\documentclass[12pt]{book} \textwidth=1.2\textwidth \textheight=1.3\textheight \hoffset=-.7in \voffset=-1.28in \usepackage{amsfonts} \usepackage{amsmath} \usepackage{amsopn} \usepackage{amsthm} \usepackage{amscd} \usepackage{makeidx} % exec the command "makeindex ribetnote.idx" \makeindex \author{William Stein} \title{Hecke Algebras and Modular Forms:\\ Notes derived from Ribet's 1996 Berkeley grad. course.} \font\bbb=msbm10 scaled \magstep 1 \font\bbbs=msbm10 \font\german=eufm10 scaled \magstep 1 \font\script=rsfs10 scaled \magstep 1 \newcommand{\rhouniv}{\rho_{\text{univ}}} \newcommand{\Set}{\mathbf{Set}} \newcommand{\pr}{pr} \newcommand{\ns}{N_{\Sigma}} \newcommand{\rs}{R_{\Sigma}} \newcommand{\ts}{\T_{\Sigma}} \newcommand{\wpt}{\wp_T} \newcommand{\wpr}{\wp_R} \newcommand{\nd}{\!\!\not|} \newcommand{\dbd}[1]{\langle#1\rangle} % make a diamond bracket d symbol %%%% Draw a correspondence diagram. \newcommand{\corr}[5]{\begin{array} {ccc}\\\stackrel{#4}{\swarrow}&&\stackrel{#5}{\searrow} \\#2&\end{array}} \newcommand{\corrs}{\rightarrow\rightarrow} % a correspondence, %% This should be a squiggly arrow, %% but that isn't supported on some %% installations. %%%% Draw a triangular commutative diagram \newcommand{\triCD}[6]{\begin{array}{ccc} #1&\xrightarrow{#2}\\ \searrow&\downarrow #4\\ & & #5\end{array}} \newcommand{\ltriCD}[6]{\begin{array}{ccc} #1 \\ #6\downarrow&\searrow\\ #5&\xrightarrow{#4} \end{array}} \newcommand{\onto}{\longrightarrow\!\!\!\!\rightarrow} \newcommand{\mapdown}[1]{\Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \newcommand{\presup}[1]{{}^{#1}\!} %\newcommand{\bZ}{\mbox{\bbb Z}} %\newcommand{\bF}{\mbox{\bbb F}} %\newcommand{\bA}{\mbox{\bbb A}} %\newcommand{\bQ}{\mbox{\bbb Q}} %\newcommand{\bP}{\mbox{\bbb P}} %\newcommand{\bR}{\mbox{\bbb R}} %\newcommand{\bC}{\mbox{\bbb C}} %\newcommand{\bsC}{\mbox{\bbbs C}} %\newcommand{\bsQ}{\mbox{\bbbs Q}} %\newcommand{\bsZ}{\mbox{\bbbs Z}} 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\newcommand{\rholamf}{\rho_{f,\lambda}} \newcommand{\rholamfbar}{\overline{\rho}_{\lambda,f}} \newcommand{\rhom}{\rho_{\m}} \newcommand{\rholambdabar}{\overline{\rho}_{\lambda}} \newcommand{\pico}{\pic^0} \newcommand{\fln}{\F_{\ell^{\nu}}} \newcommand{\flnbar}{\overline{\fnl}} \newcommand{\fl}{\F_{\ell}} \newcommand{\galql}{\Gal(\Qlbar/\Ql)} \newcommand{\sofl}{\sO_{f,\lambda}} \DeclareMathOperator{\Val}{Val} \DeclareMathOperator{\new}{new} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\WD}{WD} \DeclareMathOperator{\Char}{char} \DeclareMathOperator{\cond}{cond} \DeclareMathOperator{\Alb}{Alb} \DeclareMathOperator{\Tan}{Tan} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\sl2}{SL_2} \DeclareMathOperator{\sl2z}{SL_2(\bZ)} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\gl}{GL} \DeclareMathOperator{\tr}{Tr} \DeclareMathOperator{\trace}{trace} \DeclareMathOperator{\imag}{Im} % image or imaginary part. cute, eh? \DeclareMathOperator{\real}{Re} % image or imaginary part. cute, eh? \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Div}{Div} \DeclareMathOperator{\Mor}{Mor} \DeclareMathOperator{\Cot}{Cot} \DeclareMathOperator{\spec}{spec} \DeclareMathOperator{\Sym}{Sym} \DeclareMathOperator{\frob}{frob} \DeclareMathOperator{\Frob}{Frob} \DeclareMathOperator{\ver}{ver} \DeclareMathOperator{\Ver}{Ver} \DeclareMathOperator{\tors}{tors} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\Tate}{Tate} \DeclareMathOperator{\pic}{Pic} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\PS}{PS} \DeclareMathOperator{\ab}{ab} \DeclareMathOperator{\st}{st} \DeclareMathOperator{\Ind}{Ind} \DeclareMathOperator{\Ann}{Ann} \newcommand{\aut}{\Aut} %%%% Theoremstyles \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{prop}[thm]{Proposition} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{question}[thm]{Question} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \newtheorem{conj}[thm]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[thm]{Remark} \newtheorem{example}[thm]{Example} \newtheorem{exercise}[thm]{Exercise} \begin{document} \frontmatter \maketitle \chapter{Preface} {\bfseries Disclaimer: } These notes record some of what I saw in Ken Ribet's course on Modular Forms and Hecke Operators given at U.C. Berkeley during the Spring semester 1996. They are still {\em very rough} as I wrote them during my first semester of graduate school before I knew any real mathematics. The participants in the course were: Amod Agashe, Matt Baker, Jim Borger, Kevin Buzzard, Bruce Caskel, Robert Coleman, Jan\'{o}s Csirik, Annette Huber, David Jones, David Kohel, Loic Merel, David Moulton, Andrew Ogg, Arthur Ogus, Jessica Polito, Ken Ribet, Saul Schleimer, Lawren Smithline, William Stein, Takahashi, Wayne Whitney, and Hui Zui. I wish to thank David Moulton, Joe Wetherall, and Kevin Buzzard who helped me in preparing these notes, Arthur Ogus who asked a lot of stimulating questions during the class, and of course Ken Ribet who sees clearly. William Stein, Spring 1996, Berkeley, CA, {\tt was@math.berkeley.edu} \tableofcontents %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mainmatter %%%%%%%%%%%%%%%%%%%%%%%% %% Lecture 1, January 17 \chapter{Introduction} The main objects of study in this course are: \begin{itemize} \item Modular Forms \item Hecke Algebras \item Modular Curves\index{modular curves} \item Jacobians\index{Jacobian} \item Abelian Varieties \end{itemize} \section{Two Dimensional Galois Representations} \index{Galois representations} The geometric objects of study are elliptic curves and more generally algebraic curves of arbitrary genus. These in turn give rise via the Jacobian construction to higher dimensional abelian varieties. These geometric objects in turn give rise to Galois representations. When studying elliptic curves, the natural tool in the characteristic zero situation is to present the elliptic curve as $\bC/\sL$ for some lattice $\sL$ in $\bC$. To construct $\sL$ fix a non-zero holomorphic differential $\omega$ of $E$ over $\bC$ and construct $\sL$ as \begin{equation*} \bigg\{\int_{\gamma} \omega \quad \bigg| \quad \gamma\in H_1(E(\bC),\bZ)\bigg\}. \end{equation*} \subsection{Finite Fields (Weil, Tate)} In the 1940's, Weil study the analogous situation for elliptic curves defined over a finite field $k$. He desperately wanted to find an algebraic way to describe the above correspondence. He was able to find an algebraic definition of $\sL/n\sL$, where $n\geq 1$ and $(n,\Char k)=1$, which is as follows. Let $E[n]=\{P\in E(\overline{k}) : nP = 0\} = (\frac{1}{n} \sL) / \sL \isom \sL / n \sL$. Now fix a prime $\ell$, we let $E[\ell^\infty]=\{P\in E(\overline{k}) : \ell^{\nu}P = 0, \text{ some } \nu \geq 1\} = \cup_{\nu=1}^{\infty} E[\ell^{\nu}]$. Tate obtained an analogous construction by defining a rank 2 free $\Zl$-module $\Tatel E:=\varprojlim E[\ell^{\nu}]$ (the map from $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$ is multiplication by $\ell$). To see that the rank is $2$, check that the $\bZ/\ell^{\nu}\bZ$-module structure of $E[\ell^{\nu}]$ is compatible with the maps $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$). See \cite{silverman1} (III, 7). Then $V_{\ell}(E)=T_{\ell}(E)\tensor\bQ_{\ell}$ is a two dimensional vector space over $\bQ_{\ell}$. This gives the first nontrivial example of $\ell$-adic \'{e}tale cohomology. \subsection{Galois Representations (Taniyama, Shimura, Mumford-Tate)} Let $E/\bQ$ be an elliptic curve and $G=\gal(\overline{\bQ}/\bQ)$. Then $E[n]=\{P\in E(\overline{\bQ}) : nP = 0\} \isom (\bZ / n\bZ)^2$ is acted on by $G$ and this action respects the group operation so we have a Galois representation \begin{equation*} G\xrightarrow{\rho}\aut(E[n])\isom \gl_2(\bZ/n\bZ) \end{equation*} Let $K$ be the fixed field of $\ker \rho$ (note that $K$ is a number field), then since $\gal(K/\bQ)\isom G/\ker\rho \isom \imag \rho \subseteq \gl_2(\bZ/n\bZ)$ we obtain many subgroups of $\gl_2(\bZ/n\bZ)$ as Galois groups. Shimura\index{Shimura} showed that if we start with the elliptic curve \begin{equation*} E: \quad y^2+y = x^3-x^2 \end{equation*} then the image of $\rho$ is often all of $\gl_2(\bZ/n\bZ)$ and the image is ``most'' of $\gl_2(\bZ/n\bZ)$ when $E$ does not have complex multiplication. \section{Modular Forms and Galois Representations} \subsection{Cusp Forms} \index{cusp forms} Let $S_k(N)$ denote the space of cusp forms of weight $k$ and level $N$ on the congruence subgroup\index{congruence subgroup} $\Gamma_1(N)=\bigl\{\bigl(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\bigr) \in \sl2z : a\equiv 1 \pmod N, c\equiv 0 \pmod N, d\equiv 1 \pmod N \bigr\}$. Thus $S_k(N)$ is the finite dimensional vector space consisting of all holomorphic functions $f(z)$ on $\sH=\{z\in\bC : \imag(z)>0\}$ vanishing at $\infty$ and satisfying \begin{equation*} f(\frac{az+b}{cz+d})=(cz+d)^k f(z) \quad\text{for all} \quad \bigl ( \begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr) \in \Gamma_1(N). \end{equation*} Since, in particular, $f(z)=f(z+1)$ we can expand $f(z)$ as a $q$-series (this requires rigorous justification) \begin{equation*} f(z) = \sum_{n=1}^{\infty} c_n q^n. \end{equation*} A famous example is \begin{equation*} \Delta = q\prod_{n=1}^{\infty}(1-q^n)^{24} = \sum_{n=1}^{\infty} \tau(n) q^n \end{equation*} $\tau$ is called the Ramanujan function. One now knows that $\tau$ is multiplicative and satisfies $\tau(p^{\nu})=\tau(p)\tau(p^{\nu})-p^{11}\tau(p^{\nu-1})$. $\Delta$ is a normalized basis for $S_1(1)$. \subsection{Hecke Operators (Mordell)} Mordell defined, for $n\geq 1$, operators $T_n$ on $S_k(N)$ called {\em Hecke operators}\index{Hecke operator}. These proved very fruitful. The set of such operators forms an ``almost'' commuting family of endomorphisms and is hence ``almost'' simultaneously diagonalizable. The precise meaning of ``almost'' and the actual structure of the Hecke algebra $\bQ[T_1,\ldots,T_n,\ldots]$ will be studied in greater detail in the remainder of this course. Often there will exist a basis of cusp forms $f = \sum_{n=1}^{\infty} c_n q^n \in S_k(N)$ so that $f_n$ is a simultaneous eigenvector for all of the Hecke operators $T_n$ and, in fact, $T_n f = c_n f$. All of the $c_n$ will be algebraic integers and the field $\bQ(c_1,c_2,\ldots)$ will be finite over $\bQ$. A good claim can be made that the $c_n$ are often interesting integers because they exhibit remarkable properties. For example, $\tau(n) \equiv \sum_{d|n}d^{11} \pmod {691}$. How can we study the $c_n$? How can we interpret the $c_n$? We can do this by studying the connection between Galois representations and modular forms. In 1968 work was originally begun on this by Serre\index{Serre}, Shimura, Eichler and Deligne. \index{Shimura}\index{Eichler}\index{Deligne} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Lecture 2, 1/19/96 \chapter{Modular Representations and Curves} \section{Arithmetic of Modular Forms} Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is an eigenform for the Hecke operators. Then the Mellin transform associates to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} \frac{a_n}{n^s}$. Let $K=\bQ(a_1,a_2,\ldots)$, then one can show that the $a_n$ are algebraic integers and $K$ is a number field. When $k=2$ Shimura\index{Shimura} associates to $f$ an abelian variety $A_f$ over $\bQ$ of dimension $[K:\bQ]$ on which $K$ acts (see theorem 7.14 of \cite{shimura1}). \begin{example}[Modular Elliptic Curves] When all of the coefficients $a_n$ of the modular form $f$ lie in $\bQ$ then $[K:\bQ]=1$ so $A_f$ is a one dimensional abelian variety. A one dimensional abelian variety of nonzero genus is an elliptic curve. An elliptic curve isogenous to one arising via this construction is called {\em modular}. \end{example} \begin{defn} Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is a morphism $E_1\into E_2$ of algebraic groups, which has a finite kernel. \end{defn} The following conjecture motivates much of the theory. \begin{conj} Every elliptic curve over $\bQ$ is modular, that is, isogenous to a curve constructed in the above way. \end{conj} For $k\geq 2$ Serre\index{Serre} and Deligne found a way to associate to $f$ a family of $\ell$-adic representations. Let $\ell$ be a prime number and $K$ be as above, then it is well known that $K\tensor_{\bQ} \bQ_{\ell}\isom \prod_{\lambda|\ell}K_{\lambda}$. One can associate to $f$ a representation \begin{equation*} \rholf:G=\gal(\overline{\bQ}/\bQ) \rightarrow\gl(K\tensor_{\bQ}\bQ_{\ell}) \end{equation*} unramified