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Author: William A. Stein
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13\author{William Stein}
14\title{Hecke Algebras and Modular Forms:\\
15       Notes derived from Ribet's 1996 Berkeley grad. course.}
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238
239
240\begin{document}
241\frontmatter
242\maketitle
243\chapter{Preface}
244{\bfseries Disclaimer: }
245These notes record some of what I saw in Ken Ribet's course
246on Modular Forms and Hecke Operators given at U.C. Berkeley during
247the Spring semester 1996. They are still {\em very rough}
248as I wrote them during my first semester of graduate school before
249I knew any real mathematics.
250
251The participants in the course were:
252Amod Agashe,
253Matt Baker,
254Jim Borger,
255Kevin Buzzard,
257Robert Coleman,
258Jan\'{o}s Csirik,
259Annette Huber,
260David Jones,
261David Kohel,
262Loic Merel,
263David Moulton,
264Andrew Ogg,
265Arthur Ogus,
266Jessica Polito,
267Ken Ribet,
268Saul Schleimer,
269Lawren Smithline,
270William Stein,
271Takahashi,
272Wayne Whitney, and
273Hui Zui.
274
275I wish to thank David Moulton, Joe Wetherall, and Kevin Buzzard who
276helped me in preparing these notes, Arthur Ogus who asked
277a lot of stimulating questions during the class, and of course
278Ken Ribet who sees clearly.
279
280William Stein, Spring 1996, Berkeley, CA, {\tt was@math.berkeley.edu}
281\tableofcontents
282
283%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
284\mainmatter
285%%%%%%%%%%%%%%%%%%%%%%%%
286%% Lecture 1, January 17
287\chapter{Introduction}
288
289The main objects of study in this course are:
290\begin{itemize}
291\item Modular Forms
292\item Hecke Algebras
293\item Modular Curves\index{modular curves}
294\item Jacobians\index{Jacobian}
295\item Abelian Varieties
296\end{itemize}
297
298\section{Two Dimensional Galois Representations}
299
300\index{Galois representations}
301The geometric objects of study are elliptic curves and more generally
302algebraic curves of arbitrary genus. These in turn give rise via the
303Jacobian construction to higher dimensional
304abelian varieties. These geometric objects in turn give rise to
305Galois representations.
306
307When studying elliptic curves, the natural tool in the characteristic
308zero situation is to present the elliptic curve as
309$\bC/\sL$ for some lattice $\sL$ in $\bC$.
310To construct $\sL$ fix a non-zero holomorphic differential $\omega$
311of $E$ over $\bC$ and construct $\sL$ as
312\begin{equation*}
314\end{equation*}
315
316\subsection{Finite Fields (Weil, Tate)}
317In the 1940's, Weil study the analogous situation for
318elliptic curves defined over a finite field $k$. He desperately
319wanted to find an algebraic way to describe the above correspondence.
320He was able to find an algebraic definition of
321$\sL/n\sL$, where $n\geq 1$ and $(n,\Char k)=1$, which is as follows.
322Let $E[n]=\{P\in E(\overline{k}) : nP = 0\} = (\frac{1}{n} \sL) / \sL 323\isom \sL / n \sL$.
324
325Now fix a prime $\ell$, we let
326$E[\ell^\infty]=\{P\in E(\overline{k}) : \ell^{\nu}P = 0, \text{ some } 327\nu \geq 1\} = \cup_{\nu=1}^{\infty} E[\ell^{\nu}]$.
328Tate obtained an analogous construction by defining a rank 2 free
329$\Zl$-module $\Tatel E:=\varprojlim E[\ell^{\nu}]$
330(the map from $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$ is multiplication
331by $\ell$).
332To see that the rank is $2$, check that
333the $\bZ/\ell^{\nu}\bZ$-module structure of $E[\ell^{\nu}]$ is compatible
334with the maps $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$). See
335\cite{silverman1} (III, 7).
336Then $V_{\ell}(E)=T_{\ell}(E)\tensor\bQ_{\ell}$ is a two dimensional
337vector space over $\bQ_{\ell}$. This gives the first
338nontrivial example of $\ell$-adic \'{e}tale cohomology.
339
340\subsection{Galois Representations (Taniyama, Shimura, Mumford-Tate)}
341Let $E/\bQ$ be an elliptic curve and $G=\gal(\overline{\bQ}/\bQ)$.
