CoCalc Public Fileswww / Tables / congruences.tex
Author: William A. Stein
1\documentclass{article}
2\include{macros}
3\begin{document}
5\section*{Introduction}
6This is an account of the paper \cite{sturm}
7which is of importance to the explicit computation of Hecke algebras.
8Suppose $f=\sum a_n q^n$ and $g=\sum b_n q^n$ are modular
9forms in $S_k(\Gamma)$.   Suppose that the field $K$ generated
10by all of the $a_n$ and $b_n$ is a finite extension of $\Q$.
11Let $\lambda$ be a prime of $K$.  Sturm deduces
12an integer $B$, depending in a simple way on $k$ and the
13index of $\Gamma$ in $\sltwoz$, such that if $a_n-b_n\in\lambda$ for $n\leq B$
14then $a_n-b_n\in\lambda$ for all $n$.
15
16\section{Notation and terminology}
17Fix a positive integer $N$ and let
18$$\Gamma(N) = \{\abcd{a}{b}{c}{d} \in \sltwoz\, :\, 19 \abcd{a}{b}{c}{d} \con \abcd{1}{0}{0}{1}\!\!\!\!\!\mod{N}\}.$$
20A subgroup $\Gamma$ of $\sltwoz$ is a called a {\em congruence subgroup} of level $N$ if it
21contains $\Gamma(N)$ for some $N$.  Fix such a subgroup $\Gamma$.
22Let $$\h=\{z\in\C : \Im(z)>0\}$$ be the complex upper half-plane.
23For any function $f:\h\ra\C$, matrix $\gamma\in\sltwoz$, and
24integer $k$, define $f|[\gamma]_k:\h\ra\C$ by
25$$(f|[\gamma]_k)(z) = (cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right).$$
26Let $M_k(\Gamma)$ be the space of modular forms for $\Gamma$ of weight
27$k$, that is, the complex vector space of function $f:\h\ra\C$ such that
28(1) $f|[\gamma]_k=f$ for all $\gamma\in\Gamma$,
29(2) $f$ is holomorphic, and
30(3) $f|[\gamma]_k$ is bounded on $\{z:\Im(z)>1\}$ for all $\gamma\in\sltwoz$.
31If $f\in M_k(\Gamma)$ then $f(z)$ has a Fourier expansion
32$$f(z) = \sum_{\substack{n\geq 0\\n\in \frac{1}{N}\Z}} a_n q^n$$
33where $q=e^{2\pi i z}$ and $a_n=a_n(f)\in\C$.
34Let $S_k(\Gamma)$ be the subspace of cusp forms, that is,
35those $f\in M_k(\Gamma)$ such that
36$a_0(f|[\gamma]_k)=0$ for all $\gamma\in\sltwoz$.
37If $R$ is a subring of $\C$, denote by
38$M_k(\Gamma,R)$ the subset of $M_k(\Gamma)$ consisting of modular
39forms whose $q$-expansion coefficients $a_n(f)$ lie in $R$.
40
41\section{Congruences}
42Let $k\geq 1$ be an integer, $\Gamma$ a congruence subgroup,
43 and let $\mu=[\sltwoz:\Gamma]$.
44Let $K$ be a number field (considered as a subfield of $\C$)
45with ring of integers $\O$. Fix a prime ideal $\p$ of $\O$
46and let $\F=\O/\p$ denote the residue class field. Let
47$\O_{\p}\subset K$ denote the localization of $\O$ at $\p$,
48i.e., the domain obtained by inverting all elements not in $\p$.
49The valuation on $\F[[q]]$ induces a valuation $v_\p$ on $M_k(\Gamma,\O_\p)$ via
50the natural map $M_k(\Gamma,\O_\p)\ra \F[[q]]$.
51Thus $$v_\p(f) = \min\{ n \,:\, a_n(f) \not\con 0 \!\pmod{\p}\}.$$
52
53\begin{theorem}\label{theorem}
54If $f\in M_k(\Gamma,\O_{\p})$ and
55$\displaystyle v_\p(f) > \frac{\mu{}k}{12},$
56then
57$f\con 0\mod{\p}$.
58\end{theorem}
59We recall some notation and a lemma before giving the proof.
60Let
61\begin{eqnarray*}
62\Delta(q) &=& q - 24q^2 + 252q^3 - 1472q^4 +  4830q^5 + \cdots \in S_{12}(\sltwoz)\\
63      j(q)&=& \frac{1}{q} + 744 + 196884q + 21493760q^2+864299970q^3 + \cdots \in K(X(1))
64\end{eqnarray*}
65Note that $j(q)$ is an automorphic function for $\sltwoz$ of weight $0$
66whose only pole is at infinity (i.e., $j|[\gamma]_0=j$ for all $\gamma\in\sltwoz$).
