%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1%% ribetofficial.tex --- Scribe notes for Ken Ribets Spring 1996 course. %%2%% %%3%% This file should compile with standard latex, the amssymb package, %%4%% and the diagrams package. The diagrams package is contained in the %%5%% file "diagrams.tex". Obtain it and put it in the same directory %%6%% as this file. %%7%% %%8%% This file is being maintained by William Stein ([email protected]).%%9%% This file was assembled by Lawren Smithline ([email protected]).%%10%% %%11%% Last modification date: 9/25/96. %%12%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1314\documentclass[12pt]{report}1516%% Uncomment the following two lines if your system doesn't17%% define the \mathfrak font.18%\font\german=eufm10 scaled \magstep 119%\def\mathfrak{\german}2021\input diagrams22\usepackage{amssymb}23\marginparwidth 0pt24\oddsidemargin 0pt25\evensidemargin 0pt26\marginparsep 0pt27\topmargin 0pt28\textwidth 6.5in29\textheight 8.5 in30\def\nn{{\rm N}} % Roman N in math mode31\def\aa{{\cal A}}32\def\rr{{\cal R}}33\def\flt{F{\sc l}T} % Fermat's little theorem34\def\pf{{\sc Proof.\ }}35\def\st{\ \cdot\backepsilon\cdot\ } % ``such that'' symbol36\def\vs{\vspace{4pt}}37\def\A{{\bf A}} % Adele ring38\def\C{{\bf C}} % complex nos.39\def\Z{{\bf Z}} % integers40\def\F{{\bf F}} % field41\def\P{{\bf P}} % projective land424344\def\N{{\mathfrak N}} % index of an ideal in a number ring (norm)45\def\H{{\mathfrak H}} % the complex upper halfplane46\def\R{{\bf R}} % reals47\def\Q{{\bf Q}} % rationals48\def\O{{\cal O}} % ring of integers49\def\T{{\bf T}} % Hecke algebra50\def\G{{\bf G}} % group scheme51\def\X{{X(N)}} % Modular curve52\def\E{{E_j}}53\def\K{{\overline{K}}}54\def\ff{{\cal F}} % Modular function field5556\def\Qp{\Q_p} % p-adic numbers57\def\Zp{\Z_p} % p-adic integers5859\def\0{{\bf 0}} % zero structure60\def\a{\alpha} % your basic Greeks61\def\b{\beta}62\def\c{\gamma}63\def\d{\delta}64\def\e{\epsilon}65\def\f{\zeta}66\def\G{\Gamma}67\def\ka{\kappa}68\def\la{\lambda}69\def\t{\tau}70\def\om{\omega}71\def\Om{\Omega}72\def\rh{{\overline % rho bar73{\rho_E}}}74\def\rhla{{\rho_\lambda}}75\def\th{{\theta}}76\def\g{\gamma}7778\def\p{{\mathfrak p}} % Gothic p, a prime ideal79\def\q{{\mathfrak q}} % Gothic q80\def\M{{\mathfrak m}} % maximal ideal81\def\Tm{{\T_\M}} % T localized at m82\def\ia{{\mathfrak a}}83\def\ib{{\mathfrak b}}84\def\si{\sigma}85\def\cent{{\bf C}} % centralizer86\def\qed{\hfill $\blacksquare$\smallskip}8788\def\tate{{\rm Ta}} % Tate module89\def\tatel{{{\rm Ta}_\l}} % l-adic Tate module90\def\tatem{{{\rm Ta}_\M}} % m-adic Tate module91\def\tatels{{{\rm Ta}_\l^*}} % contrav. l-adic T. m.92\def\tatems{{{\rm Ta}_\M^*}} % contrav. m-adic T. m.9394\def\ve{\varepsilon} % epsilon the character9596\def\lan{\langle} % angle brackets97\def\ran{\rangle}98\def\<{{\langle}} % < bracket99\def\>{{\rangle}} % > bracket100101\def\l{\ell} % script l as in l-adic.102\def\Ql{{\Q_\l}} % l-adic numbers103\def\Zl{{\Z_\l}} % fewer l-adic numbers104\def\sigmaonf{105{}^\sigma\!f} % sigma acting on f, from upper left hand corner106\def\GL{{\rm GL}} % general linear group107\def\SL{{\rm SL}} % special linear group108\def\gal{{\cal G}al} % Galois group109\def\GQ{\gal(\overline\Q/\Q)}110% abs Galois gp of Q111\def\GQp{\gal(\overline\Qp/\Qp)}112% abs Galois gp of Qp113\def\modgp{\SL_2(\Z)} % modular group114\def\abcd{\left( % 2 x 2 matrix a b // c d115\begin{array}{cc}116a&b\\c&d\end{array}\right)}117\def\isom{\cong}118\def\tensor{\otimes}119120\def\tr{{\rm tr}}121\def\frob{{\rm frob}}122\def\mod{\ {\rm mod}\,}123\def\im{{\rm im}}124\def\ord{{\rm ord}}125\def\rank{{\rm rank}}126\def\aut{{\rm Aut}}127\def\Hom{{\rm Hom}}128\def\plim{{\displaystyle\lim_{\longleftarrow}\,}} % Projective limit129\def\dlim{{\displaystyle\lim_{\longrightarrow}\,}} % Direct limit130\def\plimr{{\displaystyle\lim_{131\buildrel\longleftarrow\over r}\,}} % Projective limit over r132\def\Div{{\rm Div}}133\def\det{{\rm det }} % Determinant134\def\endo{{\rm End}}135\def\Cot{{\rm Cot}} % contangent space136\def\ver{{\rm ver}} % Vershibung endmorphism137138\def\inj{\hookrightarrow}139\def\into{\rightarrow}140\def\onto{\twoheadrightarrow}141\def\isomap{{\buildrel \sim\over\rightarrow}}142\newarrow{To} ---->143\newarrow{Line} -----144145\newtheorem{thm}{Theorem}146\newtheorem{dfn}{Definition}147\newtheorem{prop}{Proposition}148\newtheorem{lem}{Lemma}149\newtheorem{cor}{Corollary}150\newtheorem{res}{Result}151\newtheorem{eg}{Example}152\newtheorem{claim}{Claim}153\newtheorem{exercise}{Exercise}154\newtheorem{remark}{Remark}155\newtheorem{conj}{Conjecture}156\newtheorem{note}{Note}157158\begin{document}159\title{Scribe notes for Ken Ribet's Math 274}160\author{}161\date{\null}162\maketitle163\section*{January 17, 1996}164\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}165\bigskip166167\noindent168Here are some topics to be discussed in this course:169170\begin{tabular}{l}171Galois representations and modular forms, \\172Hecke algebras, \\173modular curves, and174Jacobians (Abelian varieties).175\end{tabular} \\176This lecture is a brief overview of some connection between these concepts,177and also an exercise in name-dropping.178179We can describe an an elliptic curve, or the Jacobian of a higher genus180curve, or abelian variety using a lattice. For $E$, the lattice is $L =181H_1(E(\C),\Z) \hookrightarrow \C,$ by the map $\c \mapsto \int_\c182\omega.$183184Weil considered curves over a finite field, $k$ of characteristic $p$.185There is an algebraic definition of $L/nL$ for $n \geq 1, \ \gcd(n, p) = 1.$186$$E[n] = \{ P \in E(\bar k) : nP = 0\} = \textstyle\frac1n L/L = L/nL.$$187For example, let $n = \l^\nu$ for $\nu \geq 1$, and $\l$ a prime different188from $p.$189190Weil further considered the limit $$E[\l^\infty] = \bigcup_{\nu = 1}^\infty191E[\l^\nu].$$192Tate oberserved there is a map $E[\l^\nu] \stackrel{\l}\rightarrow193E[\l^{\nu-1}],$ and so of course this inverse limit, $$194\lim_\leftarrow E[\l^\nu] = T_\l(E)$$ is called the Tate module.195196As $E[n]$ is free of rank 2 over $\Z / n\Z$, so is $T_\l(E)$ free of rank 2197over $\Z_\l$. Also, $V_l(e) = T_\l(E) \otimes \Q_\l$ is a 2 dimensional198vector space over $\Q_\l$. This is the first example of $\l$-adic \'etale199cohomology. For topological space $X$, $X \mapsto H_{\mathaccent 19200et}^i(X/\bar k, \Q_\l).$201202Here are the names of some cool folks: Taniyama Shimura Mumford Tate.203204Now, elliptic curve $E/\Q$ gets an action of $G = \gal(\bar \Q/\Q)$,205and so does $E[n]$. I.e. $\si(P+Q) = \si(P) + \si(Q).$ So we have a206homomorphism $\rho: G \rightarrow {\rm Aut}(E[n]) = GL_2(\Z/n\Z).$207Since we have an exact sequence $$1 \rightarrow \ker \rho \rightarrow G208\rightarrow {\rm im}\, \rho \rightarrow 0,$$209we get a tower of fields210$$\bar \Q \rightarrow K \rightarrow \Q,$$ and ${\cal G}al(K/\Q) =211{\rm im}\, \rho \subset GL_2( \Z/n\Z).$212\smallskip213214And now for something completely different. We can also get to these215Galois representations via modular forms. Let $k$ be the weight, such as2162. Let $N$ be the level. The complex vector space $S_k(N)$ is the set of217cusp forms on $\Gamma_1(N)$, a finite dimensional vector space, namely, the218set of holomorphic functions $f$ on $\H$ such that219$$ f((az+b)/(cz+d)) = (cz+d)^kf(z)$$ for $$\left( \begin{array}{cc} a & b220\\221c & d \end{array} \right) \in \Gamma_1(N), \ \ a,d \equiv 1, c \equiv 0 (N).$$222Such an $f$ has a power series (or Fourier series) expansion in $q =223\exp(2\pi i z)$: $$f(z) =224\sum_1^\infty c_n q^n.$$225Here is a famous example observed by Ramanujan, and proved by Mordell using226(his) Hecke operators:227$$ q\prod_1^\infty (1 - q^n)^{24} = \sum_1^\infty \tau(n)q^n,$$228for Ramanujan's $\tau$ function. Now, $\tau(n)\tau(m) =\tau(nm)$ for229$\gcd(n,m) = 1$. Also, there is a recurrence for prime powers.230Amusingly, the normalized basis element of $S_{12}(1)$ is $$\Delta =231\sum_1^\infty \tau(n)\exp(2\pi i nz).$$ Even more amusingly, $$\tau(n)232\equiv \sum_{d \mid n} d^{11} \ \ (691).$$233\smallskip234235Experience and Shimura have shown that there exist $f \in S_k(N)$ such that236$$T_n(f) =c_n \cdot f$$ for all $n \geq 1$ for some scalars $c_n$, and237$$f = \sum c_n q^n$$ for the same $c_n$, and that these $c_n$ are algebraic238integers in a finitely generated number field. That is, $[\Q(c_n: n\geq 1):239\Q]$ is finite.240241How can we study and interpret this? We start with the Hecke ring,242$$\Q[T_n] \subseteq End(S_k(N)).$$243Serre in 1968 said there should be Galois representations attached to forms244of arbitrary weight. Deligne constructed them. In a broad stroke, one can245say that we get between Galois representations and modular forms via246Frobenius elements.247248Next time, we continue with the semihistorical overview.249\section*{January 19, 1996}250\noindent{Scribe: William Stein, \tt <was@math>}251\bigskip252253\noindent {\bf\large Modular Representations and Modular Curves} \smallskip254255\subsection*{Arithmetic of Modular Forms}256Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is257an eigenform for the Hecke operators. The Mellin transform associates258to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} n^{-s}{a_n}$.259Let $K=\Q(a_1,a_2,\ldots)$. One can show that the $a_n$ are algebraic260integers and $K$ is a number field. When $k=2$, $f$ is associated261to $f$ an abelian variety262$A_f$ over $\Q$ of dimension $[K:\Q]$, and $A_f$ has a $K$ action. (See263Shimura,264{\em Introduction to the Arithmetic Theory of Automorphic Functions},265Theorem 7.14.)266267\begin{eg}[Modular Elliptic Curves]268If $a_n\in\Q$ for all $n$, then $K=\Q$ and $[K:\Q]=1$. In this case, $A_f$269is a one dimensional abelian variety, which is an elliptic curve, since270it has nonzero genus.271An elliptic curve arising in this way is called modular.272\end{eg}273274\begin{dfn}275Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is276a morphism $E_1\into E_2$ of algebraic groups, which has a277finite kernel.278\end{dfn}279280The following conjecture motivates much of the theory.281282\begin{conj}283Every elliptic curve over $\Q$ is modular,284that is, isogenous to a curve constructed in the above way.285\end{conj}286287For $k\geq 2$, Serre and Deligne found a way to associate to $f$ a family288of $\l$-adic representations. Let $\l$ be a prime number and $K$ be as289above. It is well known that $$K\otimes_{\Q} \Q_{\l}\isom290\prod_{\la|\l}K_{\la}.