342Then $E[n]=\{P\in E(\overline{\bQ}) : nP = 0\} \isom (\bZ / n\bZ)^2$
343is acted on by $G$ and this action respects the group operation
344so we have a Galois representation
345\begin{equation*}
346G\xrightarrow{\rho}\aut(E[n])\isom \gl_2(\bZ/n\bZ)
347\end{equation*}
348Let $K$ be the fixed field of $\ker \rho$ (note that $K$ is a number
349field), then since $\gal(K/\bQ)\isom G/\ker\rho \isom \imag \rho \subseteq 350\gl_2(\bZ/n\bZ)$ we obtain many subgroups of $\gl_2(\bZ/n\bZ)$ as Galois
352\begin{equation*}
354\end{equation*}
355then the image of $\rho$ is often all of $\gl_2(\bZ/n\bZ)$
356and the image is most'' of $\gl_2(\bZ/n\bZ)$ when $E$
357does not have complex multiplication.
358
359\section{Modular Forms and Galois Representations}
360
361
362\subsection{Cusp Forms}
363\index{cusp forms}
364Let $S_k(N)$ denote the space of cusp forms
365of weight $k$ and level
366$N$ on the congruence subgroup\index{congruence subgroup} $\Gamma_1(N)=\bigl\{\bigl(\begin{smallmatrix} 367a&b\\ 368c&d 369\end{smallmatrix}\bigr) \in \sl2z : 370a\equiv 1 \pmod N, c\equiv 0 \pmod N, d\equiv 1 \pmod N \bigr\}$.
371Thus $S_k(N)$ is the finite dimensional vector space consisting of
372all holomorphic functions $f(z)$ on $\sH=\{z\in\bC : \imag(z)>0\}$
373vanishing at $\infty$ and satisfying
374\begin{equation*}
377\in \Gamma_1(N).
378\end{equation*}
379Since, in particular, $f(z)=f(z+1)$ we can
380expand $f(z)$ as a $q$-series (this requires rigorous justification)
381\begin{equation*}
382f(z) = \sum_{n=1}^{\infty} c_n q^n.
383\end{equation*}
384
385A famous example is
386\begin{equation*}
387\Delta = q\prod_{n=1}^{\infty}(1-q^n)^{24} = \sum_{n=1}^{\infty} \tau(n) q^n
388\end{equation*}
389$\tau$ is called the Ramanujan function.
390One now knows that $\tau$ is multiplicative and satisfies
391$\tau(p^{\nu})=\tau(p)\tau(p^{\nu})-p^{11}\tau(p^{\nu-1})$.
392$\Delta$ is a normalized basis for $S_1(1)$.
393
394\subsection{Hecke Operators (Mordell)}
395
396Mordell defined, for $n\geq 1$, operators $T_n$ on $S_k(N)$ called
397{\em Hecke operators}\index{Hecke operator}.
398These proved very fruitful. The set of such
399operators forms an almost'' commuting family of endomorphisms and
400is hence almost'' simultaneously diagonalizable. The precise meaning
401of almost'' and the actual structure of the Hecke algebra
402$\bQ[T_1,\ldots,T_n,\ldots]$
403will be studied in greater detail in the remainder of this course.
404Often there will exist a basis of cusp forms $f = \sum_{n=1}^{\infty} c_n q^n 405\in S_k(N)$ so that $f_n$ is a simultaneous eigenvector for all of the
406Hecke operators $T_n$ and, in fact, $T_n f = c_n f$. All of
407the $c_n$ will be algebraic integers and the field
408$\bQ(c_1,c_2,\ldots)$ will be finite over $\bQ$.
409
410A good claim can be made that the $c_n$ are often interesting integers
411because they exhibit remarkable properties. For example,
412$\tau(n) \equiv \sum_{d|n}d^{11} \pmod {691}$.  How can we study the $c_n$?
413How can we interpret the $c_n$? We can do this by studying the connection
414between Galois representations and modular forms. In 1968 work was originally
415begun on this by Serre\index{Serre}, Shimura, Eichler and Deligne.