at all primes $p\nd \ell N$. For $\rholf$ to be unramified we mean that for all primes $P$ lying over $p$, the inertia group of the decomposition group at $P$ is contained in the kernel of $\rholf$. The decomposition group $D_P$ at $P$ is the set of those $g\in G$ which fix $P$. Let $k$ be the residue field $\sO/P$ where $\sO$ is the ring of all algebraic integers. Then the inertia group $I_P$ is the kernel of the map $D_P\rightarrow \gal(\overline{k}/k)$. Now $I_P\subset D_P \subset \gal(\overline{\bQ}/\bQ)$ and $D_P / I_P$ is cyclic (being isomorphic to a subgroup of the Galois group of a finite extension of finite fields) so it is generated by a Frobenious automorphism $\frob_p$ lying over $p$. One has \begin{align*} \tr(\rholf(\frob_p))& = a_p\in K \subset K\tensor \bQ_{\ell}\\ &\text{and}\\ \det(\rholf) &= \chi_{\ell}^{k-1}\varepsilon \end{align*} where $\chi_{\ell}$ is the $\ell$th cyclotomic character and $\varepsilon$ is the Dirichlet character associated to $f$. There is an incredible amount of ``abuse of notation'' packed into this statement. First, the Frobenius $\frob_P$ (note $P$ not $p$) is only well defined in $\gal(K/\bQ)$ (so I think an unstated result is that $K$ must be Galois), and then $\frob_p$ is only well defined up to conjugacy. But this works out since $\rholf$ is well-defined on $\gal(K/\bQ)$ (it kills $\gal(\overline{\bQ}/K)$) and the trace is well-defined on conjugacy classes ($\tr(AB)=\tr(BA)$ so $\tr(ABA^{-1})=Tr(B)$). \section{Characters} Let $f\in S_k(N)$, then for all $\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr) \in \sl2z$ with $c\equiv 0 \mod{N}$ we have \begin{equation*} f(\frac{az+b}{cz+d}) = (cz+d)^k \varepsilon(d) f(z) \end{equation*} where $\varepsilon:(\bZ/N\bZ)^*\rightarrow \bC^*$ is a Dirichlet character mod $N$. If $f$ is an eigenform for the so called ``diamond-bracket operator'' $\dbd{d}$ so that $f|\dbd{d} = \varepsilon(d) f$ then $\varepsilon$ actually takes values in $K$. Led $\varphi_N$ be the mod $N$ cyclotomic character so that $\varphi_N: G \rightarrow (\bZ/N\bZ)^*$ takes $g\in G$ to the automorphism induced by $g$ on the $N$th cyclotomic extension $\bQ(\Mu_N)$ of $\bQ$ (where we identify $\gal(\bQ(\Mu_N)/\bQ)$ with $(\bZ/N\bZ)^*$). Then what we called $\varepsilon$ above in the formula $\det(\rho_{\ell})=\chi_{\ell}^{k-1}\varepsilon$ is really the composition \begin{equation*} G\xrightarrow{\varphi_N}(\bZ/N\bZ)^*\xrightarrow{\varepsilon} \bC^*. \end{equation*} For each positive integer $\nu$ we consider the $\ell^{\nu}$th cyclotomic character on $G$, \begin{equation*} \varphi_{\ell^{\nu}}:G\rightarrow (\bZ/\ell^{\nu}\bZ)^*. \end{equation*} Putting these together gives the $\ell$-adic cyclotomic character $$\chi_{\ell}:G\into\bZ_{\ell}^{*}.$$ \section{Parity Conditions} Let $c\in\gal(\overline{\bQ}/\bQ)$ be complex conjugation. Then $\varphi_N(c)=-1$ so $\varepsilon(c) = \varepsilon(-1)$ and $\chi_{\ell}^{k-1}(c) = (-1)^{k-1}$. Now let $\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} -1&0\\0&-1\end{smallmatrix}\bigr)$, then for $f\in S_k(N)$, $$f(z) = (-1)^k\varepsilon(-1)f(z)$$ so $(-1)^k\varepsilon(-1) = 1$ thus $$\det(\rholf(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$ Thus the $\det$ character is odd so the representation $\rholf$ is odd. \begin{remark}[Vague Question] How can one recognize representations like $\rholf$ ``in nature''? Mazur\index{Mazur} and Fontaine have made relevant conjectures. The Shimura-Taniyama conjecture can be reformulated by saying that for any representation $\rho_{\ell,E}$ comming from an elliptic curve $E$ there is $f$ so that $\rho_{\ell,E}\isom \rholf$. \end{remark} \section{Conjectures of Serre (mod $\ell$ version)} \index{Serre} Suppose $f$ is a modular form, $\ell\in\Z$ prime, $\lambda$ a prime lying over $\ell$, and the representation $$\rho_{\lambda,f}:G\rightarrow \gl_2(K_{\lambda})$$ (constructed by Serre-Deligne) is irreducible. Then $\rho_{\lambda,f}$ is conjugate to a representation with image in $\gl_2(\sO_{\lambda})$, where $\sO_{\lambda}$ is the ring of integers of $K_{\lambda}$. Reducing mod $\lambda$ gives a representation $$\overline{\rho}_{\lambda,f}:G\rightarrow\gl_2(\bF_{\lambda})$$ which has a well-defined trace and det, i.e., the det and trace don't depend on the choice of conjugate representation used to obtain the reduced representation. One knows from representation theory that if such a representation is semisimple then it is completely determined by its trace and det (more precisely, the characteristic polynomials of all of its elements -- see chapter ??). Thus if $\overline{\rho}_{\lambda,f}$ is irreducible (and hence semisimple) then it is unique in the sense that it does not depend on the choice of conjugate. \section{General remarks on mod $p$ Galois representations} \index{Galois representations} %% By Joe Wetherall [[This section was written by Joseph Loebach Wetherell.]] First, what are semi-simple and irreducible representations? Remember that a representation $\rho$ is a map from a group $G$ to the endomorphisms of some vector space $W$ (or a free module $M$ if we are working over a ring instead of a field, but let's not worry about that for now). A subspace $W'$ of $W$ is said to be invariant under $\rho$ if $\rho$ takes $W'$ back into itself. (The point is that if $W'$ is invariant, then $\rho$ induces representations on both $W'$ and $W/W'$.) An irreducible representation is one where the only invariant subspaces are ${0}$ and $W$. A semi-simple representation is one where for every invariant subspace $W'$ there is a complementary invariant subspace $W''$ -- that is, you can write $\rho$ as the direct sum of $\rho|_{W'}$ and $\rho|_{W''}$. Another way to say this is that if $W'$ is an invariant subspace then we get a short exact sequence $$0\into\rho|_{W/W'}\into\rho\into\rho|_{W'}\into 0.$$ Furthermore $\rho$ is semi-simple if and only if every such sequence splits. Note that irreducible representations are semi-simple. One other fact is that semi-simple Galois representations are uniquely determined (up to isomorphism class) by their trace and determinant. Now, since in the case we are doing, $G = \galq$ is compact, it follows that the image of any Galois representation $\rho$ into $\gl_2(K_{\lambda})$ is compact. Thus we can conjugate it into $\gl_2(\sO_{\lambda})$. Irreducibility is not needed for this. Now that we have a representation into $\gl_2(\sO_{\lambda})$, we can reduce to get a representation $\overline{\rho}$ to $\gl_2(\bF_{\lambda})$. This reduced representation is not uniquely determined by $\rho$, since we had a choice of conjugators. However, the trace and determinant are invariant under conjugation, so the trace and determinant of the reduced representation are uniquely determined by $\rho$. So we know the trace and determinant of the reduced representation. If we also knew that it was semi-simple, then we would know its isomorphism class, and we would be done. So we would be happy if the reduced representation is irreducible. And in fact, it is easy to see that if the reduced representation is irreducible, then $\rho$ must also be irreducible. Now, it turns out that all $\rho$ of interest to us will be irreducible; unfortunately, we can't go the other way and claim that $\rho$ irreducible implies the reduction is irreducible. \section{Serre's Conjecture} \index{Serre} Serre has made the following conjecture which is still open at the time of this writing. \begin{conj}[Serre] All irreducible representation of $G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$ complex conjugation, are of the form $\overline{\rho}_{\lambda,f}$ for some representation $\rho_{\lambda,f}$ constructed as above. \end{conj} \begin{example} Let $E/\bQ$ be an elliptic curve and let $\sigma_{\ell}:G\rightarrow\gl_2(\bF_{\ell})$ be the representation induced by the action of $G$ on the $\ell$-torsion of $E$. Then $\det \sigma_{\ell} = \varphi_{\ell}$ is odd and $\sigma_{\ell}$ is usually irreducible, so Serre's conjecture\index{Serre's conjecture} would imply that $\sigma_{\ell}$ is modular. From this one can, assuming Serre's conjecture, prove that $E$ is modular. \end{example} \begin{defn} Let $\sigma:G\rightarrow \gl_2(\bF)$ ($\bF$ is a finite field) be a represenation of the Galois group $G$. The we say that the {\em representions $\sigma$ is modular} if there is a modular form $f$, a prime $\lambda$, and an embedding $\bF\hookrightarrow \overline{\bF}_{\lambda}$ such that $\sigma\isom\overline{\rho}_{\lambda,f}$ over $\overline{\bF}_\lambda$. \end{defn} \section{Wiles' Perspective} Suppose $E/\bQ$ is an elliptic curve and $\rho_{\ell,E}:G\rightarrow\gl_2(\bZ_{\ell})$ the associated $\ell$-adic representation on the Tate module $T_{\ell}$. Then by reducing we obtain a mod $\ell$ representation $$\overline{\rho}_{\ell,E}=\sigma_{\ell,E}:G \rightarrow \gl_2(\bF_{\ell}).$$ If we can show this representation is modular for infinitely many $\ell$ then we will know that $E$ is modular. \begin{thm}[Langland's and Tunnel] If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they are modular. \end{thm} This is proved by using the fact that $\gl_2(\bF_2)$ and $\gl_2(\bF_3)$ are solvable so we may apply ``base-change''. \begin{thm}[Wiles] If $\rho$ is an $\ell$-adic representation which is irreducible and modular mod $\ell$ with $\ell>2$ and certain other reasonable hypothesis are satisfied, then $\rho$ itself is modular. \end{thm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Lecture 3, 1/21/96 \chapter{Modular Forms} Our goal is to explain modular forms as functions of lattices or of elliptic curves. Good references are Serre \cite{serre2} \index{Serre} and Katz \cite{antwerp}. \section{Cusp Forms} First suppose $N=1$, then we must define $S_k=S_k(1)$. Let $\Gamma_1(1)=\sl2z$, then $S_k$ consists of all functions $f$ holomorphic on the upper half plane $\sH$ and such that for all $\abcd\in\sl2z$ one has $$f(\frac{a\tau+b}{c\tau+d})=(c\tau+d)^k f(\tau),$$ and $f$ vanishes at infinity. Thus, in particular, $f(\tau+1)=f(\tau)$ and so $f$ passes to a well defined function of $q=e^{2\pi i\tau}$. So $f(q)$ is a function on $\{z:0<|z|<1\}$ and the condition that $f(\tau)$ vanishes at infinity is that $f(q)$ extends to a holomorphic function on $\{z:|z|<1\}$ and $f(0)=0$. In this case, we may write $f(q)=\sum_{n=1}^{\infty}a_n q^n$. \section{Lattices} A lattice $L\subset \bC$ is a subring $L=\bZ\omega_1 +\bZ\omega_2$ for which $\omega_1, \omega_2\in \bC$ are lineary independent over $\bR$. Without loss, we may assume that $\omega_1 / \omega_2 \in \sH$. Let $$\sR=\{\text{lattices in $\bC$}\}=\{(E,\omega): \text{ $E$ is an elliptic curve, } \omega\in\Omega_E^{1}\}$$ and $$M=\{(\omega_1,\omega_2):\omega_1,\omega_2\in\bC, \imag(\omega_1/\omega_2)>0\}.$$ There is a left action of $\sl2z$ on $M$ $$\abcd:(\omega_1,\omega_2)\mapsto (a\omega_1+b\omega_2,c\omega_1+d\omega_2)$$ and $\sl2z\backslash M\isom \sR$. \section{Relationship With Elliptic Curves} There is a map $L\mapsto \bC/L$ from lattices to complex tori which, by Weierstrass theory, correspond to elliptic curves defined over $\bC$ along with a distinguished differential $\omega=dz$. % huh? Conversely, if $E/\bC$ is an elliptic curve, we can obtain the corresponding lattice by fixing a differential $\omega$ and taking the lattice to be the image of the map $$H_1(E(\bC),\bZ)\xrightarrow{\text{integration}}\bC$$ which takes $\gamma\in H_1$ to $\int_{\gamma}\omega\in\bC$. There is a map $M/\bC\into\sH$ defined by $(\omega_1,\omega_2) \mapsto \omega_1/\omega_2$. This gives an isomorphism $$\sR/\bC^*=(\sl2z\backslash M)/\bC^* \xrightarrow{ ~ } \sl2z\backslash\sH$$ and $$\sR/\bC^*=\{\text{ elliptic curves /$\bC$ (without differentials)}\}.$$ If $f:\sH\into\bC$ we define $F:M\into\bC$ by $F(\omega_1,\omega_2)=f(\omega_1/\omega_2)$. Suppose now that $F$ is a lattice function and sattisfies the homogeneity % (spelling!?) condition $F(\lambda L)=\lambda^{-k} F(L)$. Then \begin{align*} f(\frac{a\tau+b}{c\tau+d})&=F(\bZ\frac{a\tau+b}{c\tau+d}+\bZ)\\ &= F((c\tau+d)^{-1}(\bZ(a\tau+b)+\bZ(c\tau+d)))\\ &= (c\tau+d)^k F(\bZ(a\tau+b)+\bZ(c\tau+d))\\ &= (c\tau+d)^k F(\bZ+\tau\bZ)\\ &= (c\tau+d)^k f(\tau) \end{align*} so functions of lattices with the homogeneity condition come from functions $f\in M_k$. Thus, if $f\in M_k$ and $F$ is the corresponding lattice function then $$F(\bZ\omega_1+\bZ\omega_2)=F(\omega_2(\bZ+\bZ \frac{\omega_1}{\omega_2})) =\omega_2^{-k}F(\bZ+\bZ \frac{\omega_1}{\omega_2}) =\omega_2^{-k}f( \frac{\omega_1}{\omega_2}), $$ so we can recover $F$ from $f$. %% I don't understand the stuff with H^0(E,\Omega_E^{1}) being a %% 1 dimensional vector space... \section{Hecke Operators} Define a map $T_n$ from the free abelian group generated by all $\bC$-lattices into itself by $$T_n(L)= \sum_{(L:L')=n} L'.