$$291One can associate to $f$ a family of representations292$$293\rho_{\l,f}:G=\gal(\overline{\Q}/\Q)294\rightarrow\GL(K\otimes_{\Q}\Q_{\l})295$$296unramified at all primes $p\not|\l N$.297By unramified we mean that for all primes $P$ lying over $p$,298the inertia group of the decomposition group at $P$ is contained299in the kernel of $\rho$. (The decomposition group $D_P$ at $P$ is the300set of those $g\in G$ which fix $P$ and the inertia group301is the kernel of the map $D_P\rightarrow302\gal(\O/P)$, where $\O$ is the ring of all algebraic integers.)303304305Now $I_P\subset D_P \subset \gal(\overline{\Q}/\Q)$ and306$D_P / I_P$ is cyclic, since it is isomorphic to a subgroup of the307galois group of a finite extension of finite fields.308So $D_P / I_P$309is generated by a Frobenious automorphism $\frob_p$ lying over $p$.310We have311$$312\tr(\rho_{\l,f}(\frob_p)) = a_p\in K \subset K\otimes \Q_{\l} $$313and314\begin{equation}\label{detrho}315\det(\rho_{\l}) = \chi_{\l}^{k-1}\ve,316\end{equation}317where $\chi_{\l}$ is the $\l$th cyclotomic character and318$\ve$ is a Dirichlet character.319320\subsection*{Characters}321Let $f\in S_k(N)$. For all322$\abcd \in \modgp$ with $c\cong 0 \mod{N}$ we have323$$324f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k \ve(d) f(z),325$$326where $\ve:(\Z/n\Z)^*\rightarrow \C^*$327is a Dirichlet character mod $N$. If $f$ is an eigenform for328the diamond bracket operator $<d>$, (so that329$f|<d> = \ve(d) f$)330then $\ve$ actually takes values in $K$.331332Let $\phi_n$ be the mod $n$ cyclotomic character.333The map $\phi_n: G \rightarrow (\Z/n\Z)^*$ takes $g\in G$ to334the automorphism induced by $g$ on the $n$th cyclotomic335extension $\Q(\mu_n)$ of $\Q$, where we identify336$\gal(\Q(\mu_n)/\Q)$ with $(\Z/n\Z)^*$.337The $\ve$ appearing in (\ref{detrho})338is really the composition339$$340G\stackrel{\phi_n}\longrightarrow(\Z/n\Z)^*341\stackrel{\ve}\longrightarrow \C^*.342$$343344For each positive integer $\nu$ we consider the $\l^{\nu}$th345cyclotomic character on $G$,346$$347\phi_{\l^{\nu}}:G\rightarrow (\Z/\l^{\nu}\Z)^*.348$$349Putting these together give a map350$$351\phi_{\l^{\infty}}=\lim_{\stackrel\longleftarrow\nu}352\phi_{\l^{\nu}}:G\stackrel{\chi_{\l}}\longrightarrow\Z_{\l}^{*}.353$$354355\subsection*{Parity Conditions}356357Let $c\in\gal(\overline{\Q}/\Q)$ be complex conjugation.358We have $\phi_n(c)=-1$, so $\ve(c) = \ve(-1)$ and359$\chi_{\l}(c) = (-1)^{k-1}$. Let360$$\abcd361=\left(\begin{array}{cc} -1&0\\0&-1 \end{array}\right).$$362For $f\in S_k(N)$,363$$f(z) = (-1)^k\ve(-1)f(z),$$364so $(-1)^k\ve(-1) = 1$. Thus,365$$\det(\rho_{\l}(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$366The $\det$ character is odd so the representation367$\rho_{\l}$ is odd.368369\begin{remark} (Vague Question) How can one recognize representations370like $\rho_{\l,f}$ ``in nature''? Mazur and Fontaine have made371relevant conjectures. The Shimura-Taniyama conjecture can be reformulated372by saying that for any representation $\rho_{\l,E}$ comming373from an elliptic curve $E$ there is $f$ so that374$\rho_{\l,E}\isom \rho_{\l,f}$.375\end{remark}376377\subsection*{Conjectures of Serre (mod $\l$ version)}378Suppose $f$ is a modular form, $\l$ a rational prime,379$\la$ a prime lying over $\l$, and the representation380$$\rho_{\la,f}:G\rightarrow \GL_2(K_{\la})$$381(constructed by Serre-Deligne) is irreducible.382Then $\rho_{\la,f}$ is conjugate to a representation383with image in $\GL_2(\O_{\la})$, where $\O_{\la}$384is the ring of integers of $K_{\la}$.385Reducing mod $\l$ gives a representation386$$\overline{\rho}_{\la,f}:G\rightarrow\GL_2(\F_{\la})$$387which has a well-defined trace and det, i.e., the det and trace388don't depend on the choice of conjugate used to reduce mod389$\la$. One knows from representation theory that if390such a representation is semisimple then it is completely determined391by its trace and det. Thus if $\overline{\rho}_{\la,f}$ is irreducible392it is unique in the sense that it doesn't depend on the choice393of conjugate.394395We have the following conjecture of Serre which remains open.396\begin{conj}[Serre]397All irreducible representation of398$G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$399complex conjugation, are of the form $\overline{\rho}_{\la,f}$400for some representation $\rho_{\la,f}$ constructed as above.401\end{conj}402403\begin{eg}404Let $E/\Q$ be an elliptic curve and let405$\sigma_{\l}:G\rightarrow\GL_2(\F_{\l})$ be406the representation induced by the action of $G$407on the $\l$-torsion of $E$. Then $\det \sigma_{\l} = \phi_{\l}$408is odd and $\sigma_{\l}$ is usually irreducible, so Serre's conjecture409would imply that $\sigma_{\l}$ is modular. From this one can, under Serre's410conjecture, prove that $E$ is modular.411\end{eg}412413\begin{dfn}414Let $\sigma:G\rightarrow \GL_2(\F)$ ($\F$ is a finite field)415be a represenation of the galois group $G$. The we say that the416{\em representions $\sigma$ is417modular} if there is a modular form $f$, a prime $\la$, and an embedding418$\F\hookrightarrow \overline{\F}_{\la}$ such that419$\sigma\isom\overline{\rho}_{\la,f}$ over420$\overline{\F}_\la$.421\end{dfn}422423\subsection*{Wile's Perspective}424425Suppose $E/\Q$ is an elliptic curve and426$\rho_{\l,E}:G\rightarrow\GL_2(\Z_{\l})$427the associated $\l$-adic representation on the428Tate module $T_{\l}$. Then by reducing429we obtain a mod $\l$ representation430$$\overline{\rho}_{\l,E}=\sigma_{\l,E}:G431\rightarrow \GL_2(\F_{\l}).$$432If we can show this is modular for infinitely many $\l$433then we will know that $E$ is modular.434435\begin{thm}[Langlands and Tunnel]436If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they437are modular.438\end{thm}439440This is proved by using the fact that $\GL_2(\F_2)$ and441$\GL_2(\F_3)$ are solvable so we may apply ``base-change''.442443\begin{thm}[Wiles]444If $\rho$ is an $\l$-adic representation which is irreducible445and modular mod $\l$ with $\l>2$ and certain other reasonable446hypothesis are satisfied, then $\rho$ itself is modular.447\end{thm}448449%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%450451\section*{January 22, 1996}452\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}453(Note. These were done about two months after the fact, since the assigned454person didn't. So they are terse. It was a pretty easy lecture.)455\bigskip456457Today, we have a limited goal: to explain modular forms as functions on458lattices -- or elliptic curves. (See Serre's {\em Course in Arithmetic},459or Katz's paper in the Proceedings of the Antwerp Conference in the460Springer LNM series.)461462Let the level $N = 1$. Consider a weight $k$ cusp form $f$. For $\t \in463\H$, we have $$f( a\t+b / c\t+d) = (c\t+d)^k f(\t).$$ So $f(\t+1) =464f(\t)$. By the map $\t \mapsto q= \exp(2\pi i \t)$, we map $\H$ to the465punctured disc $ \{ z : 0 < |z| < 1\}$. We abuse notation and think of $f$466as a function on this disc. Since $f$ is a cusp form, $f$ extends to 0,467and $f(0) = 0$. So we have a $q$-expansion $$ f = \sum _{n=1}^{\infty} a_n468q^n. $$469470\subsection*{Lattices inside $\C$}471472Let $L = \Z\om_1 \oplus Z\om_2$. We may assume that $\om_1 / \om_2 \in473\H$. Let ${\mathfrak R}$ be the set of lattices in $\C$. $\SL_2 \Z$ acts474on the left of $M = \{ (\om_1, \om_2) : \om_1, \om_2 \in \H\}$ by475multiplication of the column vector. This action fixes the lattice.476477Here is the relation with elliptic curves. A lattice $L$ determines a478complex torus $\C / L$. There is a Weierstrass $\wp$ function on this479torus. Consider an elliptic curve $E$ over $\C$. There is a lattice given480by the inclusion $H_1(E(\C),\Z) \hookrightarrow \C$. Choose a nonzero $\om481\in H^0(E, \Om'_E).$ Then $\c \in H_1$ maps to $\int_\c \om \in \C$.482483So maybe we should think of ${\mathfrak R}$ as the set of pairs $\{ (E,484\om) \}.$485486We have a map $M / \C^\times \into \H$ sending $(\om_1, \om_2)$ to $\om_1 /487\om_2.$ Now take the quotient on the left by $\SL_2 \Z$:488$$ {\mathfrak R}/\C^{times} = \SL_2 \Z \backslash M / \C^\times489\longrightarrow \SL_2 \Z \backslash \H.$$490But this just is the space of elliptic curves over $\C$. So $f:\H491\rightarrow \C$ which is a modular form and $F:M \rightarrow \C$ satisfying492$F(\la L) = \la^{-k}F(L)$ amount to the same thing by a simple calculation.493494\subsection*{Hecke Operators}495496Let $F$ be a function on lattices. Define the Hecke operator $T_n$ as497$$T_n F (L) = \sum_{L' \subset L, (L:L') = n} F(L) n^{k-1}.$$498499The essential case on elliptic curves is for $n = \l$ a prime. In this500case, the $L'$ correspond to the $\l+1$ subgroups of order $\l$ of $(\Z /501\l\Z)^ 2$.502%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%503\section*{January 24, 1996}504\noindent{Scribe: William Stein, \tt <was@math>}505\bigskip506507\noindent {\bf \large More On Hecke Operators}\smallskip508509We consider modular forms $f$ on $\Gamma_1(1)=\modgp$, that510is, holomorphic functions on $\H\cup\{\infty\}$ which satisfy511$$f(\tau)=f(\frac{a\tau+b}{c\tau+d})(c\tau+d)^{-k}$$512for all $\abcd\in\modgp$. Using a Fourier expansion we write513$$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$514and say $f$ is a cusp form if $a_0=0$.515There is a correspondence between modular forms $f$ and516lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$517given by $F(\Z\tau+\Z)=f(\tau)$.518519\subsection*{Explicit Description of Sublattices}520The $n$th Hecke operator $T_n$ of weight $k$ is defined by521$$T_n(L)=n^{k-1}\sum_{{L'\subset L,\ (L:L')=n}} L'.