416\index{Shimura}\index{Eichler}\index{Deligne}
417
418%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
419
420%% Lecture 2, 1/19/96
421
422\chapter{Modular Representations and Curves}
423
424\section{Arithmetic of Modular Forms}
425Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is
426an eigenform for the Hecke operators.  Then the Mellin transform associates
427to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} \frac{a_n}{n^s}$.
428Let $K=\bQ(a_1,a_2,\ldots)$, then one can show that the $a_n$ are algebraic
429integers and $K$ is a number field. When $k=2$ Shimura\index{Shimura}
430associates to $f$ an abelian variety $A_f$ over $\bQ$ of
431dimension $[K:\bQ]$ on which $K$ acts (see
432theorem 7.14 of \cite{shimura1}).
433
434\begin{example}[Modular Elliptic Curves]
435
436When all of the coefficients $a_n$ of the modular form $f$ lie in $\bQ$
437then $[K:\bQ]=1$ so $A_f$ is a one dimensional abelian variety.
438A one dimensional abelian variety of nonzero genus is an elliptic curve.
439An elliptic curve isogenous to one arising via this construction is
440called {\em modular}.
441\end{example}
442
443\begin{defn}
444
445Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is
446a morphism $E_1\into E_2$ of algebraic groups, which has a
447finite kernel.
448\end{defn}
449
450The following conjecture motivates much of the theory.
451
452\begin{conj}
453Every elliptic curve over $\bQ$ is modular,
454that is, isogenous to a curve constructed in the above way.
455\end{conj}
456
457For $k\geq 2$ Serre\index{Serre} and Deligne found a way to associate to $f$ a family
458of $\ell$-adic representations. Let $\ell$ be a prime number and $K$ be as
459above, then it is well known that $K\tensor_{\bQ} \bQ_{\ell}\isom 460\prod_{\lambda|\ell}K_{\lambda}$. One can associate to $f$ a representation
461\begin{equation*}
462\rholf:G=\gal(\overline{\bQ}/\bQ)
463\rightarrow\gl(K\tensor_{\bQ}\bQ_{\ell})
464\end{equation*}
465unramified at all primes $p\nd \ell N$.
466For $\rholf$ to be unramified we mean that for all primes $P$ lying over $p$,
467the inertia group of the decomposition group at $P$ is contained
468in the kernel of $\rholf$. The decomposition group $D_P$ at $P$ is the
469set of those $g\in G$ which fix $P$. Let $k$ be the residue
470field $\sO/P$ where $\sO$ is the ring of all algebraic integers.
471Then the inertia group $I_P$ is the kernel of the map $D_P\rightarrow 472\gal(\overline{k}/k)$.
473
474Now $I_P\subset D_P \subset \gal(\overline{\bQ}/\bQ)$ and
475$D_P / I_P$ is cyclic (being isomorphic to a subgroup of the
476Galois group of a finite extension of finite fields)
477so it is generated by a Frobenious automorphism $\frob_p$ lying over $p$.
478One has
479\begin{align*}
480\tr(\rholf(\frob_p))& = a_p\in K \subset K\tensor \bQ_{\ell}\\
481&\text{and}\\
482\det(\rholf) &= \chi_{\ell}^{k-1}\varepsilon
483\end{align*}
484where $\chi_{\ell}$ is the $\ell$th cyclotomic character and
485$\varepsilon$ is the Dirichlet character associated to $f$.
486There is an incredible amount
487of abuse of notation'' packed into this statement.  First, the Frobenius
488$\frob_P$ (note $P$ not $p$) is only well defined in $\gal(K/\bQ)$
489(so I think an unstated result is that $K$ must be Galois), and
490then $\frob_p$ is only well defined up to conjugacy.
491But this works out since $\rholf$ is
492well-defined on $\gal(K/\bQ)$
493(it kills $\gal(\overline{\bQ}/K)$) and the trace
494is well-defined on conjugacy classes
495($\tr(AB)=\tr(BA)$ so $\tr(ABA^{-1})=Tr(B)$).
496
497
498\section{Characters}
499Let $f\in S_k(N)$, then for all
500$\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr) 501\in \sl2z$ with $c\equiv 0 \mod{N}$ we have
502\begin{equation*}
503f(\frac{az+b}{cz+d}) = (cz+d)^k \varepsilon(d) f(z)
504\end{equation*}
505where $\varepsilon:(\bZ/N\bZ)^*\rightarrow \bC^*$
506is a Dirichlet character mod $N$. If $f$ is an eigenform for
507the so called diamond-bracket operator'' $\dbd{d}$ so that
508$f|\dbd{d} = \varepsilon(d) f$
509then $\varepsilon$ actually takes values in $K$.
