic function on the projective plane,
so it must be a constant, hence $0$, so that 
$f^{12}=R(j)\Delta^k.$  
Each of the coefficients of $R(j)$ must lie in $\p$,
otherwise $v_\p(R(j))\leq 0$ which would be a contradiction
because then $v_\p(\frac{f^{12}}{\Delta^k} - R(j)) \leq 0$. 
Thus $f^{12} = R(j) \Delta^k \con 0\pmod{\p}$ and the same
holds for $f$ since $(\O/\p)[[q]]$ is a domain.

Note that the same argument works with $\O$ replaced by
its localization at $\p$.  Thus we can allow denominators
in the $q$-expansion of $f$ as long as they do not lie
in $\p$. 

{\em Case two: $\Gamma$ arbitrary.}
For any $\gamma\in\sltwoz$, Lemma~\ref{lemma} implies that
$f|[\gamma]_k \in M_k(\Gamma(N),K(\zeta_n))$. 
Let $\gP|\p$ be a prime ideal of the ring of integers of $K(\zeta_n)$.
Let $\pi$ be a uniformizer for $\gP$.
For any $\gamma$, there is a power $\pi^n$ of $\pi$ 
so that the coefficients of
the $q$-expansion of $\pi^n f|[\gamma]_k$ are $\gP$-integral
and  $\pi^n f|[\gamma]_k\not\con 0\pmod{\gP}$.
To do this choose $n$ minimally (possibly negative!)
so that $\pi^n f|[\gamma]_k$ is $\gP$-integral.
If $\pi^n f|[\gamma]_k\con 0\pmod{\gP}$
then all coefficients are divisible by $\pi$ so $n$ was not chosen
minimally.

Write 
$$\sltwoz = \bigcup_{i=1}^{\mu} \Gamma \gamma_i$$
with $\gamma_1=\abcd{1}{0}{0}{1}$, and let the $n_i$
be chosen for $\gamma_i$ as in the previous paragraph.
Then
$$F = f \cdot \prod_{i=2}^{\mu} \pi^{n_i} f|[\gamma_i]_k\in M_{k\mu}(\sltwoz)$$
and 
$$v_{\gP}(F)\geq v_{\gP}(f) = v_{\p}(f) > \frac{\mu}{12}k.$$
By case 1, $F\con 0\pmod{\gP}$, so since 
$\pi^{n_i} f|[\gamma_i]_k\not\con 0\pmod{\gP}$
for each $i\geq 2$, it follows that $f\con 0\pmod{\p}$ as claimed.
\end{proof}

\begin{remark}
For cusp forms the proof of Theorem 1 yields the following {\em slightly} 
improved bound:
If $f \in S_k(\Gamma,\O_\p)$ and 
$$v_\p(f) > \frac{\mu}{12}k - \frac{\mu-1}{N}$$
then $f\con 0\pmod{\p}$. 

\begin{proof}
Consider the proof of case two. 
Note that the $q$-expansion of an element of $M_k(\Gamma(N))$
is in powers of $q^{1/N}$.
Since $f$ is a cusp form, we know that 
$v_{\gP}(f|[\gamma_i]_k)\geq \frac{1}{N}$ for each $i$.
If we insure that
$$v_\p(f) > \frac{\mu}{12}k - \frac{\mu-1}{N}$$
then
 $$v_{\gP}(F)\geq v_{\gP}(f) + \frac{\mu-1}{N} = v_{\p}(f) + \frac{\mu - 1}{N} > \frac{\mu}{12}k$$
and the proof goes through.
\end{proof}
\end{remark}


\section{Applications}
Let $\Gamma$ be a congruence subgroup of level $N$ and index $\mu$, let
$k$ be a positive integer, and set $r:=\frac{k}{12}\mu.$
\begin{proposition}\label{determine}
Suppose
$$f=\sum_{\substack{n>r\\n\in \frac{1}{N}\Z}} a_n q^n\in M_k(\Gamma)$$ 
then $f=0$.  Stated differently, an element of $M_k(\Gamma)$ is
determined by its Fourier coefficients $a_n$ for $n\leq r$. 
\end{proposition}
\begin{proof}
Let $s=q^{1/N}$. 
We must show that the composite map 
$$M_k(\Gamma)\hookrightarrow\C[[s]]\into\C[[s]]/(sq^r)$$
is injective.  
Let $K$ be a number field with ring of integers $\O$
such that $M_k(\Gamma) = M_k(\Gamma,\O)\tensor\C$. 
Because $\C$ is flat over $\O$, it suffices 
to show that the map $\Phi:M_k(N;\O)\into\O[[s]]/(sq^r)$ is injective. 
(We are just using that a linear map which is injective over $\O$ induces
an injective map over $\C$.)
Suppose $g=\sum b_n q^n \in M_k(N;\O)$ and
$\Phi(g)=0$ so that $b_n=0$ for $n\leq r$.
Thus if $\p$ be a prime of $\O$, 
$b_n\con 0\pmod{\p}$ for any $n\leq r$. 
By Theorem~\ref{theorem}, it follows that $g\con 0\pmod{\p}$. 
Repeating this argument for all $\p$ of $\O$ shows that 
the coefficients of $g$ are divisible by all primes $\p$,
i.e., they are $0$. 
\end{proof}

Now suppose $\Gamma=\Gamma_0(N)$ or $\Gamma_1(N)$ so that
we have a Hecke algebra $\T$ which acts as a commutative 
ring of  linear endomorphism on $M_k(\Gamma)$. 
\begin{theorem}\label{bound}
The Hecke algebra is generated as a $\Z$-module 
by $T_1,\ldots,T_r.$
\end{theorem} 
For the proof, see \cite{stein}.

\begin{thebibliography}{HHHHHHH}         

\bibitem[Shimura]{shimura} G. Shimura, {\em Introduction to the Arithmetic
Theory of Automorphic Functions}, Princeton University Press, 1994

\bibitem[Stein]{stein} W. Stein, {\em Generating the Hecke algebra as a $\Z$-module}

\bibitem[Sturm]{sturm} J. Sturm, {\em On the Congruence of Modular Forms}.
Number theory (New York, 1984--1985), 275--280, Lecture Notes in
Math., 1240, Springer, Berlin-New York, 1987.
\end{thebibliography} \normalsize\vspace*{1 cm}
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