$$522What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and523$$L'/nL\subset L/nL=(\Z/n\Z)^2.$$524Write $L=\Z\om_1+\Z\om_2$, let $Y_2$ be the cyclic subgroup525of $L/L'$ generated by $\om_2$ and let $d=\#Y_2$. Let526$Y_1=(L/L')/Y_2$. $Y_1$ is generated by the image527of $\om_1$ so it is a cyclic group of order $a=n/d$.528We want to exhibit a basis of $L'$. Let529$\om_2'=d\om_2\in L'$ and use the fact that $Y_1$ is530generated by $\om_1$ to write $a\om_1=\om_1'+b\om_2$531for $b\in\Z$ and $\om_1'\in L'$. Since $b$ is only532well-defined modulo $d$ we may assume $0\leq b\leq d-1$.533Thus534$$535\left(\begin{array}{c}\om_1'\\ \om_2'\end{array}\right)536=537\left(\begin{array}{cc}a&b\\0&d\end{array}\right)538\left(\begin{array}{cc}\om_1\\ \om_2\end{array}\right)539$$540and the change of basis matrix has determinent $ad=n$.541Since542$$\Z\om_1'+\Z\om_2'\subset L' \subset L=\Z\om_1+\Z\om_2$$543and $(L:\Z\om_1'+\Z\om_2')=n$ (since the change of basis matrix has544determinent $n$) and $(L:L')=n$ we see that $L'=\Z\om_1'+\Z\om_2'$.545546Thus there is a one-to-one correspondence between sublattices $L'\subset L$547of index $n$ and matrices548$\bigl(\begin{array}{cc}a&b\\0&d\end{array}\bigr)$549with $ad=n$ and $0\leq b\leq d-1$.550In particular, when $n=p$ is prime there $p+1$ of these. In general, the551number of such sublattices equals the sum of the positive divisors552of $n$.553554\subsection*{Action of Hecke Operators on Modular Forms}555Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular556form with corresponding lattice function $F$. How can we describe the557action of the Hecke operator $T_n$ on $f(\tau)$? We have558$$\begin{array}{ll}559T_nF(\Z\tau+\Z) & = n^{k-1}\displaystyle\sum_{560\stackrel{\stackrel{\stackrel{a,b,d}{ab=n}}{0\leq b<d}}\null561}562F((a\tau+b)\Z + d\Z)\smallskip \\563& = n^{k-1}\displaystyle\sum d^{-k} F(\frac{a\tau+b}{d}\Z+\Z)\smallskip \\564& = n^{k-1}\displaystyle\sum d^{-k} f(\frac{a\tau+b}{d})\smallskip \\565& = n^{k-1}\displaystyle\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\smallskip \\566& = n^{k-1}\displaystyle\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}}567\frac{1}{d}\displaystyle\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\smallskip \\568& = n^{k-1}\displaystyle\sum_{\stackrel{ad=n}{m'\geq 0}\null}569d^{1-k} c_{dm'}e^{2\pi i a m'570\tau}\smallskip \\571& = \displaystyle\sum_{{ad=n, \ m'\geq 0}} a^{k-1} c_{dm'}q^{am'}.572\end{array}$$573In the second to the last expression we574let $m=dm'$, $m'\geq 0$, then used the fact that the575sum576$\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$577is only nonzero if $d|m$.578579Thus580$$T_nf(q)=\sum_{{ad=n, \ m\geq 0}} a^{k-1}c_{dm} q^{am}$$581and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is582$$\sum_{{a|n, \ a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$583584\begin{remark}585When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong586to the $\Z$-module generated by the $c_m$.587\end{remark}588589\begin{remark}590Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is591$$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$592Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic593since its original definition is as a finite sum of holomorphic594functions.)595\end{remark}596597\begin{remark}598Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is599just $c_n$. As an immediate corollary we have the600following important result.601\end{remark}602603\begin{cor}604Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient605of $q$ for all $n\geq 1$, then $f=0$.606\end{cor}607608\begin{remark}609When $n=p$ is prime we get an interesting formula for the610action of $T_p$ on the $q$-expansion of $f$.611One has612$$T_p f = \sum_{\mu\geq 0} \sum_{{a|n, \ a|\mu}}a^{k-1}613c_{\frac{n\mu}{a^2}} q^{\mu}. $$614Since $n=p$ is prime either $a=1$ or $a=p$. When615$a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$616and when $a=p$, we can write $\mu=p\lambda$ and we get617terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$.618Thus619$$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+620p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$621\end{remark}622623624625626627\section*{January 26, 1996}628\noindent{Scribe: Amod Agashe, \tt <amod@math>}629\bigskip630631Following the notation of the last few lectures, let $M_k$ denote632the space of modular forms of weight $k$ for $SL_2(\Z)$ and $S_k$633denote the subspace of cusp forms.634635Then we have:636637\begin{prop}. $M_k$ is a finite dimensional $\C$-vector space and is generated638by modular forms having the coefficients of their Fourier expansion in $\Q$.639\end{prop}640{\sc Sketch of Proof}. (For details, refer Serre's ``A course in641Arithmetic'' or Lang's ``Introduction to modular forms''.) \smallskip642643\noindent644The key ingredient that goes into proving finite dimensionality is645the following result, which can be obtained by contour integration:646647Let $f\in M_k$ and let $D=\{z\in \C :Im(z)>0, \mid z\mid \geq 1,648\mid Re(z)\mid \leq 1/2\}$ be the fundamental domain for $SL_2(\Z)$. Then649$$\sum_{p \in D\cup\infty} \frac{1}{e_p} ord_p(f) = \frac{k}{12} $$650where651$$ e_p = \frac{1}{2} \# \{\gamma \in SL_2(\Z) : \gamma p =p \}652= \frac{1}{2} \# Aut(E_p)$$653Here, the latter equality follows from the observation that the654category of elliptic curves over $\C$ with isogenies is the same655as the category of lattices in $\C$ upto homothety with maps being656multiplication by elemets of $\C$. One can657show that the invertible maps that preserve the lattice $\Z \oplus658\Z p$ are in one-to-one correspondence with the set659$\{\gamma \in SL_2(\Z) : \gamma p =p \}$ and hence the latter equality.660661In particular,662$$ e_p = \left\{ \begin{array}{ll}6632 & \mbox{if $p=i$} \\6643 & \mbox{if $p=\sqrt[3]{-1}$} \\6651 & \mbox{otherwise}666\end{array}667\right. $$668669Using this formula and relating the dimensions of $M_k$ and670$S_k$, one can show that $M_k$ is finite dimensional and also671explicitly calculate its dimension.672673To get a basis with Fourier coefficients in $\Q$, first observe674that $M_k$ is generated by the set of Eisenstein series $G_k$ for all $k$.675As a function on the complex upper half plane,676the Eisenstein series $G_k$ for $k \in \Z$ and $k>1$ is given by677$$G_k(\tau) = \sum_{(m,n) \in \Z^2 \backslash (0,0)678}679\frac{1}{(m\tau+n)^k}$$680One can then show that681$$G_k(\tau) = \frac{1}{2}\zeta(1-k) + \sum_{k=1}^\infty \sigma_{k-1}(n)q^n$$682where $q=e^{2\pi i \tau}$, $\zeta$ is the Riemann zeta function and683$$\sigma_k(n) = \sum_{d\mid n}d^k$$684There is a theorem due to Euler which states that $\zeta(1-k)= -\frac{b_k}{k}$685where $b_k$ are the Bernoulli numbers defined by the following power series686expansion:687$$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{b_kx^k}{k!} $$688The constant term of $G_k$ is thus $-b_k/{2k}$, which is rational.689Thus the Fourier expansion of $G_k$ has rational coefficients and thus690we have found a basis with rational coefficients.691\bigskip692693Next, let $V$ be a subspace of $M_k$ which is stable under the694action of all the Hecke operators $T_n$. For example, observing695that $T_n(G_k)=\sigma_{k-1}(n)G_k$, we see that $V=\C(G_k)$ is696one such subspace.697698Let $\T=\T(V)$ = $\C$-algebra generated by the $T_n$'s inside $End(V)$699= $\C$-vector space generated by the $T_n$'s inside $End(V)$.700The latter equality of sets follows because the product of two Hecke701operators can be expressed as a linear combination702of finitely many Hecke operators.703704For $k>0$, we define a bilinear map705$\T \times V \rightarrow \C$ by706$$(T,f) \mapsto a_1(f\mid T).$$707708\begin{prop}. The induced maps $\T \rightarrow Hom(V,\C)$ and709$V \rightarrow Hom(\T, \C)$ are isomorphisms.710\end{prop}711\pf712We first show that the maps are injective.\\713Injectivity of the second map:714$$\begin{array}{l}715f \in V \mapsto 0 \\716\Rightarrow a_1(f \mid T) = 0 \ \forall T\in \T \\717\Rightarrow a_1(f \mid T_n) = 0 \ \forall n \\718\Rightarrow a_n(f) = 0 \ \forall n \geq 1 \\719\Rightarrow f\mbox{ is constant} \\720\Rightarrow f=0\mbox{ if }k>0721\end{array}$$722Injectivity of the first map:723$$\begin{array}{l}724T \in \T \mapsto 0 \\725\Rightarrow a_1(f \mid T)=0 \ \forall f\in V \\726\Rightarrow a_1((f\mid T_n)\mid T) = 0 \ \forall f\in V \\727\Rightarrow a_1((f\mid T)\mid T_n) = 0 \ \forall f \in V \\728\Rightarrow a_n(f\mid T) = 0 \ \forall n>0, \forall f\in V \\729\Rightarrow f\mid T = 0 \ \forall k>0, \forall f\in V \\730\Rightarrow T = 0731\end{array}$$732733In the fourth line, $f$ is replaced by $f \mid T_n$.734Next observe that $V$, being a subspace of $M_k$, is finite dimensional.735Hence we have from the injectivities of both the above maps that each map736is actually an isomorphism. \smallskip737738A map $\phi: M \rightarrow N$ of $\T$-modules is said to be $\T$-equivariant739if $\phi (Tm) = T\phi(m) \ \forall m\in M$.740741\begin{prop}.742The isomorphisms $\T \cong Hom(V,\C)$743and $V \cong Hom(\T, \C)$744as defined above are $\T$-equivariant.745\end{prop}746\pf747Consider the first map.748Here is the $\T$-module structure on $Hom(V,\C)$.749Given $\psi \in Hom(V,\C)$, i.e. $\psi: V \rightarrow \C$,750define $T\psi: V \rightarrow \C$ by $(T\psi) (f)=\psi (f\mid T)$.751Let $\beta$ denote the map $\T \rightarrow Hom(V, \C)$.752Then given $T'\in \T$ and $T \in \T$,753we have to show that $\beta (T(T'))=T(\beta (T'))$. Let $f \in V$.754Now $(\beta (TT'))(f) = a_1(f\mid TT')$,755while $(T(\beta T'))(f) = \beta (T') (f\mid T) = a_1((f\mid T) \mid T')756= a_1(f\mid TT')$.757Thus $(\beta (TT'))(f)=(T(\beta (T')))(f) \ \forall f\in V$ and we758are done.759760\noindent Next consider the second map.761We define the $\T$-module structure on $Hom(\T,\C)$.762Given $\phi \in Hom(\T,\C)$, i.e. $\phi: \T \rightarrow \C$,763define $T\phi: \T \rightarrow \C$ by $(T\phi) (T')=\phi (TT')$.764Let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$.765Then given $f\in V$ and $T \in \T$,766we have to show that $\alpha (T(f))=T(\alpha (f))$. Let $T' \in \T$.767Now $$(\alpha (Tf))(T') = a_1((f\mid T) \mid T') = a_1(f\mid TT'),$$768while $$(T(\alpha f))(T') = \alpha (f) (TT') = a_1(f\mid TT').