510
511Led $\varphi_N$ be the mod $N$ cyclotomic character so that
512$\varphi_N: G \rightarrow (\bZ/N\bZ)^*$ takes $g\in G$ to
513the automorphism induced by $g$ on the $N$th cyclotomic
514extension $\bQ(\Mu_N)$ of $\bQ$ (where we identify
515$\gal(\bQ(\Mu_N)/\bQ)$ with $(\bZ/N\bZ)^*$).
516Then what we called $\varepsilon$ above in the formula
517$\det(\rho_{\ell})=\chi_{\ell}^{k-1}\varepsilon$
518is really the composition
519\begin{equation*}
520G\xrightarrow{\varphi_N}(\bZ/N\bZ)^*\xrightarrow{\varepsilon} \bC^*.
521\end{equation*}
522
523For each positive integer $\nu$ we consider the $\ell^{\nu}$th
524cyclotomic character on $G$,
525\begin{equation*}
526\varphi_{\ell^{\nu}}:G\rightarrow (\bZ/\ell^{\nu}\bZ)^*.
527\end{equation*}
528Putting these together gives the $\ell$-adic cyclotomic character
529$$\chi_{\ell}:G\into\bZ_{\ell}^{*}.$$
530
531\section{Parity Conditions}
532
533Let $c\in\gal(\overline{\bQ}/\bQ)$ be complex conjugation.
534Then $\varphi_N(c)=-1$ so $\varepsilon(c) = \varepsilon(-1)$ and
535$\chi_{\ell}^{k-1}(c) = (-1)^{k-1}$. Now let
536$\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr) 537= 538\bigl(\begin{smallmatrix} -1&0\\0&-1\end{smallmatrix}\bigr)$,
539then for $f\in S_k(N)$,
540$$f(z) = (-1)^k\varepsilon(-1)f(z)$$
541so $(-1)^k\varepsilon(-1) = 1$ thus
542$$\det(\rholf(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$
543Thus the $\det$ character is odd so the representation
544$\rholf$ is odd.
545
546\begin{remark}[Vague Question] How can one recognize representations
547like $\rholf$ in nature''? Mazur\index{Mazur} and Fontaine have made
548relevant conjectures. The Shimura-Taniyama conjecture can be reformulated
549by saying that for any representation $\rho_{\ell,E}$ comming
550from an elliptic curve $E$ there is $f$ so that
551$\rho_{\ell,E}\isom \rholf$.
552\end{remark}
553
554\section{Conjectures of Serre (mod $\ell$ version)}
555\index{Serre}
556Suppose $f$ is a modular form, $\ell\in\Z$  prime,
557$\lambda$ a prime lying over $\ell$, and the representation
558$$\rho_{\lambda,f}:G\rightarrow \gl_2(K_{\lambda})$$
559(constructed by Serre-Deligne) is irreducible.
560Then $\rho_{\lambda,f}$ is conjugate to a representation
561with image in $\gl_2(\sO_{\lambda})$, where $\sO_{\lambda}$
562is the ring of integers of $K_{\lambda}$.
563Reducing mod $\lambda$ gives a representation
564$$\overline{\rho}_{\lambda,f}:G\rightarrow\gl_2(\bF_{\lambda})$$
565which has a well-defined trace and det, i.e., the det and trace
566don't depend on the choice of conjugate representation used to
567obtain the reduced representation.
568One knows from representation theory that if
569such a representation is semisimple then it is completely determined
570by its trace and det (more precisely, the characteristic polynomials
571of all of its elements -- see chapter ??).
572Thus if $\overline{\rho}_{\lambda,f}$ is irreducible (and hence semisimple)
573then it is unique in the sense that it does not depend on the choice
574of conjugate.
575
576\section{General remarks on mod $p$ Galois representations}
577\index{Galois representations}
578
579%% By Joe Wetherall
580[[This section was written by Joseph Loebach Wetherell.]]