$$ Then if $F$ is a function on lattices define $T_nF$ by $$(T_nF)(L)=n^{k-1}\sum_{(L:L')=n}F(L').$$ Since $(n,m)=1$ implies $T_nT_m=T_{nm}$ and $T_{p^k}$ is a polynomial in $\bZ[T_p]$ the essential case to consider is $n$ prime. Suppose $L'\subset L$ with $(L:L')=n$, then $L/L'$ is killed by $n$ so $nL\subset L'\subset L$ and $$L'/nL\subset L/nL\isom (\bZ/n\bZ)^2.$$ Thus the subgroups of $(\bZ/n\bZ)^2$ of order $n$ correspond to the sublattices $L'$ of $L$ of index $n$. When $n=\ell$ is prime there are $\ell+1$ such subgroups. (The subgroups correspond to nonzero vectors in $\bF_{\ell}$ modulo scalar equivalence and there are $\frac{\ell^2-1}{\ell-1}$ of them.) Suppose $L'\subset L$ is a sublattice of index $\ell$ and let $L''=\ell^{-1}L'$. Note that $\ell L\subset L'$ so $L \subset \ell^{-1} L'=L''$ and $L$ is a sublattice of $L''$ of index $\ell$. Thus, assuming $F$ satisfies the homogeneity condition $F(\lambda L)=\lambda^{-k}F(L)$, $$\ell^{k-1}\sum_{L'}F(L') = \frac{1}{\ell}\sum_{L''}F(L'')$$ which helps explain the extra factor of $n^{k-1}$ in our definition of $T_n F$ -- we are ``averaging'' over the sublattices (note that there are $\ell+1$ terms yet we divide by $\ell$ so we aren't exactly averaging). We now give a geometric description of the $\ell$th Hecke operator\index{Hecke operator}. Let $L\subset L''$ be lattices with $(L'':L)=\ell$ and let $E=\bC/L$, $E''=\bC/L''$ be the elliptic curves corresponding to $L$, $L''$, respectively. Then $E[\ell]=\frac{1}{\ell}L/L$ contains $H=L''/L$ which may be thought of as a line [Ed: I don't know why!]. Then the Hecke operator is $$E\mapsto \frac{1}{\ell}\sum_{\text{lines }H\subset E[\ell]} E/H.$$ Let $\hat{\pi}$ be the isogeny dual to $\pi:E\into E/H$. Then in terms of pairs $(E,\omega)$ we have $$(E,\omega)\mapsto \frac{1}{\ell} \sum_{H\subset E[\ell], \#H=\ell}(E/H,\pi_{*}\omega) =\ell^{k-1}\sum_{H\subset E[\ell]} (E/H,\hat{\pi}^{*}(\omega)).$$ % Lecture 4, 1/24/96 We consider modular forms $f$ on $\Gamma_1(1)=\sl2z$, that is, holomorphic functions on $\sH\cup\{\infty\}$ which satisfy $$f(\tau)=f(\frac{a\tau+b}{c\tau+d})(c\tau+d)^{-k}$$ for all $\abcd\in\sl2z$. Using a Fourier expansion we write $$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$ and say $f$ is a cusp form if $a_0=0$. There is a correspondence between modular forms $f$ and lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$ given by $F(\bZ\tau+\bZ)=f(\tau)$. \section{Explicit Description of Sublattices} The $n$th Hecke operator $T_n$ of weight $k$ is defined by $$T_n(L)=n^{k-1}\sum_{\substack{L'\subset L\\(L:L')=n}} L'.$$ What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and $$L'/nL\subset L/nL=(\bZ/n\bZ)^2.$$ Write $L=\bZ\omega_1+\bZ\omega_2$, let $Y_2$ be the cyclic subgroup of $L/L'$ generated by $\omega_2$ and let $d=\#Y_2$. Let $Y_1=(L/L')/Y_2$, then $Y_1$ is generated by the image of $\omega_1$ so it is a cyclic group of order $a=n/d$. We want to exhibit a basis of $L'$. Let $\omega_2'=d\omega_2\in L'$ and use the fact that $Y_1$ is generated by $\omega_1$ to write $a\omega_1=\omega_1'+b\omega_2$ for some integer $b$ and some $\omega_1'\in L'$. Since $b$ is only well-defined modulo $d$ we may assume $0\leq b\leq d-1$. Thus $$ \Bigl(\begin{matrix}\omega_1'\\ \omega_2'\end{matrix}\Bigr) = \Bigl(\begin{matrix}a&b\\0&d\end{matrix}\Bigr) \Bigl(\begin{matrix}\omega_1\\ \omega_2\end{matrix}\Bigr) $$ and the change of basis matrix has determinent $ad=n$. Since $$\bZ\omega_1'+\bZ\omega_2'\subset L' \subset L=\bZ\omega_1+\bZ\omega_2$$ and $(L:\bZ\omega_1'+\bZ\omega_2')=n$ (since the change of basis matrix has determinent $n$) and $(L:L')=n$ we see that $L'=\bZ\omega_1'+\bZ\omega_2'$. Thus there is a one-to-one correspondence between sublattices $L'\subset L$ of index $n$ and matrices $\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr)$ with $ad=n$ and $0\leq b\leq d-1$. In particular, when $n=p$ is prime there $p+1$ of these. In general, the number of such sublattices equals the sum of the positive divisors of $n$. \section{Action of Hecke Operators on Modular Forms} Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular form with corresponding lattice function $F$. How can we describe the action of the Hecke operator $T_n$ on $f(\tau)$? We have \begin{align*} T_nF(\bZ\tau+\bZ) & = n^{k-1}\sum_{\substack{a,b,d\\ ab=n\\ 0\leq b<d}} F((a\tau+b)\bZ + d\bZ)\\ & = n^{k-1}\sum d^{-k} F(\frac{a\tau+b}{d}\bZ+\bZ)\\ & = n^{k-1}\sum d^{-k} f(\frac{a\tau+b}{d})\\ & = n^{k-1}\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\\ & = n^{k-1}\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}} \frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\\ & = n^{k-1}\sum_{\substack{ad=n\\m'\geq 0}}d^{1-k} c_{dm'}e^{2\pi i a m' \tau}\\ & = \sum_{\substack{ad=n\\m'\geq 0}} a^{k-1} c_{dm'}q^{am'}. \end{align*} In the second to the last expression we let $m=dm'$, $m'\geq 0$, then used the fact that the sum $\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$ is only nonzero if $d|m$. Thus $$T_nf(q)=\sum_{\substack{ad=n\\m\geq 0}} a^{k-1}c_{dm} q^{am}$$ and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is $$\sum_{\substack{a|n\\ a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$ \begin{remark} When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong to the $\bZ$-module generated by the $c_m$. \end{remark} \begin{remark} Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is $$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$ Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic since its original definition is as a finite sum of holomorphic functions.) \end{remark} \begin{remark} Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is $\sum_{a|1}1^{k-1}c_n=c_n$. As an immediate corollary we have the following important result. \end{remark} \begin{cor} Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient of $q$ for all $n\geq 1$, then $f=0$. \end{cor} \begin{remark} When $n=p$ is prime we get an interesting formula for the action of $T_p$ on the $q$-expansion of $f$. One has $$T_p f = \sum_{\mu\geq 0} \sum_{\substack{a|n\\a|\mu}}a^{k-1} c_{\frac{n\mu}{a^2}} q^{\mu}. $$ Since $n=p$ is prime either $a=1$ or $a=p$. When $a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$ and when $a=p$, we can write $\mu=p\lambda$ and we get terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$. Thus $$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+ p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$ \end{remark} %% Lecture 5, 1/26/96 \chapter{Embedding Hecke Operators in the Dual} \section{The Space of Modular Forms} Let $\Gamma=\Gamma_1(1)=\sl2z$ and for $k\geq 0$ let \begin{align*} M_k&=\{f=\sum_{n=0}^{\infty}a_n q^n : \text{$f$ is a modular form for $\Gamma$}\}\\ &\subset S_k=\{f=\sum_{n=1}^{\infty}a_n q^n\}\end{align*} These are finite dimensional $\bC$-vector spaces whose dimensions are easily computed. Furthermore, they are generated by familiar elements (see Serre \cite{serre2} or Lang \cite{lang1}.) \index{Serre} The main tool is the formula $$\sum_{p\in D\union\{\infty\}} \frac{1}{e(p)}\ord_p(f) = \frac{k}{12}$$ where $D$ is the fundamental domain for $\Gamma$ and $$e(p)=\begin{cases} 1&\text{otherwise}\\ 2&\text{if $p=i$}\\ 3&\text{if $p=\rho$} \end{cases}$$ One can alternatively define $e(p)$ as follows. If $p=\tau$ and $E=\bC/(\bZ\tau+\bZ)$ then $e(p)=\frac{1}{2}\#\aut(E)$. For $k\geq 4$ we define the {\em Eisenstein series}\index{Eisenstein series} $G_k$ by $$G_k(q)=\frac{1}{2}\zeta(1-k)+\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n,$$ then the map $$\tau\mapsto\sum_{\substack{(m,n)\neq(0,0)\\m,n\in\bZ}} \frac{1}{(m\tau+n)^k}$$ differs from $G_k$ by a constant (no proof). Also, $\zeta(1-k)\in\bQ$ and one may say, {\em symbolically} at least, ``$\displaystyle \zeta(1-k)=\sum_{d=1}^{\infty} d^{k-1} = \sigma_{k-1}(0)$.'' The {\em $n$th Bernoulli number $B_n$} is defined by the equation $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty} \frac{B_nx^n}{n!}.$$ One can show that $\zeta(1-k)=-\frac{B_k}{k}$ so the constant %% How?? Reference??? coefficient of $G_k$ is $-\frac{B_k}{2k}$ which is rational. \section{Inner Product}\index{inner product} In what follows we assume $k\geq 2$ to avoid trivialities.. The Hecke operators $T_n$ acts on the space $M_k$. Fix a subspace $V\subset M_k$ which is stable under the action of the $T_n$. Let $\bT(V)$ be the $\bC$-algebra generated by the endomorphism $T_n$ acting on $V$ and note that $\bT(V)$ is actually a finite dimensional $\bC$-vector space since it is a subspace of $End(V)$ and $V$ is finite dimensional. Recall that $\bT$ is commutative. There is a bilinear form \begin{align*} \bT\times V &\into \bC \\ \langle T,f\rangle & \mapsto a_1(f|T) \end{align*} where $f|T=\sum_{n=0}^{\infty}a_n(f|T)q^n$. We thus get maps \begin{align*} V\into\Hom(\bT,\bC)=\bT^{*}\\ \bT\into \Hom(V,\bC)=V^{*}. \end{align*} \begin{thm} The above maps are isomorphisms. \end{thm} \begin{proof} It just remains to show each map is injective. Then since a finite dimensional vector space and its dual have the same dimension the result follows. First suppose $f\mapsto 0\in\Hom(\bT,\bC)$, then $a_1(f|T)=0$ for all $T\in\bT$ so, in particular, $a_n=a_1(f|T_n)=0$ for all $n\geq 1$. Thus $f$ is a constant, but since $k\geq 2$ this implies $f=0$ (otherwise $f$ wouldn't transform correctly with respect to the action of the modular group). Next suppose $T\mapsto 0\in \Hom(V,\bC)$, then $a_1(f|T)=0$ for all $f\in V$. Substiting $f|T_n$ for $f$ and using the commutativity of $\bT$ we have \begin{align*} a_1((f|T_n)|T)&=0 && \text{for all $f$, $n\geq 1$}\\ a_1((f|T)|T_n)&=0 && \text{by commutativity}\\ a_n(f|T)&=0 && \text{$n\geq 1$}\\ f|T&=0 && \text{since $k\geq 2$, as above} \end{align*} Thus $T=0$ which completes the proof. \end{proof} \begin{remark} The above isomorphisms are {\em $\bT$-equivariant}. $\Hom(\bT,\bC)$ is a $\bT$-module if we let $T\in\bT$ act on $\varphi\in\Hom(\bT,\bC)$ by $(T\cdot\varphi)(T')=\varphi(TT')$. If $\alpha:V\into\Hom(\bT,\bC)$ is the above isomorphism (so $\alpha:f\mapsto\varphi_f:=(T'\mapsto a_1(f|T'))$) then equivariance is the statement that $\alpha(Tf)=T\alpha(f).$ This follows since \begin{align*} \alpha(Tf)(T')&=\varphi_{Tf}(T')=a_1(Tf|T')=a_1(f|T'T)\\ &=\varphi_{f}(T'T)=T\varphi(T')=T\alpha(f)(T'). \end{align*} \end{remark} \section{Eigenforms} \index{eigenforms} We continue to assume that $k\geq 2$. \begin{defn} A modular form $f\in M_k$ is an {\em eigenform for $\bT$} if $f|T_n=\lambda_n f$ for all $n\geq 1$ and some complex numbers $\lambda_n$. \end{defn} Let $f$ be an eigenform, then $a_n(f)=a_1(f|T_n)=\lambda_n a_1(f)$ so if $a_1(f)=0$ then $a_n(f)=0$ for all $n\geq 1$ so since $k\geq 2$ this would imply $f=0$. Thus $a_1(f)\neq 0$ and we may as well divide through by $a_1(f)$ to obtain the {\em normalized eigenform} $\frac{1}{a_1(f)}f$. We thus assume that $a_1(f)=1$, then the formula becomes $a_n(f)=\lambda_n$ and so $f|T_n = a_n(f) f$, for all $n\geq 1$. \begin{thm} Let $f\in V$ and let $\psi$ be the image of $f$ in $\Hom(\bT,\bC)$, thus $\psi(T)=a_1(f|T)$. Then $f$ is a normalized eigenform iff $\psi$ is a ring homomorphism. \end{thm} \begin{proof} First suppose $f$ is a normalized eigenform so $f|T_n=a_n(f)f$. Then \begin{align*} \psi(T_nT_m) &=a_1(f|T_nT_m)=a_m(f|T_n)\\ &=a_m(a_n(f)f)=a_m(f)a_n(f)\\ &=\psi(T_n)\psi(T_m), \end{align*} so $\psi$ is a homomorphism. Conversely, assume $\psi$ is a homomorphism. Then $f|T_n=\sum a_m(f|T_n)q^m$, so to show that $f|T_n=a_n(f)f$ we must show that $a_m(f|T_n)=a_n(f)a_m(f)$. Recall that %% remark 3.3 $\psi(T_n)=a_1(f|T_n)=a_n$, thus \begin{align*} a_n(f)a_m(f) &= a_1(f|T_n)a_1(f|T_m) = \psi(T_n)\psi(T_m) \\ & = \psi(T_n T_m) = a_1(f|T_n|T_m) \\ & = a_m(f| T_n) \end{align*} as desired. \end{proof} %% Lecture 6, 1/29/96 \chapter{Rationality and Integrality Questions} \section{Review} In the previous lecture we looked at subspaces $V \subset M_k \subset \bC[[q]]$, $(k\geq 4)$, and considered the space $\bT=\bT(V)=\bC[\ldots,T_n,\ldots]\subset\End_{\bsC}V$ of Hecke operators on $V$. We defined a pairing $\bT\cross V\into \bC$ by $(T,f)\mapsto a_1(f|T)$ and showed this pairing is nondegenerate and that it induces isomorphisms $\bT\isom\Hom(V,\bC)$ and $V\isom\Hom(\bT,\bC)$. \section{Integrality} Fix $k\geq 4$ and let $S=S_k$ be the space of weight $k$ cusp forms with respect to the action of $\sl2z$. Let \begin{align*} S(\bQ)& =S_k\intersect \bQ[[q]]\\ S(\bZ)& = S_k\intersect \bZ[[q]]. \end{align*} \begin{thm} There is a $\bC$-basis of $M_k$ consisting of forms with integral coefficients. \end{thm} \begin{proof} This is seen by exhibiting a basis. Recall that for all $k\geq 4$ $$G_k=-\frac{b_k}{2k}+\sum_{k=1}^{\infty}\sum_{d|k}d^{k-1}q^n$$ is the $k$th Eisenstein series\index{Eisenstein series} which is a modular form of weight $k$ and $$E_k=-\frac{2k}{b_k}\cdot G_k=1+\cdots$$ is its normalization. Since the Bernoulli numbers $b_2,\ldots,b_8$ have $1$ as numerator (this isn't always the case, $b_{10}=\frac{5}{66}$) we see that $E_4$ and $E_6$ have coefficients in $\bZ$ and constant term $1$. Furthermore one shows by dimension and independence arguments that the forms $$\{E_4^aE_6^b|4a+6b=k\}$$ form a basis for $M_k$. \end{proof} \section{Victor Miller's Thesis} Let $d=\dim_{\bsC}S_k$, then Victor Miller showed in his thesis (see \cite{lang2}, ch. X, theorem 4.4) that there exists $$f_1,\ldots,f_d\in S_k(\bZ) \quad\text{such that}\quad a_i(f_j)=\delta_{ij}$$ for $1\leq i,j\leq d$. The $f_i$ clearly form a basis. \begin{prop} Let $R=\bZ[\ldots,T_n,\ldots]\subset End(S_k)$, then $R=\bigoplus_{i=1}^{d} \bZ T_i$. \end{prop} \begin{proof} To see that $T_1,\cdots,T_d\in \bT=\bT(S_k)$ are linearly independent over $\bC$ suppose $\sum_{i=1}^{d} c_i T_i = 0$, then $$0=a_1(f_j|\sum c_i T_i)=\sum_{i}c_i a_i(f_j) = \sum_{i} c_i \delta_{ij} = c_j.