$$769Thus $$(\alpha (Tf))(T')=(T(\alpha f))(T') \ \forall T'\in \T$$770and we are done.771772\begin{dfn}.773An element $f$ of $M_k$ is said to be an eigenform if it is an eigenfunction774for all the Hecke operators.\\775i.e. $f\mid T_n = \lambda_n f$ for some $\lambda_n \in \C \ \forall n\geq 1$.776\end{dfn}777Let $f$ is an eigenform with eigenvalues $\lambda_n$. Then778$a_n(f) = a_1(f\mid T_n) = a_1(\lambda_n f) = \lambda_n a_1(f) \ \forall n\geq 1$. \\779Thus if $a_1(f) = 0$, then $a_n(f) = 0 \ \forall n\geq 1$. \\780If $k>0$ then this implies $f=0$.781Hence if $k>0$, then if $f\neq 0$, we can normalize $f$ to782$\frac{1}{a_1(f)}f$.783784\begin{dfn}.785An eigenform $f$ is said to be normalized if $a_1(f) = 1$.786\end{dfn}787788If $f$ is a normalized eigenform with eigenvalues $\lambda_n$,789then $a_n(f)=\lambda_n$ and $f\mid T_n = \lambda_n f = a_n(f)f$.790791% Given $f \in M_k$ define $\phi_f: \T \rightarrow \C$ by792% $\phi_f(T) = a_1(f\mid T)$. Note that if $\alpha$ denotes the map793% $V \rightarrow Hom(\T, \C)$ induced by the bilinear pairing mentioned794% before, then $\phi_f$ is just $\alhpa(f)$.795% Then we have: \\796797Again, let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$798induced by the bilinear pairing mentioned earlier. Then if $f \in V$,799we have the map $\alpha(f): \T \rightarrow \C$.800801\begin{prop}.802Let $f$ be an eigenform.803Then $f$ is normalized $\Leftrightarrow$ $\alpha(f)$ is a ring homomorphism.804\end{prop}805\pf806If $f=0$ then the statement is trivial. So assume $f\neq 0$.807Then as discussed above, $a_1(f)\neq 0$. Also recall808from the same discussion that if $f\neq 0$ is an eigenform, then809$$f\mid T_n = \frac{a_n(f)}{a_1(f)}f.$$ For ease of notation, let810$\alpha_f$ denote the map $\alpha(f)$. So811$$\begin{array}{l}812\alpha_f(T_nT_m) \\ = a_1(f\mid T_nT_m) \\ = a_1((f\mid T_n)T_m)813\\ = a_m(f\mid T_n) \\ = a_m((a_n(f)/a_1(f))f) \\ = a_m(f)a_n(f)/a_1(f).814\end{array}$$815We have $$\alpha_f(T_n)\alpha_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m)816= a_n(f) a_m(f).$$817The following are equivalent:818819\begin{tabular}{l}820$\alpha_f$ is a ring homomorphism, \\821$\alpha_f(T_nT_m) = \alpha_f(T_n)\alpha_f(T_m) \ \forall T_n,T_m$, \\822$a_1(f) = 1$, and \\823$f$ is normalized.824\end{tabular} \\825The first implication follows because $\T$ is generated by the $T_n$'s.826827% We will only prove the ``$\Rightarrow$'' part and leave the reverse828% implication as an exercise. \\829% Assuming that $f$ is normalized, we have to show that830% $(\alpha(f))(T_nT_m) = (\alpha(f))(T_n)\phi_f(T_m) \ \forall n,m \geq 1$.831% Now $(\alpha(f))(T_nT_m) = a_1(f\mid T_nT_m) = a_1((f\mid T_n)T_m)832% = a_m(f\mid T_n) = a_m(a_n(f)f) = a_m(f)a_n(f)$. \\833% While $(\alpha(f))(T_n)\phi_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m)834% = a_n(f) a_m(f)$. \\835% Thus the two are equal and we are done.836837838\section*{January 29, 1996}839\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}840\bigskip841842Today we'll consider questions of rationality and integrality.843(References: Serre: {\em A Course in Arithmetic} and Lang: {\em844Introduction to Modular Forms}.) Let845$S=S_k$ be the space of cusp forms of weight $k$. Let846\[ S(\Q) = S_k \cap \Q [[ q ]] \]847and848\[ S(\Q) \supseteq S(\Z) = S_k \cap \Z [[ q ]]. \]849The following fact is easy to prove using explicit formul\ae: $S_k$850has a $\C$-basis consisting of forms with integral coefficients (see851Victor Miller's construction below).852853Recall that for all even $k\geq4$, there is an Eisenstein series854\[ G_k = \frac{-b_k}{2k} +855\sum^\infty_{n=1} \left ( \sum_{d|n} d^{k-1} \right) q^n, \]856which is a modular form of weight $k$. Renormalize this to obtain857\[ E_k = \frac{2k}{-b_k}\cdot G_k = 1 + \cdots. \]858The first few Bernoulli numbers of even positive index are $b_2=1/6$,859$b_4=-1/30$, $b_6=1/42$, $b_8=-1/30$, $b_{10}=5/66$, $b_{12}=-691/2730$.860The fact that the first four of these have numerator 1 is closely861related to the arithmetic of cyclotomic fields.862863The modular forms $E_4$ and $E_6$ have $q$-expansions with constant864terms equal to $1$, and all coefficients in $\Z$. The functions865$E_4^aE_6^b$ with $4a+6b=k$ form a basis for $M_k$.866It is easy to see that they are modular forms.867From the formula that the (weighted) number of zeros of any modular form868of weight $k$ is $k/12$, we deduce that $E_4$ has a simple zero at869$\rho$, and $E_6$ hasn't got a zero at $\rho$. Hence $E_4^aE_6^b$ has a870simple zero of order $a$ at $\rho$ so these expressions are linearly871independent over $\C$.872To873show that they span $M_k$, consider the following modular form of weight874$12$:875\[ \Delta = (E_4^3 - E_6^2)/1728. \]876Here the coefficient of $q$ is a simple number: $1$. $\Delta$ has a877simple zero at $\infty$. Since a cusp form of weight $k$ has $k/12$878zeros, it follows that (since879the weighting $e_\infty$ is $1$), that $\Delta$ does not vanish880anywhere on $\H$. Therefore $S_{k+12}=\Delta\cdot M_k$.881Since $E_k(i\infty)=1\neq0$, it follows that882$M_k=E_k\cdot\C\oplus S_k=E_k\cdot\C\oplus\Delta\cdot M_{k-12}$. Hence883${\rm dim}\,M_k={\rm dim}\,M_{k-12}+1$. Again using the fact that $f\in884M_k$ has $k/12$ zeros, we quickly deduce that the dimensions of $M_0$,885$M_2$, $M_4$, $M_6$, $M_8$, $M_{10}$ are $1$,$0$,$1$,$1$,$1$,$1$886respectively. (e.g., for $k=4$ any modular form must have just a simple887zero at $\rho$. So for any $f\in M_4$, it is the case that888$f(\tau)-f(i\infty)E_4(\tau)$ vanishes at889$\tau=i\infty$ and hence is identically zero. So $E_4$ spans $M_4$.) Thus890we have determined ${\rm dim}\,M_k$ for all $k\geq0$, and it is easy to891see that this number is equal to the number of solutions to $4a+6b=k$892for $a,b\geq0$. Hence the $E_4^aE_6^b$ span $M_k$ and therefore we have893proved that they form a basis.894895The following construction comes from the first page of Victor Miller's896thesis. Let $d={\rm dim}_\C\, S_k$. Then there exist $f_1,\ldots,f_d\in897S(\Z)$ such that $a_i(f_j)=\delta_{ij}$ for $1\leq i,j\leq d$.898To show this, recall that $E_4\in M_4$ and $E_6\in M_6$ have899$q$-expansions with coefficients in $\Z$. $\Delta\in S_{12}$ has900constant coefficient $0$, and the coefficient of $q$ is $1$. Also901$\Delta\in S_{12}(\Z)$, as can be seen for example from the formula902\[ \Delta = q\prod_{n=1}^\infty (1-q^n)^{24}. \]903Now pick $a,b\geq0$ so that $14\geq 4a+6b\equiv k \pmod{12}$, with904$a=b=0$ when $k\equiv0\pmod{12}$. Note that then $12d+6a+4b=k$ by our905previous result on the dimension of $M_k$ (and the fact that the906dimension of $S_k$ is one less than that for $k\geq12$). Hence the907functions908\[ g_j = \Delta^j E_6^{2(d-j)+a}E_4^b \]909for $1\leq j\leq d$ will be cusp forms of weight $k$. By our previous910remarks on the coefficients of $\Delta$, $E_6$, $E_4$, we have $g_j\in911S_k(\Z)$ and912\[ a_i(g_j) = \delta_{ij} \]913for $i\leq j$. A straightforward elimination now yields the914$f_1,\ldots,f_d$ with the stated properties.915It is clear that these $f_1,\ldots,f_d$ are916linearly independent over $\C$, hence they form a basis of $S_k$.917918If you take $T_1,\ldots,T_d\in\T=\T(S_k)$, they are also linearly919independent: for given any linear relation920\[ \sum_{i=1}^d c_iT_i=0, \]921apply this to $f_j$ and look at the first coefficient922\[ 0 = a_1\left(f_j\left|\sum_{i=1}^d c_iT_i\right.\right)923= \sum_{i=1}^d c_ia_i(f_j) = \sum_{i=1}^d c_i\delta_{ij} = c_j, \]924hence the linear relation given in the first place was trivial, so925$T_1,\ldots,T_d$ form a basis for $\T(S_k)$, since they are linearly926independent, and ${\rm dim}_\C\,\T={\rm dim}_\C\, V=d$.927928Let ${\cal R}=\Z[\ldots T_n\ldots]\subseteq {\rm End}(S_k)$.929\begin{claim}930\[ {\cal R}=\bigoplus_{i=1}^d\Z T_i. \]931\end{claim}932933\pf Since the $T_i$ form a basis of $\T$,934we have $T_n =\sum_{i=1}^d c_{n_i}T_i$ with $c_{n_i}\in\C$. We need935to check that $c_{n_i}\in\Z$. With the $f_j$ as above, consider936$$937\begin{array}{lll}938a_n(f_j) & = a_1(f_j|T_n)939\\ & = a_1\left(f_j\left|\sum_{i=1}^d c_{n_i}T_i\right.\right)940\\ & = \sum_{i=1}^d c_{n_i}a_1(f_j|T_i)941\\ & = \sum_{i=1}^d c_{n_i}a_i(f_j)942\\ & = \sum_{i=1}^d c_{n_i}\delta_{ij} & = c_{n_j}.943\end{array}$$944Hence $c_{n_i}\in\Z$. \qed945946${\cal R}$ is called the {\em integral Hecke algebra}. It is a finite947$\Z$-module of rank $d$. We still have (from the last lecture) a948pairing949\begin{eqnarray*}950S(\Z)\times{\cal R} & \rightarrow & \Z \\951(f,T) & \mapsto & a_1(f|T).952\end{eqnarray*}953Now $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)\cong\Z^d$ by the954argument given before. Therefore $S(\Z)$ is a free $\Z$-module of955finite rank. But it also contains the $f_i$, so $S(\Z)\cong\Z^d$.956957What is $S(\Z)$ as an ${\cal R}$-module?958959\begin{exercise}960The map $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)$ is in fact an961isomorphism of $\T$-modules.962\end{exercise}963964{\sc Hint.