581
582First, what are semi-simple and irreducible representations?  Remember
583that a representation $\rho$ is a map from a group $G$ to the endomorphisms of
584some vector space $W$ (or a free module $M$ if we are working over a ring
585instead of a field, but let's not worry about that for now).  A subspace $W'$
586of $W$ is said to be invariant under $\rho$ if $\rho$ takes $W'$ back into itself.
587(The point is that if $W'$ is invariant, then $\rho$ induces representations on
588both $W'$ and $W/W'$.)  An irreducible representation is one where the only
589invariant subspaces are ${0}$ and $W$.  A semi-simple representation is one
590where for every invariant subspace $W'$ there is a complementary invariant
591subspace $W''$ -- that is, you can write $\rho$ as the direct sum of $\rho|_{W'}$
592and $\rho|_{W''}$.
593
594Another way to say this is that if $W'$ is an invariant subspace then we get a
595short exact sequence $$0\into\rho|_{W/W'}\into\rho\into\rho|_{W'}\into 0.$$
596Furthermore $\rho$ is
597semi-simple if and only if every such sequence splits.
598
599Note that irreducible representations are semi-simple.
600
601One other fact is that semi-simple Galois representations
602are uniquely determined (up to isomorphism class) by their trace
603and determinant.
604
605Now, since in the case we are doing, $G = \galq$ is
606compact, it follows that the image of any Galois representation $\rho$ into
607$\gl_2(K_{\lambda})$ is compact.  Thus we can conjugate it into
608$\gl_2(\sO_{\lambda})$.  Irreducibility is not needed for this.
609
610Now that we have a representation into $\gl_2(\sO_{\lambda})$, we can reduce
611to get a representation $\overline{\rho}$ to $\gl_2(\bF_{\lambda})$.  This
612reduced representation is not uniquely determined by $\rho$, since we had a
613choice of conjugators.  However, the trace and determinant are invariant
614under conjugation, so the trace and determinant of the reduced
615representation are uniquely determined by $\rho$.
616
617So we know the trace and determinant of the reduced representation.  If we
618also knew that it was semi-simple, then we would know its isomorphism class,
619and we would be done.  So we would be happy if the reduced representation is
620irreducible.  And in fact, it is easy to see that if the reduced
621representation is irreducible, then $\rho$ must also be irreducible.  Now, it
622turns out that all $\rho$ of interest to us will be irreducible;
623unfortunately, we can't go the other way and claim that $\rho$ irreducible
624implies the reduction is irreducible.
625
626\section{Serre's Conjecture}
627\index{Serre}
628Serre has made the following conjecture which is still open at
629the time of this writing.
630\begin{conj}[Serre]
631All irreducible representation of
632$G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$
633complex conjugation, are of the form $\overline{\rho}_{\lambda,f}$
634for some representation $\rho_{\lambda,f}$ constructed as above.
635\end{conj}
636
637\begin{example}
638Let $E/\bQ$ be an elliptic curve and let
639$\sigma_{\ell}:G\rightarrow\gl_2(\bF_{\ell})$ be
640the representation induced by the action of $G$
641on the $\ell$-torsion of $E$. Then $\det \sigma_{\ell} = \varphi_{\ell}$
642is odd and $\sigma_{\ell}$ is usually irreducible, so
643Serre's conjecture\index{Serre's conjecture}
644would imply that $\sigma_{\ell}$ is modular. From this one can, assuming
645Serre's conjecture, prove that $E$ is modular.
646\end{example}
647
648\begin{defn}
649Let $\sigma:G\rightarrow \gl_2(\bF)$ ($\bF$ is a finite field)
650be a represenation of the Galois group $G$. The we say that the
651{\em representions $\sigma$ is
652modular} if there is a modular form $f$, a prime $\lambda$, and an embedding
653$\bF\hookrightarrow \overline{\bF}_{\lambda}$ such that
654$\sigma\isom\overline{\rho}_{\lambda,f}$ over
655$\overline{\bF}_\lambda$.
656\end{defn}
657
658\section{Wiles' Perspective}
659
660Suppose $E/\bQ$ is an elliptic curve and
661$\rho_{\ell,E}:G\rightarrow\gl_2(\bZ_{\ell})$
662the associated $\ell$-adic representation on the
663Tate module $T_{\ell}$. Then by reducing
664we obtain a mod $\ell$ representation
665$$\overline{\rho}_{\ell,E}=\sigma_{\ell,E}:G 666\rightarrow \gl_2(\bF_{\ell}).$$
667If we can show this representation is modular for infinitely many $\ell$
668then we will know that $E$ is modular.