$$ From the isomorphism $\bT\isom\Hom(S_k,\bC)$ we know that $\dim_{\bsC}\bT=d$, so we can write any $T_n$ as a $\bC$-linear combination $$T_n=\sum_{i=1}^{d}c_{n_i}T_i,\quad c_{n_i}\in\bC.$$ But $$\bZ\ni a_n(f_j)=a_1(f_j|T_n)=\sum_{i=1}^{d}c_{n_i}a_1(f_j|T_i) =\sum_{i=1}^{d}c_{n_i}a_i(f_j) = c_{n_j}$$ so the $c_{n_i}$ all lie in $\bZ$ which completes the proof. \end{proof} Thus $R$ is an integral Hecke algebra of finite rank $d$ over $\bZ$. We have a map \begin{align*} S(\bZ)\cross R & \into \bZ \\ (f,T) & \mapsto a_1(f|T) \end{align*} which induces an embedding $$S(\bZ)\hookrightarrow\Hom(R,\bZ)\isom \bZ^d.$$ \begin{exercise} Prove that the map $S(\bZ)\hookrightarrow\Hom(R,\bZ)$ is in fact an isomorphism of $\bT$-modules. [Hint: Show the cokernel is torsion free.] \end{exercise} \section{Petersson Inner Product}\index{inner product} The main theorem is \begin{thm} The $T_n\in\bT(S_k)$ are all diagonalizable over $\bC$. \end{thm} To prove this we note that $S_k$ supports a non-degenerate positive definite Hermitean inner product (the Petersson inner product) $$(f,g)\mapsto\langle f,g\rangle\in\bC$$ such that $\langle f|T_n,g\rangle =\langle f,g|T_n\rangle$. We need some background facts. \begin{defn} An operator $T$ is {\em normal} if it commutes with its adjoint, thus $TT^{*}=T^{*}T$. \end{defn} $T_n$ is clearly normal since $T_n^{*}=T_n$, \begin{thm} A normal operator is diagonalizable. \end{thm} Thus each $T_n$ is diagonalizable. \begin{thm} A commuting family of semisimple (=diagonalizable) operators can be simultaneously diagonalized. \end{thm} Since the $T_n$ commute this implies $S_k$ has a basis consisting of normalized eigenforms $f$. Their eigenvalues are real since \begin{align*} a_n(f)\langle f,f\rangle &=\langle a_n(f)f,f\rangle =\langle f|T_n,f\rangle\\ &=\langle f,a_n(f)f\rangle =\overline{a_n(f)}\langle f,f\rangle. \end{align*} \begin{exercise} The coefficients $a_n$ of the eigenforms are totally real algebraic integers. [Hint: The space $S_k$ is stable under the action of $\aut(\bC)$ on coefficients: if $f=\sum_{n=1}^{\infty}c_n q^n\in S_k$ and $\sigma\in\aut(\bC)$ then $\sigma(f)=\sum_{n=1}^{\infty}\sigma(c_n)q^n$ is again in $S_k$ (check this by writing $f$ in terms of a basis $f_1,\ldots,f_d\in S(\bZ)$). Next use the fact that $f$ is an eigenform iff $\sigma(f)$ is an eigenform.] %%% I have absolutely no idea how to do this!!! \end{exercise} Let $$\sH=\{x+iy : x, y\in \bR, \text{ and } y>0\}$$ be the upper half plane. Then the volume form $\frac{dx\wedge dy}{y^2}$ is invariant under the action of $$\gl_2^{+}(\bR)=\{M\in\gl_2(\bR) | \det(M)>0\}.$$ If $\alpha=\abcd\in\gl_2^{+}(\bR)$ then $\abcd$ acts on $\sH$ by $$\bigabcd:\quad z\mapsto\frac{az+b}{cz+d}$$ and one has $$\imag(\frac{az+b}{cz+d})=\frac{\det(\alpha)}{|cz+d|^2}y.$$ Differentiating $\frac{az+b}{cz+d}$ gives \begin{align*} d(\frac{az+b}{cz+d})&= \frac{a(cz+d)dz-c(az+b)dz}{(cz+d)^2}\\ &=\frac{(ad-bc)dz}{(cz+d)^2}\\ &= \frac{det(\alpha)}{(cz+d)^2}dz \end{align*} Thus, under the action of $\alpha$, $dz\wedge d{\overline z}$ takes on a factor of $$\frac{\det(\alpha)^2}{(cz+d)^2(c\overline{z}+d)^2} =\Bigl(\frac{\det(\alpha)}{|cz+d|^2}\Bigr)^2.$$ \begin{defn} The {\em Petersson inner product} of forms $f,g\in S_k$ is defined by $$<f,g>=\int_{\Gamma\backslash\sH}(f(z)\overline{g(z)}y^k) \frac{dx\wedge dy}{y^2},$$ where $\Gamma=\sl2z$. \end{defn} Integrating over $\Gamma\backslash\sH$ can be taken to mean integrating over a fundamental domain for the action of $\sH$. Showing that the operators $T_n$ are self-adjoint with respect to the Petersson inner product is a harder computation than Serre \cite{serre2} \index{Serre} might lead one to believe --- it takes a bit of thought. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Lecture 7, 1/31/96 \chapter{Modular Curves} \index{modular curves} % define various gammas \section{Cusp Forms} Recall that if $N$ is a positive integer we define the congruence subgroups $\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by \begin{align*} \Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\ \Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\ \Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv \bigl(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\bigr) \pmod{N}\}. \end{align*} Let $\Gamma$ be one of the above subgroups. One can give a construction of the space $S_k(\Gamma)$ of cusp forms of weight $k$ for the action of $\Gamma$ using the language of algebraic geometry. Let $X_{\Gamma}=\overline{\Gamma\backslash\H^{*}}$ be the compactifaction of the upper half plane (union the cusps) modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure of Riemann surface and $S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where $\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$. This works since an element of $H^0(X_{\Gamma},\Omega^1)$ is a differential form $f(z)dz$, holomorphic on $\H$ and the cusps, which is invariant with respect to the action of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then $$d(\gamma(z))/dz=(cz+d)^{-2}$$ so $$f(\gamma(z))d(\gamma(z))=f(z)dz$$ iff $f$ satisfies the modular condition $$f(\gamma(z))=(cz+d)^{2}f(z).$$ There is a similar construction of $S_k$ for $k>2$. \section{Modular Curves}\index{modular curves} One knows that $\sl2z\backslash\sH$ parameterizes isomorphism classes of elliptic curves. The other congruence subgroups also give rise to similar parameterizations. Thus $\Gamma_0(N)\backslash\sH$ parameterizes pairs $(E,C)$ where $E$ is an elliptic curve and $C$ is a cyclic subgroup of order $N$, and $\Gamma_1(N)\backslash\H$ parameterizes pairs $(E,P)$ where $E$ is an elliptic curve and $P$ is a point of exact order $N$. Note that one can also give a point of exact order $N$ by giving an injection $\bZ/N\bZ\hookrightarrow E[N]$ or equivalently an injection $\Mu_N\hookrightarrow E[N]$ where $\Mu_N$ denotes the $N$th roots of unity. $\Gamma(N)\backslash\sH$ parameterizes pairs $(E,\{\alpha,\beta\})$ where $\{\alpha,\beta\}$ is a basis for $E[N]\isom(\bZ/N\bZ)^2$. The above quotients spaces are called {\em moduli spaces} for the {\em moduli problem} of determining equivalence classes of pairs ($E + $ extra structure). \index{$\Gamma(N)$-structures} \section{Classifying $\Gamma(N)$-structures} \begin{defn} Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve} $E/S$ is a proper smooth curve $$\begin{CD} E \\ @VfVV \\ S\end{CD}$$ with geometrically connected fibers all of genus one, give with a section ``0''. \end{defn} Loosely speaking, proper is a generalization of projective and smooth generalizes nonsingularity. See Hartshorne \cite{hartshorne}, chapter III, section 10, for the precise definitions. \begin{defn} Let $S$ be any scheme and $E/S$ an elliptic curve. A {\bfseries $\Gamma(N)$-structure} on $E/S$ is a group homomorphism $$\varphi:(\bZ/N\bZ)^2\into E[N](S)$$ whose image ``generates'' $E[N](S)$. \end{defn} A good reference is chapter 3 of Katz and Mazur \cite{katzmazur}. Define a functor from the category of $\Q$-schemes to the category of sets by sending a scheme $S$ to the set of isomorphism classes of pairs $$(E, \Gamma(N)\text{-structure})$$\index{$\Gamma(N)$-structures} where $E$ is an elliptic curve defined over $S$ and isomorphisms (preserving the $\Gamma(N)$-structure) are taken over $S$. An isomorphism preserves the $\Gamma(N)$-structure if it takes the two distinguished generators to the two distinguished generators in the image (in the correct order). \begin{thm} For $N\geq 4$ the functor defined above is representable and the object representing it is the modular curve $X$ corresponding to $\Gamma(N)$. \end{thm} What this means is that given a $\Q$-scheme $S$, the set $X(S)=\Mor_{\Q\text{-schemes}}(S,X)$ is isomorphic to the image of the functor's value on $S$. There is a natural way to map a pair $(E,\Gamma(N)\text{-structure})$ \index{$\Gamma(N)$-structures} to an $N$th root of unity. If $P,Q$ are the distinguished basis of $E[N]$ we send the pair $(E,\Gamma(N)\text{-structure})$ to $$e_N(P,Q)\in\Mu_N$$ where $e_N:E[N]\cross E[N]\into \Mu_N$ is the Weil pairing. For the definition of this pairing see chapter III, section 8 of Silverman \cite{silverman1}. The Weil pairing is bilinear, alternating, non-degenerate, Galois invariant, and maps surjectively onto $\Mu_N$. %% 2/2/96 \section{More on Integral Hecke Operators} We are considering the algebra of integral Hecke operators $\T=\T_{\Z}$ on the space of cusp forms $S_k(\C)$ with respect to the action of the full modular group $\sl2z$. Our goal is to see why $\T\isom\Z^d$ where $d=\dim_{\bsC}S_k(\C)$. Suppose $A\subset\C$ is any {\em subring} of $\C$ and recall that $$\T_A=A[\ldots,T_n,\ldots]\subset \End_{\C}S_k.$$ We have a natural map $$\T_A\tensor_A\C\into\T_C$$ but we do not yet know that it is an isomorphism. \section{Complex Conjugation} We have a conjugate linear map on functions $$f(\tau)\mapsto \overline{f(-\overline{\tau})}.$$ Since $\overline{(e^{-2\pi i\overline{\tau}})}=e^{2\pi i \tau}$, it follows that $$\sum_{n=1}^{\infty} a_n q^n \mapsto \sum_{n=1}^{\infty}\overline{a_n}q^n$$ so it is reasonable to call this map ``complex conjugation''. Furthermore, if we know that $$S_k(\C)=\C\tensor_{\bsQ}S_k(\Q)$$ then it follows that complex conjugation takes $S_k(\C)$ into $S_k(\C)$. To see this note that if we have the above equality then every element of $S_k(\C)$ is a $\C$-linear combination of elements of $S_k(\Q)$ and conversely, and it is clear that the set of such $\C$-linear combinations is invariant under the action of complex conjugation. \section{Isomorphism in the Real Case} \begin{prop} $\T_{\R}\tensor_{\bsR}\C \isom \T_{\C}$, as $\C$-vector spaces. \end{prop} \begin{proof} Since $S_k(\R)=S_k(\C)\intersect\R[[q]]$ and since theorem 5.1 assures us that there is a $\C$-basis of $S_k(\C)$ consisting of forms with integral coefficients, we see that $S_k(\R)\isom \R^d$ where $d=\dim_{\bsC}S_k(\C)$. (Any element of $S_k(\R)$ is a $\C$-linear combination of the integral basis, hence equating real and imaginary parts, an $\R$-linear combination of the integral basis, and the integral basis stays independent over $\R$.) By considering the explicit formula for the action of the Hecke operators $T_n$ on $S_k$ (see section 3) we see that $\T_{\R}$ leaves $S_k(\R)$ invariant, thus $$\T_{\R}=\R[\ldots,T_d,\ldots]\subset \End_{\R}S_k(\R).$$ In section 4 we defined a ``perfect'' pairing $$\T_{\C}\times S_k(\C)\into \C$$ which allowed us to show that $\T_{\C}\isom S_k(\C).$ By restricting to $\R$ we again get a perfect pairing so we see that $\T_{\R}\isom S_k(\R) \isom {\R}^d$ which implies that $$\T_R\tensor_{\bsR}\C \xrightarrow{\sim}\T_{\C}.