67The Fourier coefficients of both $\Delta$
68and $j$ lie in $\Z$.
69
70For any positive integer $n$, let $\zeta_n=e^{2\pi i/n}$ be a primitive
71$n$th root of unity.
72
73\begin{lemma}\label{lemma}
74Let $N$ be a positive integer.
75Then any $\gamma\in\sltwoz$ induces a linear isomorphism
76$$[\gamma]_k: M_k(\Gamma(N))\ra M_k(\Gamma(N))$$
77which preserves $M_k(\Gamma(N),\Q(\zeta))$.
78\end{lemma}
79\begin{proof}
80That $[\gamma]_k$ leaves $M_k(\Gamma(N))$ invariant
81follows because $\Gamma(N)$ is a normal subgroup of $\sltwoz$:
82if $\alp\in\Gamma(N)$ then there is $\alp'\in\Gamma(N)$
83such that $\gamma\alp=\alp'\gamma$, so for any
84$f\in M_k(\Gamma(N))$,
85$$(f|[\gamma]_k)|[\alp]_k = f|[\gamma\alp]_k 86 = f|[\alp'\gamma]_k 87 = (f|[\alp']_k)|[\gamma]_k 88 = f|[\gamma]_k.$$
89Furthermore, $[\gamma]_k$ is an isomorphism because
90$[\gamma^{-1}]_k$ is its inverse.
91The second part follows because $X(N)$ has
92a model over $\Q(\zeta_N)$ and $\sltwo(\Z/N\Z)$
93acts via automorphisms of the function field of $X(N)$.
94The action of $[\gamma]_k$ is induced by this action.
95This is discussed in Chapter 6 of \cite{shimura}.
96\end{proof}
97
98\begin{proof}[Proof of Theorem~\ref{theorem}]
99We assume that $f\neq 0$, otherwise we are done.
100In each case we will give the argument for $\O$ and then
101note that it works for $\O_{\p}$ as well.
102
103{\em Case one: $\Gamma=\sltwoz$.}
104In the ring $\O\{\{q\}\}$ of Laurant series,
105$$\frac{f^{12}}{\Delta^k} = 106 f^{12} \cdot \left(\frac{1}{q}+24+324q+\cdots\right)^k = \sum_{n\geq -k} b_n q^n\in\O\{\{q\}\}.$$
107Since
108$$v_\p\left(\frac{f^{12}}{\Delta^k}\right) 109 = 12v_\p(f) - kv_\p(\Delta) = 12 v_\p(f) - k > 0$$
110we see that $b_n\in\p$ for $n\leq 0$.
111By inductively subtracting off multiples
112of powers of $j(q)=\frac{1}{q}+744+\cdots$ from $\frac{f^{12}}{\Delta^k}$
113(reducing the order of the pole by at least one at each step) we
114find a polynomial $R(j)$ in $\O[j]$ such that
115$$\frac{f^{12}}{\Delta^k} - R(j) \in q\O[[q]].$$
116But this difference is a holomorphic function on the projective plane,
117so it must be a constant, hence $0$, so that
118$f^{12}=R(j)\Delta^k.$
119Each of the coefficients of $R(j)$ must lie in $\p$,
120otherwise $v_\p(R(j))\leq 0$ which would be a contradiction
121because then $v_\p(\frac{f^{12}}{\Delta^k} - R(j)) \leq 0$.
122Thus $f^{12} = R(j) \Delta^k \con 0\pmod{\p}$ and the same
123holds for $f$ since $(\O/\p)[[q]]$ is a domain.
124
125Note that the same argument works with $\O$ replaced by
126its localization at $\p$.  Thus we can allow denominators
127in the $q$-expansion of $f$ as long as they do not lie
128in $\p$.
129
130{\em Case two: $\Gamma$ arbitrary.}
131For any $\gamma\in\sltwoz$, Lemma~\ref{lemma} implies that
132$f|[\gamma]_k \in M_k(\Gamma(N),K(\zeta_n))$.
133Let $\gP|\p$ be a prime ideal of the ring of integers of $K(\zeta_n)$.
134Let $\pi$ be a uniformizer for $\gP$.
135For any $\gamma$, there is a power $\pi^n$ of $\pi$
136so that the coefficients of
137the $q$-expansion of $\pi^n f|[\gamma]_k$ are $\gP$-integral
138and  $\pi^n f|[\gamma]_k\not\con 0\pmod{\gP}$.
139To do this choose $n$ minimally (possibly negative!)
140so that $\pi^n f|[\gamma]_k$ is $\gP$-integral.
141If $\pi^n f|[\gamma]_k\con 0\pmod{\gP}$
142then all coefficients are divisible by $\pi$ so $n$ was not chosen
143minimally.