669
670\begin{thm}[Langland's and Tunnel]
671If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they
672are modular.
673\end{thm}
674
675This is proved by using the fact that $\gl_2(\bF_2)$ and
676$\gl_2(\bF_3)$ are solvable so we may apply base-change''.
677
678\begin{thm}[Wiles]
679If $\rho$ is an $\ell$-adic representation which is irreducible
680and modular mod $\ell$ with $\ell>2$ and certain other reasonable
681hypothesis are satisfied, then $\rho$ itself is modular.
682\end{thm}
683
684%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
685
686%% Lecture 3, 1/21/96
687
688\chapter{Modular Forms}
689
690Our goal is to explain modular forms
691as functions of lattices
692or of elliptic curves. Good references are Serre \cite{serre2}
693\index{Serre}
694and Katz \cite{antwerp}.
695
696
697
698\section{Cusp Forms}
699First suppose $N=1$, then we must define $S_k=S_k(1)$. Let
700$\Gamma_1(1)=\sl2z$, then $S_k$ consists of all functions
701$f$ holomorphic on the upper half plane $\sH$ and such that for all
702$\abcd\in\sl2z$
703one has
704$$f(\frac{a\tau+b}{c\tau+d})=(c\tau+d)^k f(\tau),$$
705and $f$ vanishes at infinity.
706Thus, in particular, $f(\tau+1)=f(\tau)$ and so $f$ passes
707to a well defined function of $q=e^{2\pi i\tau}$. So $f(q)$
708is a function on $\{z:0<|z|<1\}$ and the condition that $f(\tau)$
709vanishes at infinity is that $f(q)$ extends to a holomorphic
710function on $\{z:|z|<1\}$ and $f(0)=0$. In this case, we may
711write $f(q)=\sum_{n=1}^{\infty}a_n q^n$.
712
713\section{Lattices}
714A lattice $L\subset \bC$ is a subring $L=\bZ\omega_1 +\bZ\omega_2$
715for which $\omega_1, \omega_2\in \bC$ are lineary independent over $\bR$.
716Without loss, we may assume that $\omega_1 / \omega_2 \in \sH$.
717
718Let
719$$\sR=\{\text{lattices in \bC}\}=\{(E,\omega): 720\text{ E is an elliptic curve, } \omega\in\Omega_E^{1}\}$$
721
722and
723$$M=\{(\omega_1,\omega_2):\omega_1,\omega_2\in\bC, 724\imag(\omega_1/\omega_2)>0\}.$$
725There is a left action of $\sl2z$ on $M$
726$$\abcd:(\omega_1,\omega_2)\mapsto 727(a\omega_1+b\omega_2,c\omega_1+d\omega_2)$$
728and $\sl2z\backslash M\isom \sR$.
729
730\section{Relationship With Elliptic Curves}
731
732There is a map $L\mapsto \bC/L$ from lattices to
733complex tori which, by Weierstrass theory, correspond
734to elliptic curves defined over $\bC$ along with
735a distinguished differential $\omega=dz$.  % huh?
736
737Conversely, if $E/\bC$ is an elliptic curve, we can
738obtain the corresponding lattice by fixing a differential
739$\omega$ and taking the lattice to be the image of the map
740$$H_1(E(\bC),\bZ)\xrightarrow{\text{integration}}\bC$$
741which takes $\gamma\in H_1$ to $\int_{\gamma}\omega\in\bC$.
742
743There is a map $M/\bC\into\sH$ defined by $(\omega_1,\omega_2) 744\mapsto \omega_1/\omega_2$. This gives an isomorphism
745$$\sR/\bC^*=(\sl2z\backslash M)/\bC^* \xrightarrow{ ~ } \sl2z\backslash\sH$$
746and
747$$\sR/\bC^*=\{\text{ elliptic curves /\bC (without differentials)}\}.$$
748
749
750If $f:\sH\into\bC$ we define $F:M\into\bC$ by
751$F(\omega_1,\omega_2)=f(\omega_1/\omega_2)$. Suppose now
752that $F$ is a lattice function and sattisfies the homogeneity % (spelling!?)