\ } The cokernel is a torsion (in fact finite) group.965So if we show it torsion free, we are done.966967\begin{thm}968The $T_n$ are all diagonalizable on $S_k$.969\end{thm}970971$S_n$ supports a Hermitian non-degenerate inner product, the Petersson972inner product973\[ (f,g)\mapsto \langle f,g\rangle\in\C. \]974We have $\langle f,f\rangle\geq0$, with equality iff $f=0$. Furthermore975\[ \langle f|T_n,g\rangle = \langle f,g|T_n\rangle, \]976i.e., $T_n$ is self-adjoint with respect to the given inner product.977978An operator $T$ is {\em normal} if it commutes with $T^*$ (which denotes979its Hermitian transpose). Normal operators are diagonalizable (for a980proof, refer to Math H110). In our case, $T_n^*=T_n$, so this fact981applies. It is also true (same proof) that a commuting family of982semi-simple (i.e., diagonalizable) operators is simultaneously983diagonalizable.984985\pf Put together the above facts. \qed986987We can also prove that the eigenvalues are real. This depends on the988following trick. For $f\neq0$ consider989\[ a_n\langle f,f\rangle = \langle a_nf,f\rangle990= \langle f|T_n,f\rangle = \langle f,f|T_n\rangle991= \langle f,a_nf\rangle = \bar{a_n}\langle f,f\rangle. \]992$a_n\in\R$ now follows since $f\neq0$ implies993$\langle f,f\rangle\neq0$.994995\begin{exercise}996The $a_n$ are totally real algebraic integers.997\end{exercise}998999{\sc Hint.\ } The space $S_k$ is stable under the action of1000${\rm Aut}(\C)$ ``on the coefficients''. Given a cusp form1001\[f=\sum_{n=1}^\infty c_nq^n\]1002and some $\sigma\in{\rm Aut}(\C)$, define1003$\sigmaonf=\sum_{n=1}^\infty \sigma(c_n)q^n$. This function is in1004$S_k$, since $S_k$ has a basis in $S(\Q)$, which is fixed by $\sigma$.1005Then $f$ is an eigenform iff $\sigmaonf$ is an eigenform.10061007We'll use a lame definition of the {\em Petersson inner product} for1008this section. Let $z=x+iy\in\H$. Then we have a volume form1009$y^{-2}{dx\,dy}$ which is invariant under $GL^+_2(\R)$ (the subgroup1010of the general linear group of matrices of positive determinant).1011To prove this, note that1012$dz\wedge d\bar z = (dx+i\,dy)\wedge(dx-i\,dy) = -2i(dx\wedge dy)$,1013and hence1014\[ dx\,dy = dx\wedge dy = \frac{-1}{2i} dz\wedge d\bar z \]1015Then for any1016$\alpha = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)1017\in GL^+_2(\R)$ we can consider the usual action1018\[ \alpha: z \mapsto \frac{az+b}{cz+d}1019= \frac{(az+b)(c\bar z+d)}{|cz+d|^2} \]1020where the imaginary part of the result is1021$|cz+d|^{-2}(ad-bc){\rm Im}\, z = y|cz+d|^{-2}\det(\alpha).$1022As for differentials,1023\[ d\left(\frac{az+b}{cz+d}\right)1024= \frac{a(cz+d)\,dz-(az+b)c\,dz}{(cz+d)^2}1025= \frac{ad-bc}{(cz+d)^2}\,dz \]1026hence under application of $\alpha$, $dz\wedge d\bar z$ takes on a1027factor of1028$$\frac{\det(\alpha)}{(cz+d)^2}\frac{\det(\alpha)}{(c\bar z+d)^2}1029= \left( \frac{\det(\alpha)}{|cz+d|^2} \right)^2.$$1030This finally proves that the differential $y^{-2}{dx\,dy}$1031is invariant under the action of $GL^+_2(\R)$.10321033The formula for the {\em Petersson inner product} is1034\[ \langle f,g\rangle=1035\int_{SL_2(\Z)\setminus\H}\left(f(z)\overline{g(z)}1036y^k\right)\,\frac{dx\,dy}{y^2}.\]10371038This could be considered either an integral over the fundamental region1039or, noting that the integrand is invariant under $SL_2(\Z)$, an integral1040over the quotient space. One then checks that for $f,g$ cusp forms, the1041integral will converge, since they go down exponentially as $z$ tends to1042infinity. It is then clear that this inner product is Hermitian.10431044It is not immediately clear that the Hecke operators are self-adjoint.1045\section*{January 31, 1996}1046\noindent{Scribe: William Stein, \tt <was@math>}1047\bigskip10481049{\bf \large Modular Curves}\smallskip1050105110521053\subsection{Cusp Forms}1054Recall that if $N$ is a positive integer we define the congruence1055subgroups1056$\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by1057$$1058\begin{array}{cl}1059\Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\1060\Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\1061\Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv1062\bigl(\begin{array}{cc}1&0\\0&1\end{array}\bigr) \pmod{N}\}1063\end{array}1064$$10651066Let $\Gamma$ be one of the above subgroups.1067One can give a construction of the space $S_k(\Gamma)$ of cusp forms1068of weight $k$ for the action of $\Gamma$ using the language of1069algebraic geometry.1070Let $X_{\Gamma}=\Gamma\backslash\H^{*}$1071be the upper half plane (union the cusps)1072modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure1073of Riemann surface. Furthermore,1074$S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where1075$\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$.1076This works since an element of $H^0(X_{\Gamma},\Omega^1)$1077is a differential form $f(z)dz$, holomorphic on $\H$ and1078the cusps, which is invariant with respect to the action1079of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then1080$$d(\gamma(z))/dz=(cz+d)^{-2}$$1081so1082$$f(\gamma(z))d(\gamma(z))=f(z)dz$$1083iff $f$ satisfies the modular condition1084$$f(\gamma(z))=(cz+d)^{2}f(z).$$10851086There is a similiar construction when $k>2$.10871088\subsection{Modular Curves}1089$\modgp\backslash\H$ parametrizes isomorphism1090classes of elliptic curves. The other congruence subgroups also1091give rise to similiar parametrizations.1092$\Gamma_0(N)\backslash\H$ parametrizes pairs $(E,C)$ where1093$E$ is an elliptic curve and $C$ is a cyclic subgroup of order1094$N$, and $\Gamma_1(N)\backslash\H$ parametrizes pairs $(E,P)$ where1095$E$ is an elliptic curve and $P$ is a point of exact order $N$.1096Note that one can also give a point of exact order $N$ by giving1097an injection $\Z/N\Z\hookrightarrow E[N]$1098or equivalently an injection $\mu_N\hookrightarrow E[N]$1099where $\mu_N$ denotes the $N$th roots of unity.1100$\Gamma(N)\backslash\H$ parametrizes pairs $(E,\{\alpha,\beta\})$1101where $\{\alpha,\beta\}$ is a basis for1102$E[N]\isom(\Z/N\Z)^2$.11031104The above quotients spaces are {\em moduli spaces} for the1105{\em moduli problem} of determining equivalence classes of1106pairs ($E + $ extra structure).11071108\subsection{Classifying $\Gamma(N)$-structures}1109\begin{dfn}1110Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve}1111$E/S$ is a proper smooth curve1112$$\begin{array}{c} E \\ \bigl| \\ S\end{array}$$1113with geometrically connected fibers all of genus one, together with a1114section ``0''.1115\end{dfn}11161117Loosely speaking, proper generalizes the notion of projective,1118and smooth generalizes nonsingularity. See Chapter III, section 10 of1119Hartshorne's {\em Algebraic Geometry} for the precise definitions.11201121\begin{dfn}1122Let $S$ be any scheme and $E/S$ an elliptic curve.1123A {\bfseries $\Gamma(N)$-structure} on $E/S$ is1124a group homomorphism1125$$\varphi:(\Z/N\Z)^2\into E[N](S)$$1126whose image ``generates'' $E[N](S)$.1127\end{dfn}11281129See Katz and Mazur, {\em Arithmetic Moduli of1130Elliptic Curves}, 1985, Princeton University Press, especially1131chapter 3.11321133Define a functor from the category of $\Q$-schemes to the1134category of sets by sending a scheme $S$ to the1135set of isomorphism classes of pairs1136$$(E, \Gamma(N)\mbox{\rm -structure})$$1137where $E$ is an elliptic curve defined over $S$ and1138isomorphisms (preserving the $\Gamma(N)$-structure) are taken1139over $S$. An isomorphism preserves the $\Gamma(N)$-structure1140if it takes the two distinguished generators to the two1141distinguished generators in the image (in the correct order).11421143\begin{thm}1144For $N\geq 4$ the functor defined above is representable and1145the object representing it is the modular curve $X(N)$ corresponding1146to $\Gamma(N)$.1147\end{thm}11481149What this means is that given a $\Q$-scheme $S$, the1150set $X(S)=Mor_{\Q\mbox{\rm -schemes}}(S,X)$ is isomorphic to1151the image of the functor's value on $S$.11521153There is a natural way to map a pair $(E,\Gamma(N)\mbox{\rm -structure})$1154to an $N$th root of unity.1155If $P,Q$ are the distinguished basis of $E[N]$ we send1156the pair $(E,\Gamma(N)\mbox{\rm -structure})$ to1157$$e_N(P,Q)\in\mu_N$$1158where $$e_N:E[N]\times E[N]\into \mu_N$$ is the Weil pairing. For1159the definition of this pairing see chapter III, section 8 of1160Silverman's {\em The Arithmetic of Elliptic Curves}. The Weil pairing1161is bilinear, alternating, non-degenerate, galois invariant, and1162maps surjectively onto $\mu_N$.1163\section*{February 2, 1996}1164\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}1165(Note. These were done about a month after the fact, since the assigned1166person dropped the course. So they are terse.)1167\bigskip11681169Earlier, we looked at V. Miller's construction for eigenforms. (See, for1170instance, Lang X \S4 in the course references. This is a special miracle for1171$\SL_2(\Z)$.) Over $\Z$, $\T \cong \Z^d$ and $S(\Z)$ are dual, where $d =1172\dim S_k(\C)$.11731174Here is Shimura's explanation (Lang III \S5, VIII). The Hecke operator $\T_n$1175maps $S_k$ to itself. Let $A\subset \C$ be a subring and1176$$\T_A = A[\T_n : n > 0] \subseteq {\rm End}_\C S_k.$$1177Denote by $\T$, $\T_\Z$. There is a natural tensor product $\T_A \otimes_A1178\C \twoheadrightarrow \T_\C$.11791180There is also a complex conjugation automorphism of $S_k$ by1181$f \mapsto \overline{f(-\bar\tau)}$. This map is conjugate linear.1182The map $\tau \mapsto \exp(2\pi i \tau)$ becomes1183$\tau \mapsto \overline{\exp(-2\pi i \bar\tau)} = \exp(2 \pi i \tau)$.1184Say $f = \sum a_n q^n.$ Its conjugate1185$g = \sum \bar a_n q^n.$ If you know $S_k(\C) = \C \otimes_\Q S_k(\Q)$,1186then you know that modular forms can be conjugated in this sense.11871188There is an isomorphism $\T_\R \otimes_\R \C \stackrel\sim\longrightarrow1189\T_\C$ since the map is surjective and the complex dimensions on each side1190are equal.11911192Shimura (1959) exhibited the (Eichler-)Shimura isomorphism1193$$S_k(\C) \cong H^1(X_\Gamma,\R).$$1194We have $S_k(\C) = H^0(X_\Gamma, \Omega^1)$ and a map1195$H_1(X_\Gamma,\Z) \times S_k(\C) \rightarrow \C$.1196Now, $H_1(X,\Z) \cong \Z^{2d}$ embeds in $\Hom_\C(S_k(\C),\C) \cong1197\C^{2d}$ as a lattice.11981199So we have $S_k(\C) \rightarrow \Hom(H_1(X,\Z),\C))$ and1200$S_k(\C) \stackrel\sim\rightarrow \Hom(H_1(X,\Z),\R))$ as real vector1201spaces. By the Shimura isomorphism, this is isomorphic to $H^1(X,\R) \sim1202H^1_p(\Gamma,\R)$, the parabolic cohomology of $\Gamma$.12031204So $S_k(\C) \cong H^1_p(\Gamma, V_k)$, for a certain $d-1$ dimensional1205subspace $V_k$. (Let $W = \R \oplus \R$. $\Gamma$ acts by linear fractional1206transformation. Let $V_k = {\rm Sym}^{k-2} W$. There is a lattice in1207$S_k(\C)$1208corresponding to $H^1_p(\Gamma, {\rm Sym}^{k-2} \Z^2)$) We have an action1209of1210$\Gamma$ by $$f\cdot \c \mapsto \int_{\tau_0}^{\c(\tau_0)}1211f(\tau)\tau^{k-1}d\tau.