$$ \end{proof} This also shows that $S_k(\C)\isom \C\tensor_{\bsR} S_k(\R)$ so we have complex conjugation over $\R$. %% For some reason I feel something circular is going on here, but %% i can't put my finger on it... :( \section{The Eichler-Shimura Isomorphism} \index{Eichler-Shimura} Our goal in this section is to outline a homological interpretation of $S_k$. For details see chapter 6 of Lang \cite{lang2}, the original paper of Shimura \cite{shimura2}, or chapter VIII of Shimura \cite{shimura1}. How is $S_k(\C)$ sort of isomorphic to $H^1(X_{\Gamma},\R)$? Suppose $k=2$ and $\Gamma\subset\sl2z$ is a congruence subgroup, let $X_{\Gamma}=\overline{\Gamma\backslash\H}$ be the Riemann surface obtained by compactifying the upper half plane modulo the action of $\Gamma$. Then $S_k(\C)=H^0(X_{\Gamma},\Omega^{1})$ so we have a pairing $$H_1(X_{\Gamma},\Z)\cross S_k(\C)\into \C$$ given by integration $$(\gamma,\omega)\mapsto \int_{\gamma}\omega.$$ This gives an embedding $$\Z^{2d}\isom{}H_1(X_{\Gamma},\Z)\hookrightarrow \Hom_{\C}(S_k(\C),\C)\isom {\C}^d$$ of a ``lattice'' in $\C^d$. (We say ``lattice'' since there were some comments by Ribet that $Z^{2d}$ isn't a lattice because the rank might be too small since a subring of $\C^d$ having $\Z$-rank $2d$ might not spans $\C^d$ over $\C$). Passing to the quotient (and compactifying) gives a complex torus called the Jacobian\index{Jacobian} of $X_{\Gamma}.$ Again using the above pairing we get an embedding $${\C}^d\isom S_k(\C)\hookrightarrow \Hom(H_1(X_{\Gamma},\Z),\C)\isom\C^{2d}$$ which, upon taking the real part, gives $$ S_k(\C) \into \Hom(H_1(X_{\Gamma},\Z),\R) \isom H^1(X_{\Gamma},\R) \isom H^1_p(\Gamma,\R) $$ where $H^1_p(\Gamma,\R)$ denotes the {\em parabolic} group cohomology of $\Gamma$ with respect to the trivial action. It is this result, that we may view $S_k(\C)$ as the cohomology group $H^1_p(\Gamma,\R)$, that was alluded to above. Shimura\index{Shimura} generalized this for arbitrary $k\geq 2$ so that $$S_k(\C)\isom{}H_p^1(\Gamma,V_k)$$ where $V_k$ is a $k-1$ dimensional $\R$-vector space. The isomorphism is (approximately) the following: $f\in S_k(\C)$ is sent to the map $$\gamma\mapsto \real\int_{\tau_0}^{\gamma\tau_0}f(\tau){\tau}^{i}d\tau,\quad i=0,\ldots,k-2.$$ Let $W=\R\oplus\R$, then $\Gamma$ acts on $W$ by $$\begin{pmatrix}a&b\\c&d\end{pmatrix}: \begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}ax+by\\cx+dy\end{pmatrix} $$ so $\Gamma$ acts on $$V_k=\Sym^{k-2}W=W^{\otimes k-2}/S_{k-2}$$ where $S_{k-2}$ is the symmetric group on $k-2$ symbols (note that $\dim V_k = k-1$). Let $$L=H_p^1(\Gamma,\Sym^{k-2}(\Z\oplus\Z))$$ then under the isomorphism $$S_k(\C)\isom H^1_p(\Gamma,\R)$$ $L$ is a sublattice of $S_k(\C)$ of $\Z$-rank 2 which is $T_n$-stable for all $n$. Thus we have an embedding $$\T_{\Z} = \T \hookrightarrow \End L$$ and so $\T_{\R}\subset\End_{\bsR}(L\tensor\R)$ and $\T_{\Z}\tensor_{\bsZ}\R\isom \T_{\R}$ which has rank $d$. %% What is the point of this last bit?? \section{The Petterson Inner Product is Hecke Compatible}\index{inner product} \begin{thm} Let $\Gamma=\sl2z$, let $f,g\in S_k(\C)$, and let $$\langle f,g\rangle=\int_{\Gamma\backslash\H} f(\tau)\overline{g(\tau)}y^{k} \frac{dx dy}{y^2}.$$ Then this integral is well-defined and Hecke compatible, that is, $\langle f|T_n,g\rangle=\langle f,g|T_n\rangle$ for all $n$. \end{thm} \begin{proof} See Chapter 3 of Lang \cite{lang2}. \end{proof} %% 2/5/96 \chapter{Higher Weight Modular Forms} We are considering the spaces $S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$ which all have rank $d$. Each is acted upon by the Hecke algebra $\T$. We defined a Hecke compatible inner product (the Peterrson product) and used it to show that $$S_k(\Z)\isom\Hom_{\bsZ}(\T,\Z).$$ \section{Definitions of $\T$} \begin{quote} ``I may be asking you to explain something we have already discussed, but have we intrinsically defined the Hecke operators yet?'' -- Saul Schliemer \end{quote} The $T_n$ are defined as operators on $S_k(\C)$ by defining their action on modular forms and noting from explicit formulas that $S_k(\C)$ is preserved. But the $T_n$ can be thought of in other ways, for example, since $$S_k(\C)\isom H_p^1(\Gamma,\Sym^{k-2}(\R\oplus\R))$$ one may give an {\em explicit} description of the action of $T_n$ on $H_p^1$. \section{Double Cosets} Let $p$ be a prime and $$\sM_p=\{\alpha\in M_2(\Z) : \det(\alpha)=p\}.$$ Let $F:\sL\into\C$ be a function on the free abelian group of lattices and recall that $T_p$ acts on $F$ by $$(T_{p}F)(L)=p^{k-1}\sum_{\substack{L'\subset L\\(L:L')=p}}F(L').$$ One can write $\sM_p$ as a disjoint union of left cosets, $$\sM_p= \sl2z \begin{pmatrix}p&0\\0&1\end{pmatrix} \sl2z = \bigcup_{\substack{ad=p\\0\leq b<d}} \begin{pmatrix}a&b\\0&d\end{pmatrix} \sl2z.$$ Then $\sum_{(L:L')=p}L'$ may be thought of as the sum of the images of $L$ under the action of the left cosets of $\sM_p$. For a complete exposition, in greater generality, see Shimura\index{Shimura} \cite{shimura1}, especially chapter 3. \section{More General Congruence Subgroups} \begin{defn} A {\bfseries Dirichlet character} mod $N$ is a homomorphism $$\varepsilon:(\Z/n\Z)^{*}\into\C^{*}$$ extended to $\Z/N\Z$ by putting $\varepsilon(m)=0$ if $(m,N)\neq 1$. \end{defn} Fix integers $k\geq 0$ and $N\geq 1$. In this section we consider the spaces $$S_k(\Gamma_1(N),\epsilon)$$ for Dirichlet characters $\varepsilon$ mod $N$ and explicitly describe the action of the Hecke operators $$\begin{cases} T_n,&n\geq{}1\\ \dbd{d},&d\in(\Z/N\Z)^{*} \end{cases} $$ on these spaces. \begin{remark} Let $n$ be a positive integer. If $(n,N)=1$, then the $T_n$ behave like they do for $\sl2z$. In fact, the $T_n$ and $\dbd{d}$ commute and $$(f|T_n,g)=(f,g|\dbd{n}^{-1}T_n)$$ $$(f|\dbd{d},g)=(f,g|\dbd{d}^{-1})$$ so the $T_n$ (for $n$ prime to $N$) and $\dbd{d}$ are simultaneously diagonalizable. But if $(n,N)\neq{}1$ then $T_n$ may not be diagonalizable. \end{remark} \begin{defn} Let $$S_k(\Gamma_1(N)) = \{ f : f(\gamma \tau) = (c\tau+d)^{k}f(\tau) \text{ all } \gamma \in \Gamma_1(N) \}$$ where the $f$ are assumed holomorphic on $\sH\union\{\text{cusps}\}$. For each Dirichlet character $\varepsilon$ mod $N$ let $$S_k(\Gamma_1(N),\varepsilon)=\{ f : f(\gamma\tau)(c\tau+d)^{-k} = \varepsilon(d) f \text{ all } \gamma=\abcd \in \Gamma_0(N) \}.$$ \end{defn} When $\varepsilon\neq 0$ and $f\in{}S_k(\Gamma_1(N),\varepsilon)$ one calls $\varepsilon$ the {\bfseries nebentypus} of $f$. Let $d\in(\Z/N\Z)^{*}$ and let $f\in S_k(\Gamma_1(N))$. Let $\gamma=\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr) \in\Gamma_1(N)$ be a matrix whose lower right entry is congruent to $d$ mod $N$. Then we define $$f(\tau)|\dbd{d} = f(\gamma\tau)(c\tau+d)^{-k}.$$ Since $f|\dbd{d}=\varepsilon(d)f$, $S_k(\Gamma_1(N),\varepsilon)$ is the $\varepsilon(d)$ eigenspace of $\dbd{d}$ and $\dbd{d}$ is diagonalizable so one has a direct sum decomposition $$S_k(\Gamma_1(N))=\bigoplus_{\varepsilon:(\bsZ/N\bsZ)^{*}\into\bsC^{*}} S_k(\Gamma_1(N),\varepsilon).$$ If $f\in{}S_k(\Gamma_1(N),\varepsilon)$ then $$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\in\Gamma_0(N)$$ so $$f(-\tau)(-1)^{-k}=\varepsilon(-1)f(\tau)$$ so that $S_k(\Gamma_1(N),\varepsilon)=0$ unless $\varepsilon(-1)=(-1)^{k}$. Thus about half of the direct summands vanish. \section{Explicit Formulas} Let $$f=\sum_{n=1}^{\infty}a_n q^n \in S_k(\Gamma_1(N),\varepsilon)$$ and let $p$ be a prime, then $$f|T_p = \begin{cases} \displaystyle \sum_{n=1}^{\infty} a_{np}q^n + p^{k-1}\varepsilon(p) \sum_{n=1}^{\infty} a_n{}q^{pn}, &p\nd N\\ \displaystyle \sum_{n=1}^{\infty} a_{np}q^n + 0, &p|N \end{cases} $$ When $p|N$, $T_p$ is often denoted $U_p$ and called an Atkin-Lehner operator. We have the relations \begin{align*} T_mT_n&=T_{mn},\quad (m,n)=1\\ T_{p^k}&=\begin{cases} (T_p)^k, & p|N\\ ?, & p\nd N \end{cases} \end{align*} \section{Old and New Forms} \index{newform} {\bfseries Warning:} $T_p$ is not necessarily diagonalizable if $p|N$. There is an example due to Shimura\index{Shimura}, to present it we must first introduce old and new forms. Let $M$ and $N$ be positive integers such that $M|N$ and let $d|\frac{N}{M}$. If $f(\tau)\in S_k(\Gamma_1(M))$ then $f(d\tau)\in{}S_k(\Gamma_1(N))$. We thus have a map $S_k(\Gamma_1(M))\into{}S_k(\Gamma_1(N))$ for each $d|\frac{N}{M}$. Combining these gives a map $$\varphi_M:\bigoplus_{d|\frac{N}{M}}S_k(\Gamma_1(d))\into S_k(\Gamma_1(N)).$$ \begin{defn} The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the subspace generated by the images of the $\varphi_M$ for $M|N$, $M\neq N$. \end{defn} We remark that the {\bfseries new part} of $S_k(\Gamma_1(N))$ is the orthogonal complement of the old part with respect to the Petersson inner product. %% NOte; he goes on to map S_k(\Gamma_1(N))^2\into S_k(\Gamma_1(p)) %% which makes no sense to me in this framework -- see the written notes %% if it makes sense wednesday. %% 2/7/96 \chapter{New Forms} Today we discuss how the Hecke operators $T_n$ on $S_k(\Gamma_1(N))$ can fail to be diagonalizable. Let $N$ be a positive integer and $M$ a divisor of $N$. For each $d|\frac{N}{M}$ we define a map $$\alpha_{d}:S_k(\Gamma_1(M))\into S_k(\Gamma_1(N)):\quad{} f(\tau)\mapsto{}f(d\tau).$$ Note that when $T_p$ acts on the image space $S_k(\Gamma_1(N))$ we will often denote it by $U_p$. We must check that $f(d\tau)\in{}S_k(\Gamma_1(N))$. Define for $\gamma=\abcd$, $$(f|[\gamma]_k)(\tau)=\det(\gamma)^{k-1}(cz+d)^{-k}f(\gamma(\tau)).$$ Thus $f\in S_k(\Gamma_1(N))$ iff $f|[\gamma]_k(\tau)=f(\tau)$ (and $f$ is holomorphic). Now let $f(\tau)\in\Gamma_1(M)$ and let $\iota_d=\bigl(\begin{smallmatrix}d&0\\0&1\end{smallmatrix}\bigr)$. Then $f|[\iota_d]_k(\tau)=d^{k-1}f(d\tau)$ is a modular form on $\Gamma_1(N)$ since $\iota_d^{-1}\Gamma_1(M)\iota_d$ contains $\Gamma_1(N)$ (check this directly by conjugating an element of $\Gamma_1(N)$ by $\iota_d$). Moreover if $f$ is a cusp form then so is $f|[\iota_d]_k$. If $f\in S_k(\Gamma_1(M))$ is nonzero, then as $d$ varies over divisors of $\frac{N}{M}$, the various $f(d\tau)$ are linearly independent. Suppose $f\in S_k(\Gamma_1(M))$ is a normalized eigenform for all of the Hecke operators $T_n$ and $\dbd{n}$, and $p$ is a prime not dividiing $M$. Then $$f|T_p=a f \quad \text{and} \quad f|\dbd{p}=\varepsilon(p)f.$$ Assume $N=p^{r}M$ where $r$ is an integer $\geq 1$. Let $$f_i(\tau)=f(p^i\tau),$$ so $f_0,\ldots,f_r$ are the images of $f$ under the maps defined above and $f=f_0$. Consider the action of $U_p$ on the $f_i$. From previous work we have \begin{align*} f|T_p & = \sum_{n\geq 1} a_{np}q^n+\varepsilon(p)p^{k-1}\sum a_n{}q^{pn}\\ & = f_0|U_p + \varepsilon(p)p^{k-1} f_1 \end{align*} so $$f_0|U_p = f|T_p - \varepsilon(p)p^{k-}f_1 = af_0 - \varepsilon(p)p^{k-1}f_1.$$ Also $$f_1|U_p = (\sum a_n q^{pn}) | U_p = \sum a_n q^n = f_0.$$ More generally one can show that $f_i|U_p = f_{i-1}$. $U_p$ preserves the two dimensional vector space spanned by $f_0$ and $f_1$. The matrix of $U_p$ is $$A=\Bigl(\begin{matrix}a&1\\-\varepsilon(p)p^{k-1}&0\end{matrix}\Bigr)$$ which has characteristic polynomial $$\chi_A(X)=X^2 - aX + p^{k-1}\varepsilon(p).$$ \section{Connection With Galois Representations} \index{Galois representations} This leads to a striking connection with Galois representations. Let $f$ be a modular form and $E$ be the field generated over $\Q$ by the coefficients of $f$. Let $\ell$ be a prime and $\lambda$ a prime lying over $\ell$. Then one constructs a representation $$\rho_{\lambda}:\gal(\overline{\Q}/\Q)\into\GL(2,E_{\lambda}).$$ If $p\nd N\ell$, then $\rho_{\lambda}$ is unramified at $p$, so there is a Frobenious element $\frob_p\in\gal(\overline{\Q}/\Q)$. One can show that \begin{align*} \det(\rho_{\lambda}(\frob_p)) &= p^{k-1}\varepsilon(p) \\ \tr(\rho_{\lambda}(\frob_p)) & = a_p = a, \end{align*} so the characteristic polynomial of $\rho_{\lambda}(\frob_p)\in\GL_2(E_{\lambda})$ is $$X^2-a_p X + p^{k-1}\varepsilon(p).$$ \section{Semisimplicity of $U_p$} \begin{question} Is $U_p$ semisimple on the span of $f_0$ and $f_1$? \end{question} If the eigenvalues are distinct the answer is clearly yes. If the eigenvalues are the same, then $X^2-aX+p^{k-1}\varepsilon(p)$ has discriminant zero, that is, $4\varepsilon(p)p^{k-1}=a^2$ so $$a=2p^{\frac{k-1}{2}}\sqrt{\varepsilon(p)}.