144
145Write
146$$\sltwoz = \bigcup_{i=1}^{\mu} \Gamma \gamma_i$$
147with $\gamma_1=\abcd{1}{0}{0}{1}$, and let the $n_i$
148be chosen for $\gamma_i$ as in the previous paragraph.
149Then
150$$F = f \cdot \prod_{i=2}^{\mu} \pi^{n_i} f|[\gamma_i]_k\in M_{k\mu}(\sltwoz)$$
151and
152$$v_{\gP}(F)\geq v_{\gP}(f) = v_{\p}(f) > \frac{\mu}{12}k.$$
153By case 1, $F\con 0\pmod{\gP}$, so since
154$\pi^{n_i} f|[\gamma_i]_k\not\con 0\pmod{\gP}$
155for each $i\geq 2$, it follows that $f\con 0\pmod{\p}$ as claimed.
156\end{proof}
157
158\begin{remark}
159For cusp forms the proof of Theorem 1 yields the following {\em slightly}
160improved bound:
161If $f \in S_k(\Gamma,\O_\p)$ and
162$$v_\p(f) > \frac{\mu}{12}k - \frac{\mu-1}{N}$$
163then $f\con 0\pmod{\p}$.
164
165\begin{proof}
166Consider the proof of case two.
167Note that the $q$-expansion of an element of $M_k(\Gamma(N))$
168is in powers of $q^{1/N}$.
169Since $f$ is a cusp form, we know that
170$v_{\gP}(f|[\gamma_i]_k)\geq \frac{1}{N}$ for each $i$.
171If we insure that
172$$v_\p(f) > \frac{\mu}{12}k - \frac{\mu-1}{N}$$
173then
174 $$v_{\gP}(F)\geq v_{\gP}(f) + \frac{\mu-1}{N} = v_{\p}(f) + \frac{\mu - 1}{N} > \frac{\mu}{12}k$$
175and the proof goes through.
176\end{proof}
177\end{remark}
178
179
180\section{Applications}
181Let $\Gamma$ be a congruence subgroup of level $N$ and index $\mu$, let
182$k$ be a positive integer, and set $r:=\frac{k}{12}\mu.$
183\begin{proposition}\label{determine}
184Suppose
185$$f=\sum_{\substack{n>r\\n\in \frac{1}{N}\Z}} a_n q^n\in M_k(\Gamma)$$
186then $f=0$.  Stated differently, an element of $M_k(\Gamma)$ is
187determined by its Fourier coefficients $a_n$ for $n\leq r$.
188\end{proposition}
189\begin{proof}
190Let $s=q^{1/N}$.
191We must show that the composite map
192$$M_k(\Gamma)\hookrightarrow\C[[s]]\into\C[[s]]/(sq^r)$$
193is injective.
194Let $K$ be a number field with ring of integers $\O$
195such that $M_k(\Gamma) = M_k(\Gamma,\O)\tensor\C$.
196Because $\C$ is flat over $\O$, it suffices
197to show that the map $\Phi:M_k(N;\O)\into\O[[s]]/(sq^r)$ is injective.
198(We are just using that a linear map which is injective over $\O$ induces
199an injective map over $\C$.)
200Suppose $g=\sum b_n q^n \in M_k(N;\O)$ and
201$\Phi(g)=0$ so that $b_n=0$ for $n\leq r$.
202Thus if $\p$ be a prime of $\O$,
203$b_n\con 0\pmod{\p}$ for any $n\leq r$.
204By Theorem~\ref{theorem}, it follows that $g\con 0\pmod{\p}$.
205Repeating this argument for all $\p$ of $\O$ shows that
206the coefficients of $g$ are divisible by all primes $\p$,
207i.e., they are $0$.
208\end{proof}
209
210Now suppose $\Gamma=\Gamma_0(N)$ or $\Gamma_1(N)$ so that
211we have a Hecke algebra $\T$ which acts as a commutative
212ring of  linear endomorphism on $M_k(\Gamma)$.
213\begin{theorem}\label{bound}
214The Hecke algebra is generated as a $\Z$-module
215by $T_1,\ldots,T_r.$
216\end{theorem}
217For the proof, see \cite{stein}.
218
219\begin{thebibliography}{HHHHHHH}
220
221\bibitem[Shimura]{shimura} G. Shimura, {\em Introduction to the Arithmetic
222Theory of Automorphic Functions}, Princeton University Press, 1994
223
224\bibitem[Stein]{stein} W. Stein, {\em Generating the Hecke algebra as a $\Z$-module}
225
226\bibitem[Sturm]{sturm} J. Sturm, {\em On the Congruence of Modular Forms}.
227Number theory (New York, 1984--1985), 275--280, Lecture Notes in
228Math., 1240, Springer, Berlin-New York, 1987.
229\end{thebibliography} \normalsize\vspace*{1 cm}
230\end{document}
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