$$12121213Recall $\T = \T_\Z$ is a set of endomorphisms of a lattice $L$, and $\T$1214has finite rank over $\Z$. We have the inclusion1215$S_k(\Z) \hookrightarrow S_k(\C)$, or equivalently, $$\Hom_\Z(\T,\Z)1216\hookrightarrow \Hom_\Z(\T,\C) = \Hom(\T_\C,\C) = S_k(\C) = S_k(\Z)1217\otimes_\Z \C.$$12181219Here is a nifty inner product (the Petersson innner product) on $S_k(\C)$.1220For $f,g \in S_k(\C)$, let1221$$\lan f,g \ran = \int_{\Gamma \backslash \H} f(\tau)g(\tau) y^k1222\frac{dx\,dy}{y^2}.$$1223The Hecke operators are self-adjoint for $(p, N) = 1$:1224$$\lan f | T_p, g \ran = \lan f, g | T_p \ran.$$1225Indeed, for $\a \in \GL_2^+ (\R)$,1226$$\lan f | \a , g | \a \ran = \lan f,g \ran.$$12271228\section*{February 5, 1996}1229\noindent{Scribe: Shuzo Takahashi, \tt <shuzo@math>}1230\bigskip12311232We have studied actions of $\T$ on1233$S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$ where1234$\Gamma = SL_2(\Z)$. What we know so far is1235$$S_k(\Z) \simeq {\rm Hom}_{\Z}(\T,\Z).$$1236Also we have studied the Peterson product. It is Hermitian, i.e.,1237$$\lan f|T_n, g\ran = \lan f,g|T_n\ran $$1238for $T_n \in \T$ and for $f,g \in S_k(\C)$.12391240\noindent1241Note: $T_n$ defined on $S_k(\C)$ preserves $S_k(\Z)$.12421243Today we study when $\Gamma = \G_1(N)$ or $\G_0(N)$ for1244$N \geq 1$.12451246\medbreak1247\noindent1248{\bf\large 1. The Diamond Operator and the Decomposition of1249$S_k(\G_1(N))$}12501251\begin{thm}1252$\G_1(N)$ is a normal subgroup of $\G_0(N)$ and we have1253$\G_0(N)/\G_1(N) \simeq (\Z/n\Z)^*$.1254\end{thm}12551256\begin{dfn}1257The diamond operator $< > $ is defined as follows: for1258$\pmatrix{a & b \cr c & d \cr} \in \G_0(N)$, the map on $S_k(\G_1(N))$1259$$f \rightarrow f | \pmatrix{a & b \cr c & d \cr} $$1260defines an endmorphism $< d> $ of $S_k(\G_1(N))$ which depends only1261on $d \ {\rm mod}\ N$. Thus we get an action $< > $ of1262$(\Z/n\Z)^*$ on $S_k(\G_1(N))$.1263\end{dfn}12641265Since $(\Z/n\Z)^*$ is a finite group, we have the following decomposition1266theorem:12671268\begin{thm}1269$$S_k(\G_1(N)) = \bigoplus_{\e} S_k(\G_0(N),\e)$$1270where $\e$ runs over the set of characters $(\Z/n\Z)^* \rightarrow \C^*$1271and $S_k(\G_0(N),\e)$ is defined as:1272$$S_k(\G_0(N),\e) = \{ f \in S_k(\G_1(N)) : f | < d> = \e(d) f\}.$$1273We have $S_k(\G_0(N),\e) = 0$ unless $\e(-1) = (-1)^k$.1274\end{thm}12751276\medbreak1277\noindent1278{\bf\large 2. The Hecke Operators on $S_k(\G_1(N))$}12791280For $n \geq 1$, we have the operation on $S_k(\G_1(N))$ of the $n$th Hecke1281operator $T_n$. The following are basic properties:12821283\begin{thm}12841285(1) $T_n$'s commute each other and with $< d> $.12861287(2) $T_n$'s preserve $S_k(\G_0(N),\e)$.12881289(3) if $(n,N) = 1$, then $\lan f | T_n,g\ran =1290\lan f,g|{< n>}^{-1}T_n\ran $.12911292(4) $\lan f | < d> ,g\ran =1293\lan f,g|{< d>}^{-1}\ran $.12941295(5) if $(n,N) = 1$, then $T_n$ is diagonalizable.12961297(6) if $(n,N) \neq 1$, then $T_n$ is not diagonalizable.12981299\end{thm}13001301The action of $T_n$ is described in the following theorem:13021303\begin{thm}1304Let $f = \sum_{n = 1}^\infty a_n q^n \in S_k(\G_0(N),\e)$. Then13051306%$$f | T_p =1307%\cases1308%\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n1309%q^{pn}1310%&\text{if $p \not| N$;} \\1311%\sum_{n=1}^{\infty} a_{pn} q^n1312%&\text{if $p | N$.}1313%\endcases$$13141315(1) $f | T_p = \sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p)1316\sum_{n=1}^{\infty} a_n q^{pn}$.1317(Note: when $p|N$, $\e(p)$ is considered to be $0$ and $U_p$ is used instead1318of $T_p$ which is called Atkin-Lehner operator.)13191320(2) if $(n,m)=1$, then $T_{nm} = T_n T_m$.13211322(3) if $p \not| N$, then $T_{p^l} = T_{p^{l-1}}T_p - p^{k-1}< p>1323T_{p^{l-2}}$.13241325(4) if $p | N$, then $T_{p^l} = (T_p)^l$.1326\end{thm}13271328The last formula in the theorem can be proved by comparing the coefficients1329of $q^{p^{l-1}}$ in both sides of the following formal identity:1330$$\left(\sum_{n=1}^{\infty} T_n q^n\right)|T_p =1331\sum_{n=1}^{\infty} T_{pn} q^n + p^{k-1}< p>1332\sum_{n=1}^{\infty} T_n q^{pn}.$$1333For example, the coefficient of $q^p$ in the LHS is $T_p T_p$. On the other1334hand, the coefficient of $q^p$ in the RHS is1335$T_{p^2} + p^{k-1}< p> T_1$. Thus, we have1336$$T_{p^2} = (T_p)^2 - p^{k-1}< p> Id$$1337where $< p> $ should be considered to be a null map if $p | N$.13381339\medbreak1340\noindent1341{\bf\large 3. The Old Forms}13421343Suppose $M|N$. Let $f \in S_k(\Gamma_1(M))$. Then for $d$ such that1344$d | \frac{N}{M}$, $f(d\tau) \in S_k(\G_1(N))$.1345Thus we have a map1346$$\phi_M : \bigoplus_{d | \frac{N}{M}} S_k(\Gamma_1(M)) \rightarrow1347S_k(\G_1(N)).$$1348The old part of $S_k(\G_1(N))$ is defined as the subspace generated by the1349images of $\phi_M$ for $M | N$, $M \neq N$.13501351\begin{eg} $\phi_M$ is not injective. Consider the case that $k = 12$,1352$M = p$ and $N = p^2$. $S_k(\Gamma_1(p))$ contains $\Delta(\tau)$ and1353$\Delta(p\tau)$. But $\phi_p$ maps both of them to $\Delta(p\tau)$ in1354$S_k(\Gamma_1(p^2))$.1355\end{eg}13561357\begin{thm}1358Suppose $p \nmid N$. Consider $f,g \in S_k(\G_1(N))$. Then $f$ and1359$g(p\tau)$ are both in $S_k(\Gamma_1(Np))$. Then we have1360$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau))$$1361and1362$$g(p\tau) | U_p = g(\tau)$$1363where $f | T_p$ is considered in $S_k(\G_1(N))$.1364\end{thm}1365\noindent1366{\bf Proof.} Let $f = \sum_{n=1}^{\infty} a_n q^n$. Then, considering in1367$S_k(\G_1(N))$, we have1368$$1369f | T_p =1370\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n1371q^{pn}.$$1372Also, considering in $S_k(\Gamma_1(Np))$, we have1373$$f | U_p = \sum_{n=1}^{\infty} a_{pn} q^n.$$1374Thus, we have1375$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau)).$$1376Now, let $g = \sum_{n=1}^{\infty} b_n q^n$. Then1377$$g(p\tau) | U_p =1378\left(\sum_{n=1}^{\infty} b_{n/p} q^n \right) | U_p = g(\tau)$$1379where $b_{n/p} = 0$ unless $p | n$.1380138113821383\section*{February 7, 1996}1384\noindent{Scribe: Amod Agashe, \tt <amod@math>}1385\bigskip13861387We are in the process of showing that the Hecke operators $T_p$ acting1388on the space of cusp forms $S_k(\Gamma_1(N))$ are not necessarily1389semisimple if $p\mid N$.13901391Recall from last time that if $M \mid N$ then for every divisor $d$1392of $M/N$, we had a map $S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$1393given by $f(\tau) \mapsto f(d\tau)$.13941395Note that the various $f(d\tau)$'s are linearly independent over $\C$,1396because the Fourier expansion of $f(d\tau)$ starts with $q^d$.13971398Let $f$ be an eigenfuntion for all the Hecke operators $T_n$ in1399$S_k(\Gamma_1(M))$. Let $p$ be a prime not dividing $M$.1400So $f\mid T_p = af$ where $a=a_p(f)$ and $f \mid <p>=\epsilon (p)f$1401where $\epsilon (p)$ is the character associated to the modular1402form $f$. Note that one can prove that if $f$ is an eigenfunction1403for the $T_n$'s then it is an eigenfunction for the diamond operators1404also (or alternatively, make it part of the definition of eigenform).1405Let $N=p^\alpha M$ with $\alpha \geq 1$. We will look at the action of1406the $p^{th}$ Hecke operator $U_p$ in $S_k(\Gamma_1(N))$ on the1407images of $f$ under the maps described above. Let $f_i(\tau) = f(p^i \tau)$1408for $0\leq i\leq \alpha$. As we showed earlier, \\1409$$f\mid T_p = \sum a_{np}q^n + \epsilon (p)p^{k-1}\sum a_nq^{pn}.$$1410So $$af = f_0\mid U_p + \epsilon (p)p^{k-1}f_1.$$1411Thus, $$f_0\mid U_p = af_0 -\epsilon(p)p^{k-1}f_1.$$1412From last time, we have $f_1\mid U_p = f_0.$ In fact, in general,1413one can see easily that $f_i\mid U_p=f_{i-1}$ for $i\geq 1$.14141415So $U_p$ preserves the 2-dimensional space spanned by $f_0$ and $f_1$.1416The matrix of $U_p$ (acting on the right) with this basis is given1417(from the equations above) by: \smallskip1418\[ \left( \begin{array}{cc}1419a & 1 \\1420-\epsilon(p)p^{k-1} & 01421\end{array}1422\right)1423\]1424The characteristic polynomial of this matrix is $x^2-ax+p^{k-1}\epsilon(p)$.14251426There is the following striking coincidence:1427Let $E$ be the number field generated over $\Q$ by the coefficients of the1428Fourier series expansion of $f$ and let $\la$ be a prime ideal1429of $\O_E$ lying over some rational prime $l$.1430Then we have a Galois representation \smallskip1431$$\rho_\la: Gal(\overline{\Q}/\Q) \rightarrow \GL_2(E_\la)$$1432If $p\not|Nl$ then $\rho_\la$ is unramified and also1433$det\ \rho_\la(Frob_p)=\epsilon(p)p^{k-1}$ and1434$tr\ \rho_\la(Frob_p)=a_p(f)=a$. Thus the characteristic1435polynomial of $\rho_\la(Frob_p)$ is $x^2-ax+p^{k-1}\epsilon(p)$,1436the same as that of the matrix of $U_p$!14371438A question one can ask is: Is $U_p$ semisimple on the space spanned by1439$f_0$ and $f_1$? The answer is yes if the eigenvalues of $U_p$ are1440different.14411442Now, the eigenvalues are the same iff the1443discriminant of the characteristic polynomial is zero i.e.1444$a^2=4\epsilon(p)p^{k-1}$ i.e. $a=2p^{\frac{k-1}{2}}\zeta$ where1445$\zeta$ is some square root of $\epsilon(p)$.1446Here is a curious fact: the Ramanujan-Petersson conjecture proved by Deligne1447says $|a|\leq 2p^{\frac{k-1}{2}}$; thus the above equality is allowed1448by it, so we do not get any conclusion about the semisimplicity of1449$U_p$.14501451Let us now specialize to $k=2$. Weil has shown that $\rho_\la(Frob_p)$1452is semisimple. Thus if the eigenvalues of $U_p$ are equal, then1453$\rho_\la(Frob_p)$ is a scalar. Edixhoven proved that it is not.1454So the eigenvalues of $U_p$ are different and hence $U_p$ is semisimple1455in this case. So this example (for k=2) does not give