$$ Is this possible? The answer is still {\em unknown}, although it is a curious fact that the Ramanujan conjectures (proved by Delign in 1973) imply that $|a|\leq 2p^{\frac{k-1}{2}}$, so the above equality remains taunting. When $k=2$ Weil showed that $\rho_{\lambda}(\frob_p)$ is semisimple so if the eigenvalues of $U_p$ are equal then $\rho_{\lambda}(\frob_p)$ is a scalar. But Edixhoven and Coleman \cite{edixcole} show that it is not a scalar by looking at the abelian variety attached to $f$. \section{Shimura's Example of Nonsemisimple $U_p$} \index{Shimura} Let $W$ be the space spanned by $f_0, f_1$ and let $V$ be the space spanned by $f_0, f_1, f_2, f_3$. $U_p$ acts on $V/W$ by $\overline{f_2}\mapsto 0$ and $\overline{f_3}\mapsto \overline{f_2}$. Thus the matrix of the action of $U_p$ on $V/W$ is $\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$ which is nonzero and nilpotent hence not semisimple. Since $W$ is invariant under $U_p$ this shows that $U_p$ is not semisimple on $V$. %% I CAN'T SEE THIS! It seems like basic representation theory... \section{An Interesting Duality} Now suppose $N=1$ thus $\Gamma_1(N)=\sl2z$. Because of the Petersson product all the $T_n$ are diagonalizable, so $S_k=S_k(\Gamma_1(1))$ has a basis $$f_1,\ldots,f_d$$ of normalized eigenforms where $d=\dim S_k$. Let $\T=\T_{\C}$, then there is a {\em canonical} map $$\T_{\C}\hookrightarrow{}\C^d: \quad T\mapsto(\lambda_1,\ldots,\lambda_d)$$ where $f_i|T=\lambda_{i}f_i$. This map is clearly injective and we know by previous arguments that $\dim\T_{\C}=d$ so the map is an isomorphism of $\C$-vector spaces. The form $$v=f_1+\cdots+f_n$$ generates $S_k$ as a $\T$-module. Since $v$ corresponds to the vector $(1,\ldots,1)$ and $\T\isom\C^d$ acts on $S_k\isom\C^d$ componentwise this is just the statement that $\C^d$ is generated by $(1,\ldots,1)$ as a $\C^d$-module -- which is clear. Thus we have simultaneously: 1) $S_k$ is free of rank 1 over $\T$, and 2) $S_k=\Hom_{\C}(\T,\C)$ as $\T$-modules, thus $$\T\isom\Hom_{\C}(\T,\C).$$ The isomorphism sends an element of $T\in \T$ to $Tv\in S_k$. Since the identification $S_k=\Hom_{\C}(\T,\C)$ was constructed using the Petersson product it is canonical and since the choice of a normalized eigenbasis $f_1,\ldots,f_d$ is canonical we see that the isomorphism $T\isom\Hom_{\C}(\T,\C)$ is canonical. \begin{prop} $v\in S_k(\Q)$ \end{prop} \begin{proof} Let $\sigma\in\gal(\overline{\Q}/\Q)$, then if $f_i$ is a normalized eigenform so is $\sigma(f_i)$ (from the explicit formula). Thus $\sigma(f_1+\cdots+f_n)=f_1+\cdots+f_n$ for all $\sigma$ as desired. \end{proof} Now we consider the case for general $N$. Recall that we have defined maps $$S_k(\Gamma_1(M))\into S_k(\Gamma_1(N))$$ for all $M$ dividing $N$ and all divisors $d$ of $\frac{N}{M}$. \begin{defn} The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the space generated by all images of these maps with $M|N$ but $M\neq N$. The {\bfseries new part} is the orthogonal complement of the old part with respect to the Petersson product. \end{defn} There is an algebraic definition of the new part. One defines certain trace maps $$S_k(\Gamma_1(N))\into S_k(\Gamma_1(M))$$ for all $M<N$, $M|N$ which are the adjoints to the above maps (w.r.t Petersson product). Then $f$ is in the new part of $S_k(\Gamma_1(N))$ iff $f$ is killed by all of these maps. It can be shown that the $T_n$ act semisimply on $S_k(\Gamma_1(M))_{\text{new}}$ for all $M\geq 1$. Thus $S_k(\Gamma_1(M))_{\text{new}}$ has a basis of eigenforms. We have a natural map $$\bigoplus_{M|N} S_k(\Gamma_1(M))_{\text{new}}\hookrightarrow S_k(\Gamma_1(N)).$$ The image in $S_k(\Gamma_1(N))$ of an eigenform $f$ for some $S_k(\Gamma_1(M))_{\text{new}}$ is called a {\bfseries newform} of level $M_f=M$. Note that a newform is not necessarily an eigenform for the Hecke operators acting on $S_k(\Gamma_1(N))$. Let $$v=\sum_{f} f(q^{\frac{N}{M_f}})\in S_k(\Gamma_1(N))$$ where the sum is taken over all newforms $f$ of weight $k$ and some level $M|N$. This generalizes the $v$ constructed above when $N=1$ and has many of the same good properties. For example, $S_k(\Gamma_1(N))$ is free of rank $1$ over $\T$ with basis element $v$. The coefficients of $v$ lie in $\Q$, but to show this we need to know the new part of $S_k(\Gamma_1(N))$ is stable under the action of the Galois group of $\Q$. This is not easy since the new part is defined in terms of the Petersson product which is an analytic construction. Serre\index{Serre} circumvents this problem by giving an alternative definition in terms of trace maps going the other way. %% 2/9/96 \section{Observations on $T_n$} Let $\T_{\Q}=\Q[\cdots,T_n,\cdots]$ and $\Gamma=\Gamma(1)=\modgp$. Let $f_1,\ldots,f_d$ be a basis of $\Gamma$ consisting of normalized eigenforms. \begin{prop} The coefficients of the $f_i$ are totally real algebraic integers. \end{prop} \begin{proof} $\gal(\C/\Q)$ acts on $f_i$ by acting on the coefficients of its $q$-expansion. From the explicit formula in section 3.2 one sees that the set $\{f_1,\ldots,f_d\}$ is stable under the action of $\gal(\C/\Q)$. For any $i$, $a_n(f_i)$ is an eigenvalue of $T_n$ since $f_i|T_n = a_n(f_i)f_i$, and $T_n$ is self-adjoint so $a_n(f_i)$ must be real. Thus all conjugates of $a_n(f_i)$ are real and there are only finitely many since a conjugate of $a_n(f_i)$ must be $a_n(f_j)$ for some $j$, $1\leq j\leq d$. \end{proof} \begin{prop} The operators $\dbd{d}$ on $S_k(\Gamma_1(N))$ lie in $\Z[\ldots,T_n,\ldots]$. \end{prop} \begin{proof} It is enough to show $\dbd{p}\in\Z[\ldots,T_n,\ldots]$ for There is a formula relating $\dbd{p}$ and $T_p$, $$p^{k-1}\dbd{p}={T_p}^2-T_{p^2}.$$ By Dirichlet's theorem on prime's in arithmetic progression, see VIII.4 of Lang \cite{lang1}, there is another prime $q$ congruent to $p$ mod $N$. Since $p^{k-1}$ and $q^{k-1}$ are relatively prime there exist integers $a$ and $b$ so that $a p^{k-1} + b q^{k-1} = 1$. Then $$ \dbd{p}=\dbd{p}(a p^{k-1} + b q^{k-1}) = a({T_p}^2-T_{p^2}) + b({T_q}^2-T_{q^2}). $$ \end{proof} Let $\Sigma$ be a set of representatives of $\{f_1,\ldots,f_d\} \backslash \gal(\C/\Q)$. It is unknown whether or not $\#\Sigma$ can be larger than one, that is, whether the eigenforms are all conjugate under the action of Galois. Let $K_f = \Q(\ldots,a_n(f),\ldots)$ and defined a homomorphism of $\Q$-algebras $$T_{\bsQ}\into K_f : T_n\mapsto \lambda \text{ where }T_n f = \lambda f$$ Taking the product over a set of representatives of the $f_i$ yields a map $$\T_{\bsQ}\xrightarrow{\sim}\prod_{f\in\Sigma}K_f$$ which one can show is an isomorphism of $\Q$-algebras. \begin{example} Consider $S_2(\Gamma_0(N))$ with $N$ prime, then $$\T_{\bsQ}\isom E_1\times\cdots E_t$$ with the $E_i$ totally real fields. When $N=37$, that $\T_{\bsQ}\isom \Q\cross\Q$. \end{example} \chapter{Some Explicit Genus Computations} \section{Computing the Dimension of $S_k(\Gamma)$} Let $k=2$ unless otherwise noted, and let $\Gamma\subset\modgp$ be a congruence subgroup. Then $$S_2(\Gamma)=H^{0}(X_{\Gamma},\Omega^1)$$ where $$X_{\Gamma}=(\Gamma\backslash\H)\union (\Gamma\backslash\bP^1(\Q)).$$ By definition $\dim H^{0}(X_{\Gamma},\Omega^1)$ is the genus of $X_{\Gamma}$. \begin{exercise} Prove that when $\Gamma=\modgp$ then $\Gamma\backslash\bP^1(\Q)$ is a point. \end{exercise} Since $\Gamma\subset\Gamma(1)$ there is a covering $$\begin{CD} \Gamma\backslash\H @>>> X_{\Gamma} \\ @VVV @VVV \\ \Gamma(1)\backslash\H @>>> X_{\Gamma(1)} @>j>> \bP^1(\C) \end{CD}$$ which is only ramified at points above $0, 1728, \infty$ ($0$ corresponds to $i$ and $\rho$ to $1728$ under $j$). \begin{example} Suppose $\Gamma=\Gamma_0(N)$, then the degree of the covering is the index $(\modgp/\{\pm 1\}:\Gamma_0(N)/\{\pm 1\})$. A point on $Y_{\Gamma(1)}$ corresponds to an elliptic curve, whereas a points on $Y_{0}(N)$ correspond to a pair consisting of an elliptic curve and a subgroup of order $N$. \end{example} \section{Application of Riemann-Hurwitz} Now we compute the genus of $X_{\Gamma}$ by applying the Riemann-Hurwitz formula. Intuitively the Euler charcteristic should be totally additive, that is, if $A$ and $B$ are disjoint spaces then $$\chi(A\union B)=\chi(A)+\chi(B).$$ Let $X$ be a compact Riemann surface of genus $g$, then $\chi(X)=2-2g$. Since $\chi(\{\text{point}\})=1$ we should have that $$\chi(X-\{p_1,\ldots,p_n\})=\chi(X)-n\chi(1)=(2-2g)-n.$$ If we have an umramified covering $X\into Y$ of degree $d$ then $\chi(X)=d\cdot\chi(Y)$. Consider the covering $$\begin{CD}X_{\Gamma}-\{\text{points over $0,1728,\infty$}\}\\ @VVV \\ X_{\Gamma(1)}-\{0,1728,\infty\} \end{CD}$$ Since $X_{\Gamma(1)}$ has genus $0$, $X_{\Gamma(1)}-\{0,1728,\infty\}$ has Euler characteristic $2-3=-1$. If we let $g=\chi(X_{\Gamma})$ then $\chi(X_{\Gamma}-\{\text{points over $0,1728,\infty$}\} = 2-2g -n_{0} - n_{1728} - n_{\infty}$, where $n_p$ denotes the number of points lying over $p$. Thus $-d=2-2g-n_0-n_{1728}-n_{\infty}$ whence $$2g-2 = d - n_0 -n_{1728}-n_{\infty}.$$ Suppose $\Gamma=\Gamma_0(N)$ with $N>3$, then $n_0=d/3$ and $n_{1728}=d/2$ (I'm not sure why). The degree $d$ of the covering is equal to the number of unordered ordered basis of $E[N]$, thus $$d=\#\SL_2(\Z/N\Z)/2.$$ We still need to compute $n_{\infty}$. $\modgp$ acts on $\bP^1(\Q)$ if we view $\bP^1(\Q)$ as all pairs $(a,b)$ of relatively prime integers and suppose $\infty$ corresponds to $(1,0)$. The stabilizer of $(1,0)$ is the sugroup $\{\abcd \in \modgp : c=0 \}$ of upper triangular matrices. Since the points lying over $\infty$ are all conjugate by the Galois group of the covering (which is $\SL_2(\Z/N\Z)/\{\pm 1\}$), $$\text{number of cusps}=\frac{\text{order of $\SL_2(\Z/N\Z)/\{\pm 1\}$}} {\text{order of stabilizer of $\infty$}}.$$ We thus have $$2g(X(N))-2=\frac{d}{6}-\frac{d}{N}$$ where $\frac{d}{N}$ is the number of cusps. %% 2/12/96 \section{Explicit Genus Computations} Let $N>3$ and consider the modular curve $X=X(N)$. There is a natural covering map $X\into{}X(1)\xrightarrow{j}\C$. Let $d$ be the degree, then $$2g-2=d-m_0-m_{1728}-m_{\infty}$$ where $g$ is the genus of $X$ and $m_x$ is the number of points lying over $x$. Since $m_0$ is approximately $\frac{d}{3}$ and $m_{1728}$ is approximately $\frac{d}{2}$, $$2g-2=\frac{d}{6}-m_{\infty}\pm\text{ small correction factor}.$$ \section{The Genus of $X(N)$} Now we count the number of cusps of $X(N)$, that is, the size of $\Gamma(N)\backslash\bP^1(\Q)$. There is a surjective map from $\modgp$ to $\bP^1(\Q)$ given by $$\bigabcd\mapsto \bigabcd\Bigl(\begin{matrix}1\\0\end{matrix}\Bigr).$$ Let $U$ be the kernel, thus $U$ is the stabilizer of $\infty=\bigl(\begin{smallmatrix}1\\0\end{smallmatrix}\bigr)$, so $U=\{\pm\bigl(\begin{smallmatrix}1&a\\0&1\end{smallmatrix}\bigr):a\in\Z\}$. Then the cusps of $X(N)$ are the elements of $$\Gamma(N)\backslash(\modgp/U)=(\Gamma(N)\backslash\modgp)/U=\SL_2(\Z/N\Z)/U$$ which has order $$\frac{\#\SL_2(\Z/N\Z)}{2N}=\frac{d}{N}.$$ Substituting this into the above formula gives $$2g-2=\frac{d}{6}-\frac{d}{N}=\frac{d}{6N}(N-6)$$ so $$g=1+\frac{d}{12N}(N-6).$$ When $N$ is prime $$d=\frac{1}{2}\#\SL_2(\Z/N\Z)=\frac{1}{2}\cdot\frac{(N^2-1)(N^2-N)}{N-1}.$$ Thus when $N=5$, $d=60$ so $g=0$, and when $N=7$, $d=168$ so $g=3$. \section{The Genus of $X_0(N)$} Suppose $N>3$ and $N$ is prime. The covering map $X_0(N)\into X(1)$ is of degree $N+1$ since a point of $X_0(N)$ corresponds to an elliptic curve along with a subgroup of order $N$ and there are $N+1$ such subgroups because $N$ is prime. \begin{exercise} $X_0(N)$ has two cusps; they are the orbit of $\infty$ which is unramified and $0$ which is ramified of order $N$. \end{exercise} Thus $$2g-2=N+1-2-n_{1728}-n_0.$$ $n_0$ is the number of pairs $(E,C)$ (modulo isomorphism) such that $E$ has $j$-invariant $0$. So we consider $E=\C/\Z[\frac{-1+i\sqrt{3}}{2}]$ which has endomorphism ring $\End(E)=\Z[\Mu_6]$. Now $\Mu_6/{\pm 1}$ acts on the cyclic subgroups $C$ so, letting $\omega$ be a primitive cube root of unity, we have $$(E,C)\isom(E,\omega C)\isom(E,\omega^2 C).