\documentclass{article} \include{macros} \begin{document} \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} \end{document}