us an example1456of $U_p$ being not semisimple.14571458There is the following example given by Shimura which shows that the Hecke1459operator $U_p$ need not be semisimple. Let $W$ denote the space spanned1460by $f_0, f_1$ and $V$ denote the space spanned by $f_0,f_1,f_2,f_3$.1461$U_p$ preserves both spaces $W$ and $V$, so it acts on $V/W$. The action1462is given by $\overline{f_2}\mapsto \overline{f_1}=0$ and1463$\overline{f_3}\mapsto \overline{f_2}$ where the bar denotes the image1464in $V/W$. Thus the matrix of $U_p$ on the space $V/W$ is1465\[ \left( \begin{array}{cc}14660 & 1 \\14670 & 01468\end{array}1469\right)1470\]1471which is nilpotent, and in particular not semisimple. If $U_p$ were1472semisimple on $V$ then it would be semisimple on $V/W$ also; but we1473have just shown that it is not. Thus $U_p$ is not semisimple1474on $V$, and hence not on $S_2(\Gamma_1(M))$ (because $V$ is invariant1475under $U_p$).14761477\bigskip14781479We next discuss the structure of the $\C$-algebra $\T =\T_\C$ generated1480by the Hecke and diamond operators and the structure of $S_k(\Gamma_1(N))$1481as a $\T$-module.14821483First we consider the case of level $1$ i.e. $N=1$.1484Then $\Gamma_1(1)=SL_2(\Z)$. All the $T_n$'s are diagonalizable.1485$S_k=S_k(\Gamma_1(N))$ has a basis of $f_1,....,f_d$ of normalized eigenforms1486where $d=dim(S_k)$. Thus $S_k\cong \C^d$ as a $\C$-vector space.1487Then we have the $\C$-algebra homomorphism $\T\rightarrow \C^d$ given by1488$T\mapsto (\la_1,....,\la_d$) where $f_i\mid T=\la_i f_i$.1489It is injective because if the image of $T$ is zero, then it kills all1490$f_i$ i.e. all of $S_k$ i.e. it is the zero operator. The map is1491surjective because $\T$ has dimension $d$.1492%and hence it is surjective1493%when we consider both sides as $\C$-vector spaces.1494Thus as a $\C$-algbebra, $\T\cong \C^d$.1495Next, we claim that the modular form $v=f_1+...+f_d$ generates1496$S_k$ as a $\T$-module. This follows because under the1497map $S_k\cong \C^d, v\mapsto (1,....,1)$ and our1498statement is just the trivial fact that $(1,....,1)$ generates1499$\C^d$ as a $\C^d$-module (acting component-wise).15001501Thus $S_k$ is free of rank $1$ as a $\T$-module.1502We already know that $S_k\cong Hom(\T,\C)$ as $\T$-modules.1503Thus $\T\cong Hom(\T,\C)$ as $\T$-modules. In fact the isomorphism1504is canonical since the $f_i$'s are normalized.1505We remark that $v$ in fact lies in $S_k(\Q)$.15061507Next, we deal with the general case where the level is not necessarily $1$.15081509First we need to talk about newforms. Recall the maps1510$S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$1511for every divisor $d$ of $M/N$ mentioned at the beginning of this lecture.1512The old part of $S_k(\Gamma_1(N))$ is defined as the space generated by1513all the images of $S_k(\Gamma_1(M))$ for all $M\mid N, M\neq N$1514under these maps.1515The new part of $S_k(\Gamma_1(N))$ can be defined in two different ways.1516Firstly we can define it as the orthogonal complement of the old part1517with respect to the Petersson inner product.1518There is also an algbraic definition1519as follows. There are certain maps going the other way:1520$S_k(\Gamma_1(N)) \rightarrow S_k(\Gamma_1(M))$ for $M\mid N, M\neq N$.1521The new part is the space killed under all these maps. The space1522of newforms, denoted $S_k(\Gamma_1(N))_{new}$ is like $S_k(\Gamma(1))$ in the1523sense that all the $T_n$'s (including $U_p$) are semisimple and there1524is a basis consisting of newforms. A form of level $N$ is said to1525be new of level $N$ if it is in $S_k(\Gamma_1(N))_{new}$.15261527Next, one can show that the map1528$\bigoplus_{M\mid N,M\leq N} S_k(\Gamma_1(M))_{new} \rightarrow1529S_k(\Gamma_1(N))$ given by $f(\tau)\mapsto f(d\tau)$ for $d\mid \frac{N}{M}$1530is injective (See W.-C. W.Li, Newforms and functional equations,1531Math. Annalen, 212(1975), 285-315).1532Note that an eigenform in one of the subspaces of the source1533need not be an eigenfuntion for all the operators in the image.1534If $f$ is a newform, then let $M_f$ denote its level (i.e. $f$1535is new of level $M_f$). Let $S$ be the set of newforms of weight $k$1536and some level dividing $N$. Let1537$$v=\sum_{f\in S} f(\frac{N}{M_f}\tau).$$1538Then one can show that $S_k(\Gamma_1(N))$ is free of rank $1$ over1539$\T_\C$ with $v$ as the basis element. Also one can show that1540$v$ has rational coefficients.15411542\section*{February 9, 1996}1543\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}1544\vspace{.2in}15451546\subsubsection*{Final comments about Hecke algebras}15471548Recall that for the case $\Gamma=SL(2,\Z)$, if we set $f_1,\ldots,f_d$1549to be the normalized eigenforms (newforms of level 1), then they have1550possibly complex coefficients but in any case $\{f_i\}$ is finite and1551stable under automorphisms of $\C$ and all the coefficients1552$a_n(f_i)$ lie in some number field. Furthermore this field is totally1553real: to show this we used that since the set is stable under1554conjugation, it suffices to show that all the $a_n(f_i)$ are real, which1555followed by remarking that they are eigenvalues of the operators $T_n$1556which are self-adjoint with respect to the Petersson inner product.15571558More generally, let $f\in S(\Gamma_1(N))$ be a normalized eigenform of1559character $\varepsilon$. Then $a_n=\varepsilon(n)\overline{a_n}$.15601561Note that the algebra $\T_\Q$ generated by the $T_i$ over $Q$ contains1562the diamond bracket operators: the formula relating $T_{p^2}$ and1563$(T_p)^2$ tells us that the difference is $p^{k-1}\varepsilon(p)$, so1564$\varepsilon(p)\in\T_\Q$. Using Dirichlet's Theorem on primes in1565arithmetic progressions, for any $d$ relatively prime to $N$ we can find1566a prime $p\equiv d\pmod{N}$, so $\varepsilon(d)=\varepsilon(p)\in\T_\Q$.15671568If the space of modular forms has dimension 1, then it is spanned by a1569(normalized) eigenform with rational coefficients, so the eigenvalues1570are all in $\Q$.15711572The next simplest example is $k=24$, which is the smallest weight such1573that the dimension is more than one. There are two eigenforms, which1574are conjugate to each other. If $f=\sum a_nq^n$ is an eigenform, then1575$\Q(\ldots a_n\ldots)=\Q(\sqrt{144169})$. In fact $S_{24}$ is spanned1576by $\Delta^2$ and $\Delta^2|T_2$. (Note that $\Delta^2$ is definitely1577not an eigenform since its $q$-coefficient is 0.) The action of $T_2$1578on $S_{24}$ with respect to this basis is described by a two by two1579matrix of trace $1080$ and determinant $-2^{10}3^2 2221$. For high $k$,1580eigenforms tend to form a single orbit; however, no proof1581is known for this.15821583For every newform $f$, let $E_f$ be the number field generated by its1584coefficients. Let $\Sigma$ be a set of representatives for $f$'s modulo1585$Gal(\bar\Q/\Q)$. Define1586\begin{eqnarray*}1587\T_\Q & \rightarrow & E_f \\1588T & \mapsto & \lambda_T,1589\end{eqnarray*}1590with $f|T=\lambda_Tf$. Thus $T_n\mapsto a_n$ so this map is surjective.15911592In fact the induced1593\[ \T_\Q \to \prod_{f\in\Sigma} E_f \]1594is an isomorphism of $\Q$-algebras: it is injective since if $T$ dies on1595the image then it acts as zero on $f\in\Sigma$ and so on all of f, by1596the rationality of Hecke operators: ${}^\sigma\!(g|T)={}^\sigma\!g|T$.1597(And if an operator acts as zero on everything, then it is zero.)15981599Here we used the fact that there were no oldforms around.16001601For example, consider $S_2(\Gamma(N))$ for $N$ prime. Then1602$S_2(\Gamma(1))$ is empty, hence we get an isomorphism1603\[1604\T_\Q\cong E_1\times\ldots\times E_t,1605\]1606with the right hand side a1607product of totally real number fields. $t>1$ is possible, e.g. for1608$N=37$, $\T_\Q=\Q\times\Q$.16091610In general, oldforms complicate the situation.1611161216131614\subsubsection*{Final comments about Hecke algebras}16151616We'll only treat the case $k=2$. Then (for $\Gamma$ a congruence1617subgroup),1618\[ S_2(\Gamma)=H^0(X_\Gamma,\Omega^1) \]1619where $X_\Gamma=\Gamma\setminus\H\cup\Gamma\setminus\P^1(\Q)$, with1620$\Gamma\setminus\P^1(\Q)$ being the set of cusps that we need to adjoin1621to make it a compact Riemann surface.16221623Therefore1624\[ {\rm dim}\,S_2(\Gamma)= g(X_\Gamma). \]16251626{\sc Example.\ } $SL(2,\Z)\setminus\P^1(\Q)$ has just one point. The1627proof involves the Euclidean algorithm: any element of $\P^1(\Q)$ can be1628written as $\left( \begin{array}{cc} x\\y \end{array} \right)$ with $x$1629and $y$ relatively prime integers. By the Euclidean Algorithm, we can1630find $a$ and $b$ integers such that $ax+by=1$. Then1631\[1632\left(\begin{array}{cc} a & b \\ -y & x \end{array} \right)1633\left( \begin{array}{cc} x\\y \end{array} \right) =1634\left( \begin{array}{cc} 1\\0 \end{array} \right)1635\]16361637To calculate $g(X_\Gamma)$, use the following covering (recall that1638$X_\Gamma$ is the compactification of $\Gamma\setminus\H$ we obtain by1639adjoining the cusps)1640\[1641X_\Gamma \rightarrow X_{\Gamma(1)},1642\]1643keeping in mind the isomorphism1644\begin{eqnarray*}1645j:X_{\Gamma(1)} & \rightarrow & \P^1(\C) \\1646(i,\rho,\infty) & \mapsto & (1728,0,\infty)1647\end{eqnarray*}1648The only ramification in our covering1649occurs above the points $0,1728,\infty$.16501651{\sc Example.