$$ This might lead one to think that $m_0$ is $(N+1)/3$, but it may be bigger if, for example, $C=\omega{}C$. Thus we must count those $C$ so that $\omega C=C$ or $\omega^2 C=C$, that is, those $C$ which are stable under $\sO=\Z[\frac{-1+i\sqrt{3}}{2}]$. So we must compute the number of stable $\sO/N\sO$-submodules of order $N$. This depends on the structure of $\sO/N\sO$: $$\sO/N\sO = \begin{cases} \F_N\oplus\F_N&\text{if $(\frac{-3}{N})=1$ ($N$ splits)}\\ \F_{N^2} &\text{if $(\frac{-3}{N})=-1$ ($N$ stays inert)} \end{cases} $$ Since $\sO/N\sO=\F_{N^2}$ is a field it has no submodules of order $N$, whereas $\F_{N}\oplus\F_{N}$ has two $\sO/N\sO$-submodules of order $N$, namely $\F_N\oplus{}0$ and $0\oplus\F_N$. Thus $$m_0 = \begin{cases} \frac{N+1}{3} & \text{if $N\equiv 2\pmod{3}$}\\ \frac{N-1}{3}+2 & \text{if $N\equiv 1\pmod{3}$} \end{cases} $$ \begin{exercise} It is an exercise in elegance to write this as a single formula involving the quadratic symbol. \end{exercise} By similar reasoning one shows that $$m_{1728}=\begin{cases} \frac{N+1}{2} & \text{if $N\equiv 3\pmod{4}$}\\ \frac{N-1}{2}+2 & \text{if $N\equiv 1\pmod{4}$} \end{cases} $$ We can now compute the genus of $X_0(N)$ for any prime $N$. For example, if $N=37$ then $2g-2=36-(2+18)-(14)=2$ so $g=2$. Similarly, $X_0(13)$ has genus $0$ and $X_0(11)$ has genus $1$. In general, $X_0(N)$ has genus approximately $N/12$. Serre\index{Serre} constructed a nice formula for the above genus. Suppose $N>3$ is a prime and write $N=12a+b$ with $0\leq b\leq 11$. Then Serre's formula is \begin{center} \begin{tabular}{|c|cccc|}\hline $b$&1&5&7&11\\\hline $g$&$a-1$&$a$&$a$&$a+1$\\\hline \end{tabular} \end{center} \section{Modular Forms mod $p$} Let $N$ be a positive integer, let $p$ be a prime and assume $\Gamma$ is either $\Gamma_0(N)$ or $\Gamma_1(N)$. \begin{defn} Let $M_k(\Gamma,\Z)=M_k(\Gamma,\C)\intersect\Z[[q]]$, then $$M_k(\Gamma,\F_p)=M_k(\Gamma,\Z)\tensor_{\Z}\F_p$$ is the space of {\bfseries modular forms mod $p$} of weight $k$. \end{defn} Suppose $p=N$, then one has {\bfseries Serre's Equality}:\index{Serre} $$M_{p+1}(\modgp,\F_p)=M_2(\Gamma_0(p),\F_p)$$ The map from the right hand side to the left hand side is accomplished via a certain normalized Eisenstein series\index{Eisenstein series}. Recall that in $\modgp$, $$G_k=-\frac{B_k}{2k}+\sum_{n=1}^{\infty}(\sum_{d|n}d^{k-1})q^n$$ and $$E_k=1-\frac{2k}{B_k}\sum_{n=1}^{\infty}(\sum_{d|n}d^{k-1})q^n.$$ One finds $\ord_p(-\frac{B_k}{2k})$ using Kummer congruences. In particular, $\ord_p(B_{p-1})=-1,$ so $E_{p-1}\equiv 1\pmod{p}$. Thus multiplication by $E_{p-1}$ raises the level by $p-1$ but does not change the $q$-expansion mod $p$. We thus get a map $$M_2(\Gamma_0(p),\F_p)\into M_{p+1}(\Gamma_0(p),\F_p).$$ The map $$M_{p+1}(\Gamma_0(p),\F_p)\into{}M_{p+1}(\modgp,\F_p)$$ is the trace map (which is dual to the natural inclusion going the other way) and is accomplished by averaging in order to get a form invariant under $\modgp$. % 2/14/96 \chapter{The Field of Moduli} In this chapter we will study the field of definition of the modular curves\index{modular curves} $X(N)$, $X_0(N)$, and $X_1(N)$. The function field of $X(1)=\bP^1_{\Q}$ is $\Q(t)$. If $E$ is an elliptic curve given by a Weierstrass equation $y^2=4x^3-g_2x-g_3$ then $$j(E)=j(g_2,g_3)=\frac{1728g_2^3}{g_2^3-27g_3^2}.$$ The $j$ invariant determines the isomorphism class of $E$ over $\C$. Pick an elliptic curve $E/\Q(t)$ such that $j(E)=t$. In particular we could pick the elliptic curve with Weierstrass equation $$y^2=4x^3-\frac{27t}{t-1728}x-\frac{27t}{t-1728}.$$ Let $E/k$ be an arbitrary elliptic curve and $N$ a positive integer prime to $\Char{} k$. Then $E[N](\overline{k})\isom(\Z/N\Z)^2$. Let $k(E[N])$ be the field obtained by adjoining the coordinates of the $N$-torsion points of $E$. Consider the tower of fields $\overline{k}\supset k(E[N])\supset k$. There is a Galois representation on the $N$ torsion of $E$: $$\gal(\overline{k}/k)\xrightarrow{\rho_{E,N}}\aut(E[N])\isom\GL_2(\Z/N\Z)$$ and $\gal(\overline{k}/k(E[N]))=\ker(\rho_{E,N}).$ Thus the Galois group of the extension $\Q(t)(E[N])$ over $\Q(t)$ is contained in $\GL_2(\Z/N\Z)$. Let $X(N)$ be the curve corresponding to the function field $\Q(t)(E[N])$ over $\Q$. Since $\overline{\Q}\intersect\Q(t)(E[N])$ is contained in $Q(\Mu_N)$, $X(N)$ is defined over $\Q(\Mu_N)$. Composing $\rho_{E}$ with the natural map $\GL_2(\Z/N\Z)\into\GL_2(\Z/N\Z)/\{\pm 1\}$ gives a map $$\overline{\rho}_E:\gal(\overline{K}/K)\into\GL_2(\Z/N\Z)/\{\pm 1\}.$$ \begin{prop}\label{PropModSurj} $\overline{\rho}_E$ is surjective iff $\rho_E$ is surjective. \end{prop} \begin{proof} If $\overline{\rho}_E$ is surjective then either $\bigl(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\bigr)$ or its negative lies in the image of $\rho$. Thus $\bigl(\begin{smallmatrix}-1&0\\0&-1\end{smallmatrix}\bigr)$ lie in the image of $\rho$. Since $\overline{\rho}_E$ is surjective this implies that $\rho$ is surjective. The converse is trivial. \end{proof} \section{Digression on Moduli} $X_0(N)(\overline{K})$ is the set of $\overline{K}$-isomorpisms classes of pairs $(E,C)$ where $E/\overline{K}$ is an elliptic curve and $C$ is a cyclic subgroup of order $N$. $X_0(N)(K)$ is the set of isomorphism classes of pairs $(E,C)$ such that for all $\sigma\in\gal(\overline{K}/K)$, $\sigma(E,C)=(E,C)$. There is a map $$\{\text{$k$-isomorphism classes of pairs $(E,C)/K$}\}\into{}X_0(N)(K)$$ which is ``notoriously'' non-injective. Deligne and Rapaport \cite{lnm349} prove the map is surjective. When $N=1$ they observe that the map is surjective, then for $N>1$ they show that certain obstructions vanish. A related question is \begin{question} If $E/\overline{K}$ is isomorphic to all its Galois conjugates, is there $E'/K$ which is isomorphic to $E$ over $\overline{K}$? \end{question} \section{When is $\rho_E$ Surjective?} \begin{prop} Let $E_1$ and $E_2$ be elliptic curves defined over $K$ with equal $j$-invariants, thus $E_1\isom E_2$ over $\overline{K}$. Assume $E_1$ and $E_2$ do not have complex multiplication over $\overline{K}$. Then $\rho_{E_1}$ is surjective iff $\rho_{E_2}$ is surjective. \end{prop} \begin{proof} Assume $\rho_{E_1}$ is surjective. Since $E_1$ has no complex multiplication over $\overline{K}$, $\aut E_1=\{\pm 1\}$. Choose an isomorphism $\varphi:E_1\xrightarrow{\sim}E_2$ over $\overline{K}$. Then for $\sigma\in\gal(\overline{K}/K)$ we have the diagram $$\begin{CD} E_1 @>\varphi>> E_2 \\ @V=VV @V=VV\\ \presup{\sigma}E_1 @>\sigma\varphi>> \presup{\sigma}E_2 \end{CD}$$ Thus $\sigma\varphi=\pm\varphi$ for all $\sigma\in\gal(\overline{K}/K)$, so $\varphi:E_1[N]\into{}E_2[N]$ defines an equivalence $\overline{\rho}_{E_1}\isom\overline{\rho}_{E_2}$. Since $\rho_{E_1}$ is surjective this implies $\overline{\rho}_{E_2}$ is surjective which, by the previous proposition, implies $\rho_{E_2}$ is surjective. \end{proof} Let $K=\C(j)$, with $j$ transcendental over $\C$. Let $E/K$ be an elliptic curve with $j$-invariant $j$. Fix a positive integer $N$ and let $$\rho_E:\gal(\overline{K}/K)\into\GL_2(\Z/N\Z)$$ be the associated Galois representation. Then $\det\rho_E$ is the cyclotomic character which is trivial since $\C$ contains the $N$th roots of unity. Thus the image of $\rho_E$ lands inside of $\SL_2(\Z/N\Z)$. Our next theorem states that a generic elliptic curve has maximal possible Galois action on its division points. \begin{thm} $\rho_E:\gal(\overline{K}/K)\into\SL_2(\Z/N\Z)$ is surjective. \end{thm} Igusa \cite{ig1} found an algebraic proof of this theorem We will now make some comments on how an analytic proof goes. \begin{proof} Let $\C(j)=K=\sF_1$ be the field of modular functions for $\modgp$. Suppose $N\geq 3$ and let $\sF_N$ be the field of mereomorphic functions for $\Gamma(N)$. Then $\sF_N/\sF_1$ is a Galois extension with Galois group $\SL_2(\Z/N\Z)/\{\pm 1\}$. Let $E$ be an elliptic curve over $K$ with $j$-invariant $j$. We will show that $\gal(\sF_N/\sF_1)=\SL_2(\Z/N\Z)/\{\pm 1\}$ acts transitively on the $x$-coordinates of the $N$ torsion points of $E$. This will show that $\overline{\rho}_E$ maps surjectively onto $\SL_2(\Z/N\Z)/\{\pm 1\}$. Then by proposition \ref{PropModSurj}, $\rho_E$ maps surjectively onto $\SL_2(\Z/N\Z)$, as claimed. We will now construct the $x$-coordinates of $E[N]$ as functions on $\H$ which are invariant under $\Gamma(N)$. (Thus $K(E[N]/\{\pm 1\})\subset \sF_N$.) Let $\tau\in\H$ and let $L_{\tau}=\Z\tau+\Z$. Consider $\wp(z,L_{\tau})$ which gives the $x$ coordinate of $\C/L_{\tau}$ in it standard form. Define, for each nonzero $(r,s)\in((\Z/N\Z)/\{\pm{}1\})^2$, a function $$f_{(r,s)}:\H\into\C: \quad \tau\mapsto\frac{g_2(\tau)}{g_3(\tau)} \wp(\frac{r\tau+s}{N},L_{\tau}).$$ First notice that for any $\alpha=\abcd\in\modgp$, $$f_{(r,s)}(\alpha\tau)=f_{(r,s)\abcd}(\tau).$$ Indeed, $\wp$ is homogenous of degree $-2$, $g_2$ is modular of weight 4 and $g_3$ is modular of weight 6, so \begin{align*} f_{(r,s)}(\alpha\tau)& =\frac{g_2(\alpha\tau)}{g_3(\alpha\tau)} \wp(\frac{r\alpha\tau+s}{N})\\ & = (c\tau+d)^{-2}\frac{g_2(\tau)}{g_3(\tau)} \wp(\frac{ra\tau+rb+cs\tau+sd}{N(c\tau+d)})\\ & = \frac{g_2(\tau)}{g_3(\tau)}\wp(\frac{(ra+sc)\tau+rb+sd}{N}) = f_{(r,s)\alpha}(\tau) \end{align*} If $\tau\in\H$ with $g_2(\tau), g_3(\tau)\neq 0$ then the $f_{(r,s)}(\tau)$ are the $x$-coordinates of the nonzero $N$-division points of $E_{j(\tau)}$. The various $f_{(r,s)}(\tau)$ are distinct. Thus $\SL_2(\Z/N\Z)/\{\pm 1\}$ acts transitively on the $f_{(r,s)}$. The consequence is that the $N^2-1$ nonzero division points of $E_j$ have $x$-coordinates in $\overline{\sF}_N$ equal to the $f_{(r,s)}\in\sF_N$. \end{proof} %%%%%%%%%%%%%%%%%%%%% %% Trivial Observation: I just don't understand this proof! %%%%%%%%%%%%%%%%%%%%% \section{Observations} \begin{prop} If $E/\Q(\Mu_N)(t)$ is an elliptic curve with $j(E)=t$, then $\rho_E$ has image $\SL_2(\Z/N\Z)$. \end{prop} \begin{proof} Since $\Q(\Mu_N)$ contains the $N$th roots of unity $N$th cyclotomic character is trivial hence the determinent of $\rho_E$ is trivial. Thus the image of $\rho_E$ lies in $\SL_2(\Z/N\Z)$. In the other direction, there is a natural inclusion $$\SL_2(\Z/N\Z)=\gal(\C(t)(E[N])/\C(t))\hookrightarrow \gal(\Q(\Mu_N)(t)(E[N])/\Q(\Mu_N)(t)).$$ \end{proof} \begin{prop} If $E/\Q(t)$ is an elliptic curve with $j(E)=t$, then $\rho_E$ has image $\GL_2(\Z/N\Z)$ and $\overline{\Q}\intersect\Q(t)(E[N])=\Q(\Mu_N).$ \end{prop} \begin{proof} Since $\Q(t)$ contains no $N$th roots of unity, the mod $N$ cyclotomic character, and hence $det\rho_E$, is surjective onto $(\Z/N\Z)^{*}$. Since the image of $\rho_E$ already contains $SL_2(\Z/N\Z)$ it must equal $GL_2(\Z/N\Z)$. For the second assertion consider the diagram $$\begin{array}{rcccl} &\overline{\Q}&&\Q(t)(E[N])&\\ &\mid & & \mid& SL_2\\ &\Q(\Mu_N) & \subset & \Q(\Mu_N)(t)&\\ (\Z/N\Z)^{*} & \mid & & \mid &GL_2/SL_2=(\Z/N\Z)^{*}\\ &\Q & \subset & \Q(t) & \end{array}$$ \end{proof} This gives a way to view $X_0(N)$ as a projective algebraic curve over $\Q$. Let $K=\Q(t)$ and let $L=K(E[N])=\Q(\Mu_N)(t)$. Then $$H=\{\bigl(\begin{smallmatrix}*&*\\0&*\end{smallmatrix}\bigr)\} \subset\GL_2(\Z/N\Z)=\gal(L/K).$$ The fixed field $L^H$ is an extension of $\Q(t)$ of transcendence degree 1 with field of constants $\overline{\Q}\intersect L^H=\Q$, i.e., a projective algebraic curve. %2/16/96 \section{A Descent Problem}\index{descent} Consider the following exercise which may be approached in an honest or dishonest way. \begin{exercise} Suppose $L/K$ is a finite Galois extension and $G=\gal(L/K)$. Let $E/L$ be an elliptic curve, assume $\aut_L{E}=\{\pm 1\}$, and suppose that for all $g\in{}G$, there is an isomorphism $\presup{g}E\iso{}E$ over $L$. Show that there exists $E_0/K$ such that $E_0\isom{}E$ over $L$. \end{exercise} {\bfseries Caution!} The exercise is {\em false} as stated. Both the dishonet and honest approaches below work only if $L$ is a separable closure of $K$. Now: can one construct a counterexample? {\em Discussion.} First the hard, but ``honest'' way to look at this problem. For notions on descent\index{descent} see Serre\index{Serre} \cite{serre1}. By descent theory, to give $E_0$ is the same as to give a family $(\lambda_g)_{g\in{}G}$ of maps $\lambda_{g}:\presup{g}E\iso{}E$ such that $\lambda_{gh}=\lambda_{g}\circ\presup{g}\lambda_h$ where $\presup{g}\lambda_h=g\circ\lambda_h\circ{}g^{-1}$. Note that $\lambda_{g}\circ\presup{g}\lambda_h$ maps $\presup{gh}E\into E$. This is the natural condition to impose, because if $f:E_0\iso E$ and we let $\lambda_g=f\circ\presup{g}(f^{-1})$ then $\lambda_{gh}=\lambda_{g}\circ\presup{g}\lambda_h$. Using our hypothesis choose, for each $g\in G$, an isomorphism $$\lambda_{g}:\presup{g}E\iso{}E.$$ Define a map $c$ by $$c(g,h)=\lambda_g\circ\presup{g}\lambda_h\circ\lambda_{gh}^{-1}.$$ Note that $c(g,h)\in\aut E=\{\pm 1\}$ so $c$ defines an element of $$H^2(G,\{\pm 1\})\subset H^2(\gal(\overline{L}/K),\{\pm 1\}) =Br(K)[2].$$ Here $Br(K)[2]$ denotes the 2-torsion of the Brauer group $$Br(K)=H^2(\gal(\overline{L}/K),\overline{L}^{*}).$$ This probably leads to an honest proof. The dishonest approach is to note that $g(j(E))=j(E)$ for all $g\in{}G$ since all conjugates of $E$ are isomorphic and $j(\presup{g}E)=g(j(E))$. Thus $j(E)\in K$ so we can define $E_0/K$ by substituting $j(E)$ into the universal elliptic curve formula (see III.1.4 of \cite{silverman1}). This gives an elliptic curve $E_0$ defined over $K$ but isomorphic to $E$ over $\overline{K}$. %% ???This doesn't quite work??? %% I mean, is it true that if $E$ and $E_0$ have the same j-invariant %% and the $j$-invariant lies in the field of definition $L$ %% then $E$ and $E_0$ are isomorphic {\em over} $L$? No, see silverman1. %%%%%%%%%%%%%%%%%%%%%%%%%% %% 2/21/96 %%%%%%%%%%%%%%%%%%%%%%%%%% \section{Second Look at the Descent Exercise} \index{descent} Last time we talked about the following problem. Suppose $L/K$ is a Galois extension with $\Char K=0$, and let $E/L$ be an elliptic curve. Suppose that for all $\sigma\in{}G=\gal(L/K)$, $\presup{\sigma}E\isom E$ over $L$. Conclude that there is an elliptic curve $E_0/K$ such that $E_0\isom E$ over $L$. The conclusion may fail to hold if $L$ is a finite extension of $K$, but the exercise is true when $L=\overline{K}$. First we give a descent argument which holds when $L=\overline{K}$ and then give a counterexample to the more general statement. For $g,h\in{}G=\gal(L/K)$ we define an automorphism $c(g,h)\in\aut E=\{\pm 1\}$. Choose for every $g\in\gal(L/K)$ some isomorphism $$\lambda_g:\presup{g}E\iso E.$$ If the $\lambda_g$ were to all satisfy the compatibility criterion $\lambda_{gh}=\lambda_g\circ\presup{g}\lambda_h$ then by descent theory we could find a $K$-structure on $E$, that is a model for $E$ defined over $K$ and isomorphic to $E$ over $L$. Define $c(g,h)$ by $c(g,h)\lambda_{gh}=\lambda_{g}\circ\presup{g}\lambda_h$ so $c(g,h)$ measures how much the $\lambda_g$ fail to satisfy the compatibility criterion. Since $c(g,h)$ is a cocycle it defines an elements of $H^2(G,\{\pm 1\})$. We want to know that this element is trivial. When $L=\overline{K}$, the map $H^2(G,\{\pm 1\})\into H^2(G,L^{*})$ is injective. To see this first consider the exact sequence $$0\into\{\pm 1\}\into\overline{K}^{*}\xrightarrow{2}\overline{K}^{*}\into 0$$ where $2:\overline{K}^{*}\into\overline{K}^{*}$ is the squaring map. Taking cohomology yields an exact sequence $$H^1(G,\overline{K}^{*})\into{}H^2(G,\{\pm 1\}) \into{}H^2(G,\overline{K}^{*}).$$ By Hilbert's theorem 90 (\cite{serre3} Ch. X, Prop. 2), $H^1(G,\{\pm 1\})=0$. Thus we have an exact sequence $$0\into H^2(G,\{\pm 1\})\into H^2(G,\overline{K}^{*})[2]\into 0.$$ Thus $H^2(G,\{\pm 1\})$ naturally sits inside $H^2(G,L^*)$. To finish Ribet does something with differentials and $H^0(\presup{g}E,\Omega^1)$ which I don't understand. The counterexample in the case when $L/K$ is finite was provided by Kevin Buzzard\index{Buzzard} (who said Coates gave it to him). Let $L=\Q(i)$, $K=\Q$ and $E$ be the elliptic curve with Weirstrass equation $iy^2=x^3+x+1$. Then $E$ is isomorphic to its conjugate over $L$ but one can show directly that $E$ has no model over $\Q$. %%%%%%%%%%%%%%%%%%%%%%% %% Big mystery: Today Gal(F_N/F_1)=GL2, whereas before it was SL_2. %% Solution: Today we are over Q, before we were over C! %%%%%%%%%%%%%%%%%%%%%%% \section{Action of $\GL_2$} Let $N>3$ be an integer and $E/\Q(j)$ an elliptic curve with $j$-invariant $j(E)=j$. Then there is a Galois extension $$\begin{array}{c} \sF_N=\Q(j)(E[N]/\{\pm 1\})\\ |\\ \sF_1=\Q(j) \end{array} $$ with Galois group $\GL_2(\Z/N\Z)/\{\pm 1\}$. Think of $\Q(j)(E[N]/\{\pm 1\}$ as the field obtained from $\Q(j)$ by adjoining the $x$-coordinates of the $N$-torsion points of $E$. Note that this situation differs from the previous situation in that the base field $\C$ has been replaced by $\Q$. Consider $$\sF=\bigcup_{N}\sF_N$$ which corresponds to a projective system of modular curves \index{modular curves}. Let $\A_f$ be the ring of finite ad\`{e}les, thus $$\A_f=\hat{\Q}=\hat{\Z}\tensor\Q\subset\prod_{p}\Q_p.$$ $\A_f$ can be thought of as $$\{(x_p):x_p\in\Z_p \text{ for almost all $p$}\}.$$ The group $\GL_2(\A_f)$ acts on $\sF$. To understand what this action is we first consider the subgroup $\GL_2(\hat{\Z})$ of $\GL_2(\A_f)$. It can be shown that $$\sF=\Q(f_{N,(r,s)} : (r,s)\in(\Z/N\Z)^2-\{(0,0)\}, N\geq{}1)$$ where $f_{N,(r,s)}$ is a function $$f_{N,(r,s)}:\H\into\C: \quad \tau\mapsto\frac{g_2(\tau)}{g_3(\tau)} \wp(\frac{r\tau+s}{N},L_{\tau}).$$ We define the action of $\GL_2(\hat{\Z})$ on $\sF$ as follows. Let $g\in\GL_2(\hat{\Z})$, then to give the action of $g$ on $f_{N,(r,s)}$ first map $g$ into $\GL_2(\Z/N\Z)$ via the natural reduction map, then note that $\GL_2(\Z/N\Z)$ acts on $f_{N,(r,s)}$ by $$\bigl(\begin{matrix}a&b\\c&d\end{matrix}\bigr)\cdot{}f_{N,(r,s)} =f_{N,(r,s)\Bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\Bigr)} =f_{N,(ra+sc,rb+sd)}.$$ Let $E$ be an elliptic curve. Then the universal Tate module is $$T(E)=\varinjlim{}E[N]=\prod_{p}T_p(E).$$ There is an isomorphism $\alpha:\hat{\Z}^2\iso T(E)$. Via right composition $\GL_2(\hat{\Z})$ acts on the collection of all such isomorphism $\alpha$. So $\GL_2(\hat{\Z})$ acts naturally on pairs $(E,\alpha)$ but the action does nothing to $E$. One of the first important things we must do in understanding the construction of things like Shimura\index{Shimura variety} varieties is to free ourselves and allow $\GL_2(\hat{\Z})$ to act on the $E$'s as well. Let $$g=\Bigl(\begin{matrix}a&b\\c&d\end{matrix}\Bigr)\in\GL_2^{+}(\Q)$$ (thus $g$ has positive determinent). Let $\tau\in\H$ and let $E=E_{\tau}$ be the elliptic curve determined by the lattice $L_{\tau}=\Z+\Z\tau$. Let $$\alpha_{\tau}:L_{\tau}=\Z+\Z\tau\iso\Z^2$$ be the isomorphism defined by $\tau \mapsto(1,0)$ and $1\mapsto (0,1)$. Now view $\alpha=\alpha_{\tau}$ as a map $$\alpha:\Z^2\iso H_1(E(\C),\Z).$$ Tensoring with $\Q$ then gives another map (also denoted $\alpha$) $$\alpha:\Q^2\iso H_1(E,\Q).$$ Then $\alpha\circ g$ is another isomorphism $$\Q^2\xrightarrow{\alpha\circ g} H_1(E,\Q)$$ which induces an isomorphim $\Z^2\iso{}L'\subset H_1(E,\Q)$ where $L'$ is a lattice. There exists an elliptic curve $E'/\C$ and a map $\lambda\in\Hom(E',E)\tensor\Q$ which induces a map (also denoted $\lambda$) $$\lambda:H_1(E',\Z)\iso L'\subset H_1(E,\Q)$$ on homology groups. Now we can define an action on pairs $(E,\alpha)$ by sending $(E,\alpha)$ to $(E',\alpha')$. Here $\alpha'$ is the map $\alpha':\Z^2\into H_1(E^1,\Z)$ given by the composition $$\Z^2\xrightarrow{\alpha{}g}L'\xrightarrow{\lambda^{-1}}H_1(E^1,\Z).$$ In more concrete terms the action is $$g:(E_{\tau},\alpha_{\tau})\mapsto (E_{\tau}',\alpha_{\tau}')$$ where $\tau'=g\tau=\frac{a\tau+b}{c\tau+d}.$ [[ CHECK THIS SOMETIME SOON!!]] %%%%%%%%%%%%%%%%%%%%%% %% 2/24/96 \chapter{Hecke Operators as Correspondences} \section{Some Philosophy} We are studying modular forms over $\C$ and more generally over subrings $R$ of $\C$. The Hecke algebras occur naturally as operators on various spaces of modular forms. We are aiming for an arithmetic perspective. One way is to study the arithmetic of cusp forms of weight 2 for congruence subgroups like $\Gamma_0(N)$ or $\Gamma_1(N)$. These cusp forms correspond to differentials on the modular curves\index{modular curves} $X_0(N)$ and $X_1(N)$. We have constructed models for each of these over $\Q$. When $N'|N$ there is a natural map $X(N)\into{}X(N')$. Thus we get a tower of curves and a corresponding tower of number fields. $$\begin{array}{ccc} \cdots&&\cdots\\ \downarrow&&\cup\\ X(N)&&\sF_N\\ \downarrow&&\cup\\ X(N')&&\sF'_N\\ \downarrow&&\cup\\ \cdots&&\cdots \end{array}$$ Taking limits gives a curve $X=\varprojlim{}X(N)$ and a corresponding field $\sF=\varinjlim{}\sF_N$. There is an action of $\GL_2(\A_f)$ on pairs $(E,\alpha)$. By $\A_f$ we mean the ring of finite ad\`{e}les, which may be identified with the restricted product $\prod_p\Q_p$. The subscript $f$, for `finite', indicates that the infinite place is omitted. The full ring of ad\`{e}les is $\A=\A_f\cross\R$. If $g\in\GL_2(\A_f)$, then $g$ acts on pairs $(E,\alpha)$ where $E$ is an elliptic curve and $$\alpha:\hat{\Z}^2\iso{}T(E)=\prod_p T_p(E).$$ Note that $$\A_f=\hat{\Z}\tensor_{\Z}\Q=\prod\Z_p\tensor_{\Z}\Q=\hat{\Q},$$ and $T(E)$ is free of rank $2$ over $\hat{\Z}$. Let $$V(E)=T(E)\tensor_{\Z}\Q=\prod_{p}V_p(E)$$ where the product is restricted and $V_p(E)=T_p(E)\tensor_{\Z_p}\Q_p$. View $g\in\GL_2(\A_f)$ as an automorphism of $\hat{\Q}^2$. Then $\alpha\circ g$ sends $\hat{\Z}^2\subset\hat{\Q}^2$ to a lattice $T'\subset V(E)$. As a lemma, one shows that there is an elliptic curve $E'$ and a canonical map $\lambda:E'\into E$ such that the induced map $\lambda':V(E')\iso V(E)$ is an isomorphism which sends $T(E')$ maps to $T'$ in $V(E)$. Then $g$ sends the pair $(E,\alpha)$ to $(E',{\lambda'}^{-1}\circ\alpha\circ g)$. \section{Hecke Operators as Correspondences} Our goal is to think of Hecke operators ($T_n$, $\dbd{d}$) as objects defined over $\Q$. We will define the Hecke operators as correspondences. \begin{defn} Let $C_1$ and $C_2$ be curves, then a {\bfseries correspondence} $C_1\corrs{}C_2$ is a curve $C$ together with morphisms $\alpha:C\into C_1$ and $\beta:C\into C_2$. $$\corr{C}{C_1}{C_2}{\alpha}{\beta}$$ Giving a correspondence $C_1\corrs{}C_2$ the {\bfseries dual correspondence} $C_2\corrs{}C_1$ is obtained by looking at the diagram in a mirror $$\corr{C}{C_2}{C_1}{\beta}{\alpha}$$ \end{defn} The simplest case to consider is the modular curve $X_0(N)$ and Hecke operator $T_p$, where $p\nd N$. We view $T_p$ as a correspondence $X_0(N)\corrs X_0(N)$, thus there is a curve $C=X_0(pN)$ plus 2 maps $\alpha$ and $\beta$ $$\corr{X_0(pN)}{X_0(N)}{X_0(N)}{\alpha}{\beta}$$ The maps $\alpha$ and $\beta$ are degeneracy maps which forget data. To define them view $X_0(N)$ as classifying pairs $(E,C)$ where $E$ is an elliptic curve and $C\isom\Z/N\Z$ is a cyclic subgroup of order $N$. Similarly $X_0(pN)$ classifies pairs $(E,G)$ where $G=C\oplus{}D\isom\Z/N\Z\oplus\Z/p\Z$ and $D$ is cyclic of order $p$. Then \begin{align*} \alpha:(E,G)&\mapsto(E,C)\\ \beta:(E,G)&\mapsto(E/D,(C+D)/D)\end{align*} Now we now translate this into the language of complex analysis. The first map $\alpha$ corresponds to the map $$\Gamma_0(pN)\backslash\H\into\Gamma_0(N)\backslash\H$$ induced by the inclusion $\Gamma_0(pN)\hookrightarrow\Gamma_0(N)$. The second map $\beta$ is constructed by composing the map $$\Gamma_0\backslash\H\iso \Bigl(\begin{matrix}p&0\\0&1\end{matrix}\Bigr) \Gamma_0(pN) \Bigl(\begin{matrix}p&0\\0&1\end{matrix}\Bigr)^{-1} \backslash\H$$ with the map to $\Gamma_0(N)\backslash\H$ induced by the inclusion $$\Gamma_0(N) \subset \Bigl(\begin{matrix}p&0\\0&1\end{matrix}\Bigr) \Gamma_0(pN) \Bigl(\begin{matrix}p&0\\0&1\end{matrix}\Bigr)^{-1}$$ The maps $\alpha$ and $\beta$ induce maps $$\alpha^{*},\beta^{*}: H^0(X_0(N),\Omega^1)\into H^0(X_0(pN),\Omega^1).$$ We can identify $S_2(\Gamma_0(N))$ with $H^0(X_0(N),\Omega^1$ by sending the cusp form $f(\tau)$ to the holomorphic differential $f(\tau)d\tau$. Now we get maps $$\alpha^{*},\beta^{*}:S_2(\Gamma_0(N))\into{}S_2(\Gamma_0(pN)).$$ \begin{exercise} Show that $\alpha^{*}(f)=f$ thought of as a cusp form with respect to the smaller group $\Gamma_0(pN)$. Then show that $$\beta^{*}(f)=p\sum_{n=1}^{\infty}a_n{}q^{pN}.$$ (Ribet was unsure whether the factor of $p$ should be there.) \end{exercise} \section{Generalities on Correspondences} Let $X$, $Y$, and $C$ be curves and let $\alpha$ and $\beta$ be nonconstant holomorphic maps so that we have the correspondence $$\corr{C}{X}{Y}{\alpha}{\beta}$$ Then we have maps $$H^0(X,\Omega^1)\xrightarrow{\alpha^{*}}H^0(C,\Omega^1) \xrightarrow{\beta_{*}}H^0(Y,\Omega^1).$$ The composition $\beta_{*}\circ\alpha^{*}$ is a map $$H^0(X,\Omega^1)\into H^0(Y,\Omega^1).$$ Switching the roles of $X$ and $Y$ gives a map $$H^0(Y,\Omega^1)\into H^0(X,\Omega^1).$$ In this context we can identify $T_p$ by viewing it as the map $$\beta_*\circ\alpha^*:H^0(X_0(N),\Omega^1) \into{}H^0(X_0(N),|Omega^1)$$ and using the fact that $S_2(\Gamma_0(N))=H^0(X_0(N),\Omega^1)$. One should recover the explicit representation $$T_p:\sum a_n q^n\mapsto\sum a_{np}q^n+p\sum a_n q^{np}.$$ Now lets think more generally about correspondences. Suppose $\varphi:X\into Y$ is a map of curves. Let $\Gamma\subset X\cross Y$ be the graph of $\varphi$. This gives a stupid correspondence $$\corr{\Gamma}{X}{Y}{\alpha}{\beta}$$ We can reconstruct $\varphi$ since $\varphi(x)=\beta(\alpha^{-1}(x))$. More generally suppose $\alpha:\Gamma\into X$ has degree $d\geq 1$. View $\alpha^{-1}(x)$ as a divisor on $\Gamma$. Then $\beta(\alpha^{-1})$ is a divisor on $Y$. We thus get a map $$\Div