\ } Let $\Gamma=\Gamma_0(N)$. The degree of the covering1652is $(PSL(2,\Z):\Gamma_0(N)/\pm1)$ which is the number of cyclic subgroups1653of order $N$ in $SL(2,\Z/N\Z)$. We have a covering1654$Y_0(N)\to Y_{\Gamma(1)}$ where $Y_0(N)$ parametrizes elliptic curves1655$E$ with a cyclic subgroup of order $N$,1656$C\subseteq E[N]\cong\Z/N\Z\times\Z/n\Z$ up to isomorphism; and1657$Y_{\Gamma(1)}$ parametrises elliptic curves. The isomorphism1658$(E,C_1)\cong(E,C_2)$ is an automorphism $\alpha$ of $E$ with1659$\alpha:C_1\mapsto C_2$. Usually $\alpha=\pm1$, unless1660$j=1728,0,\infty$.16611662If we understand ramification, we can use the {\em Riemann-Hurwitz1663formula}. The following mnemonic way of thinking about it is due to1664N. Katz. The Euler characteristic (alternating sum of the dimensions of1665cohomology groups) should be thought of as totally additive:1666\[ \chi(A\coprod B) = \chi(A) + \chi(B). \]16671668If $X$ is a Riemann surface of genus $G$, $\chi(X)=2-2G$. A single1669point has Euler characteristic $\chi(P)=1$. Hence1670\[ \chi(X\setminus\{P_1,\ldots,P_n\})=2-2G-n. \]1671Therefore1672\[ \chi(X_{\Gamma(1)}\setminus\{0,1728,\infty\})=-1 \]1673and1674\[ \chi(X_\Gamma\setminus\{\mbox{points over $1728,0,\infty$}\})=2-2g-n \]1675if $n$ points lie over ${0,1728,\infty}$.1676If the covering map $X_\Gamma\to X_{\Gamma(1)}$ has degree $d$, then we1677can think of the top space as $d$ copies of the bottom space, so1678\[ d \cdot (-1) = 2-2g-n \]1679and therefore1680\[ 2g-2 = d-n = d-n_0 - n_{1728} -n_\infty.\]16811682{\sc Example.\ } Let $\Gamma=\Gamma(N)$. What happens over $j=0$?1683This corresponds to an elliptic curve $E$ with an automorphism $\alpha$1684of order three. Let $N>3$.16851686For any $(E,P,Q)$, we have $(E,\alpha P,\alpha Q)$ and1687$(E,\alpha^2 P,\alpha^2 Q)$ which are isomorphic to it and hence they1688are the same point on $X_\Gamma$. So $E$ only has one-third the usual1689number of points lying over it, $n_0=d/3$. (Except if the above three1690points are equal, i.e., $\alpha$ fixes $E[N]$. This can't happen since1691$N>3$.)16921693Similarly we get $n_{1728}=d/2$.16941695To determine the degree $d$, fix $E$ and count the points lying over it:1696these are all of the form $(\C/\Z\oplus\tau\Z,1/n,\tau/N)$ with Weil1697pairing $e^{2\pi i/N}$ and all such occur, so we need to1698count the number of $P,Q\in E[N]$ which form a basis of1699$E[N]$ and $e_N\langle P,Q\rangle=e^{2\pi i/N}$. This gives us the1700order of $SL(2,\Z/n\Z)$. However, (since $N\neq2$) we have to take into1701account that $(E,P,Q)\cong(E,-P,-Q)$ but they are not equal1702so the degree of the covering is1703$\#(SL(2,\Z/N\Z))/2$.17041705We also have $d=(PSL(2,\Z):\Gamma(N)/(\Gamma(N)\cap\pm1))$ since1706$\Gamma(N)\setminus SL(2,\Z)\cong SL(2,\Z/N\Z)$.17071708So we have established that1709\[ 2g-2 = d-n_0 - n_{1728} -n_\infty = d/6 - n_\infty. \]1710To determine $n_\infty$, note that $SL(2,\Z)$ acts on1711$\P^1(\Q)\ni\infty=1712\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$.1713The stabilizer $\Gamma(N)_\infty$1714of $\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$1715is $U=\pm\left( \begin{array}{cc} 1 & * \\ 0 & 1 \end{array} \right)$1716So the index1717$((\modgp/\pm1)_\infty:\Gamma(N)_\infty)=N$ is the ramification degree1718of a point over $\infty$, so $n_\infty=N/d$.17191720Hence $2g(X(N))-2=d/6-d/N$.1721\section*{February 12, 1996}1722\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}1723\vspace{.2in}17241725\def\T{{\bf T}} % Hecke algebra1726\def\qed{\hfill $\blacksquare$}17271728Plug: A useful reference for the next lecture is Andrew Ogg: {\em Rational1729points on certain elliptic modular curves} (1972).17301731From last lecture's results on easily deduces that1732\[ g(X(N)) = 1 + \frac{d}{12N}(N-6). \]17331734{\sc Example.} Let $N=5,7$. If $N$ is prime then the degree of the1735covering is $\frac{(N^2-1)(N^2-N)}{N-1}$. Therefore $N=5$ gives $d=60$1736and $g=0$ (the Galois group of the covering in this case is $A_5$.)1737Similarly, $N=7$ yields $g=3$ and $d=168$. (Remark: For $p\geq5$ prime, the1738group $SL(2,\Z/p\Z)/\pm1=L_2(p)$ is simple.)17391740{\sc Example.} What is the genus of $X_0(N)$, for $N$ a prime?17411742It is an exercise to show that there are two cusps:1743$\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)=\infty$1744and1745$\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)=0$.1746The covering $X_0(N)\to X(1)$ is easily seen to have degree $N+1$, since1747an elliptic curve has $N+1$ cyclic subgroups of order $N$. Here1748$\infty$ is unramified and $0$ has ramification index $N$.1749Hence1750\[ 2g-2 = N+1 -n_\infty-n_{1728}-n_0 \]1751with $n_\infty=2$, $n_{1728}$ approximately $d/2$ and1752$n_0$ approximately $d/3$.17531754To calculate $n_0$, we need the number of isomorphism classes $(E,C)$1755with $E$ fixed with $End(E)=\Z[\mu_6]$. $\mu_6$ acts on the1756set of $C$'s. Since $\pm1$ is acting trivially, we really have an1757action of $\mu_3$ with $(E,C)\cong(E,\zeta C)\cong(E,\zeta^2C)$ with1758$\zeta$ some third root of unity.17591760If we consider $E=\C/\O$ with $\O=\Z[(-1+i\sqrt3)/2]$, then1761$E[N]=\O/N\O$ as an $\O$-module. In fact1762\[ \O/N\O =1763\left\{ \begin{array}{ll}1764\F_N\oplus\F_N & \mbox{if N splits in $\O$} \\1765\F_{N^2} & \mbox{if N does not split in $\O$.}1766\end{array}1767\right.1768\]1769The above covers all possibilities, because $N>3$ can't ramify in $\O$.1770By Kummer's theorem, $N$ splits in $\O$ iff1771$\left(\frac{-3}{N}\right)=1$, and1772there are exactly two $\O$-stable submodules of $\O/N\O$. In the second1773case, which happens iff $\left(\frac{-3}{N}\right)=-1$, $\O/N\O$ has no1774$\O$-stable submodules. Therefore1775\[ n_0 =1776\left\{ \begin{array}{ll}1777\frac{N-1}{3} + 2 &1778\mbox{if $N$ splits in $\O$, i.e., $N\equiv1\pmod3$.}\\1779\frac{N+1}{3} &1780\mbox{if $N$ does not split in $\O$, i.e., $N\equiv2\pmod3$}1781\end{array}1782\right.1783\]1784In the first case, note that the two $\O$-stable submodules have to be1785counted separately.17861787Similarly we obtain1788\[ n_{1728} =1789\left\{ \begin{array}{ll}1790\frac{N-1}{2}+2 & \mbox{if $N\equiv1\pmod4$} \\1791\frac{N+1}{2} & \mbox{if $N\equiv3\pmod4$}1792\end{array}1793\right.1794\]1795depending on whether$N$ splits in $\Q(i)$.17961797{\sc Examples.} For $N=37$ the genus is $2$. Using the formula, we get1798$2g-2=36-(2+18)-14=2$ (so it works).17991800Using the formula we can also obtain $g(X_0(13))=0$ and $g(X_0(11))=1$.18011802It is therefore clear that the genus of $X_0(N)$ is approximately1803$N/12$. In the article by Serre in the Lecture Notes in Mathematics 3491804(Antwerp), we find the following table. Write $N=12a+b$ with $0\leq1805b\leq11$. Then1806\begin{center}1807\begin{tabular}{c|cccc}1808$b$ & $1$ & $5$ & $7$ & $11$ \\ \hline1809$g$ & $a-1$ & $a$ & $a$ & $a+1$. \\1810\end{tabular}1811\end{center}1812Hence1813\[18141 + g(X_0(p)) = {\rm dim}\,M_{p+1}(\modgp).1815\]18161817It was Serre's idea to think of ``modular forms mod $p$'', for some1818congruence subgroup1819$\Gamma\ni\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$, like1820$\Gamma_0(N)$ or $\Gamma_1(N)$. We could use our moduli theoretic1821interpretations, but instead we'll define1822\[1823M_k(\Gamma,\F_p) \subseteq \F_p[[q]].1824\]18251826By Shimura's cohomology trick, we know that $M_k(\Gamma,\Z)$ is a1827lattice in $M_k(\Gamma,\C)$. Hence we can set1828\[1829M_k(\Gamma,\F_p) = M_k(\Gamma,\Z) \otimes_{\Z} \F_p.1830\]1831Then {\em Serre's equality} states that for a prime $p$,1832\[1833M_{p+1}(\modgp,\F_p) = M_2(\Gamma_0(p),\F_p)1834\]1835in $\F_p[[q]]$. The philosophy is {\em mod $p$ forms with $p$ in the1836level can be taken to mod $p$ formswith no $p$ in the level, but of a1837higher weight}. So for example1838$M_k(\Gamma_1(p^\alpha N),\F_p)$ is a subset of1839$M_?(\Gamma_1(N),\F_p)$ of forms of some higher level.18401841Finally, consider the map from the right hand side to the left hand1842side in Serre's equality. Recall that1843\[ G_k = \frac{-B_k}{2k} + \sum^\infty_{n=1}\sigma_k(n) q^n1844\in M_k(\modgp). \]1845By Kummer, ${\rm ord}_p(B_{p-1}) = -1$, so1846\[1847E_{p-1}=1+\frac{-2(p-1)}{B_{p-1}}\sum\sigma_{p-1}(n)q^n \equiv 1 \pmod{p}.1848\]1849Hence we got the map from the right to the left:1850multiply by $E_{p-1}$ to get to $M_{p+1}(\Gamma_0(p),\F_p)$. Then take1851the trace to get to $M_{p+1}(\modgp,\F_p)$. The trace map is dual to1852the inclusion and is expressed by1853\[1854\tr(f) = \sum_{i=1}^{p+1} f | \gamma_i1855\qquad1856\gamma_i\in\Gamma_0(p)\setminus\modgp.1857\]1858\section*{February 14 and 16, 1996}1859\noindent{Scribe: Jessica Polito, \tt <polito@math>}1860\bigskip186118621863Our goal, for these two days, is to define the modular curves $\X$ over1864$\Q(\mu_N)$, and1865$X_1(N)$, and $X_0(N)$ over $\Q$.1866These notes will spell out the construction of $\X$, with some1867discussion of the construction of the other two types of curves.1868The idea comes from Shimura.18691870Let $\Q(t)$ be the function field of $\P^1/\Q$, and pick an elliptic1871curve $\E/\Q(t)$ with $j$-invariant $t$:1872$$\E: y^2 = 4x^3 - \frac{27t}{t-1728}x - \frac{27t}{t-1728}$$1873(Note that the general formula for the $j$-invariant of a curve $y^2 =18744x^3 - g_2x -g_3$ is $j = \frac{1728g_2^3}{g_2^3 - 27g_3^2}$, and1875that $j(E)$ determines the isomorphism class of the given curve $E$1876over the algebraic closure of the field of definition.)18771878By substituting in a given value $j$ for $t$, we would get a formula1879for an elliptic curve over $\Q(j)$ with $j$-invariant $j$; for $j= 0 $1880or 1728, we could pick a diferent formula for $\E$ which would give an1881isomorphic curve over $\Q(t)$, for which that substitution would make1882sense.18831884Notice that, in general, if we have an elliptic curve $E/K$, with $K$1885some field of charictaristic prime to $N$, then we can consider1886$E[N](\K)$, the set of all $N$-torsion point of $E$ defined over $\K$,1887which is isomorphic to $(\Z/N\Z)^2$. Then we let $K(E[N])$ be the1888smallest extension of $K$ over which all the points of $E[N]$ are1889defined. Notice also that $E/K$ and $N$ together define a1890representation of the Galois groups $\gal(\K/K)$ into the1891automorphisms of the $N$-torsion points of $E$, as they are defined by1892polymial equations with coefficients in $E$. We get1893$$\rho_{E,N}: \gal(\K/K) \into \aut(E[N]) \isom \GL_2(\Z/N\Z)$$1894with $\gal(\K/K(E[N])) = \ker(\rho_{E,N})$. Unsurprisingly, we will1895frequently leave out the $N$, writing simply $\rho_E$.18961897Now, to return to our construction, where $K=\Q(t)$. We will show1898that $\Q(t)(E[N])$ is the function field of a curve defined over1899