753condition $F(\lambda L)=\lambda^{-k} F(L)$.
754Then
755\begin{align*}
756f(\frac{a\tau+b}{c\tau+d})&=F(\bZ\frac{a\tau+b}{c\tau+d}+\bZ)\\
757&= F((c\tau+d)^{-1}(\bZ(a\tau+b)+\bZ(c\tau+d)))\\
758&= (c\tau+d)^k F(\bZ(a\tau+b)+\bZ(c\tau+d))\\
759&= (c\tau+d)^k F(\bZ+\tau\bZ)\\
760&= (c\tau+d)^k f(\tau)
761\end{align*}
762so functions of lattices with the homogeneity condition come
763from functions $f\in M_k$. Thus, if $f\in M_k$ and $F$ is the corresponding
764lattice function then
765$$F(\bZ\omega_1+\bZ\omega_2)=F(\omega_2(\bZ+\bZ \frac{\omega_1}{\omega_2})) 766=\omega_2^{-k}F(\bZ+\bZ 767\frac{\omega_1}{\omega_2}) 768=\omega_2^{-k}f( 769\frac{\omega_1}{\omega_2}), 770$$
771so we can recover $F$ from $f$.
772
773%% I don't understand the stuff with H^0(E,\Omega_E^{1}) being a
774%% 1 dimensional vector space...
775
776\section{Hecke Operators}
777
778Define a map $T_n$ from the free abelian group generated by all
779$\bC$-lattices
780into itself by
781$$T_n(L)= 782\sum_{(L:L')=n} L'.$$
783Then if $F$ is a function on lattices define $T_nF$ by
784$$(T_nF)(L)=n^{k-1}\sum_{(L:L')=n}F(L').$$
785
786Since $(n,m)=1$ implies $T_nT_m=T_{nm}$ and $T_{p^k}$ is a polynomial
787in $\bZ[T_p]$ the essential case to consider is $n$ prime.
788
789Suppose $L'\subset L$ with $(L:L')=n$, then $L/L'$ is killed by $n$
790so $nL\subset L'\subset L$  and
791$$L'/nL\subset L/nL\isom (\bZ/n\bZ)^2.$$
792Thus the subgroups of $(\bZ/n\bZ)^2$ of order $n$ correspond to
793the sublattices $L'$ of $L$ of index $n$. When $n=\ell$ is prime
794there are $\ell+1$ such subgroups. (The subgroups correspond to
795nonzero vectors in $\bF_{\ell}$ modulo scalar equivalence and
796there are $\frac{\ell^2-1}{\ell-1}$ of them.)
797
798Suppose $L'\subset L$ is a sublattice of index $\ell$ and let
799$L''=\ell^{-1}L'$. Note that $\ell L\subset L'$ so
800$L \subset \ell^{-1} L'=L''$ and $L$ is a sublattice of $L''$ of
801index $\ell$. Thus, assuming $F$ satisfies the homogeneity condition
802$F(\lambda L)=\lambda^{-k}F(L)$,
803$$\ell^{k-1}\sum_{L'}F(L') = \frac{1}{\ell}\sum_{L''}F(L'')$$
804which helps explain the extra factor of $n^{k-1}$ in our
805definition of $T_n F$ -- we are averaging'' over the sublattices
806(note that there are $\ell+1$ terms yet we divide by $\ell$ so
807we aren't exactly averaging).
808
809We now give a geometric description of the $\ell$th Hecke operator\index{Hecke operator}.
810Let $L\subset L''$ be lattices with $(L'':L)=\ell$ and let
811$E=\bC/L$, $E''=\bC/L''$ be the elliptic curves corresponding
812to $L$, $L''$, respectively. Then $E[\ell]=\frac{1}{\ell}L/L$
813contains $H=L''/L$ which may be thought of as a line
814[Ed: I don't know why!]. Then the Hecke operator is
815$$E\mapsto \frac{1}{\ell}\sum_{\text{lines }H\subset E[\ell]} E/H.$$
816Let $\hat{\pi}$ be the isogeny dual to $\pi:E\into E/H$.
817Then in terms of pairs $(E,\omega)$ we have
818$$(E,\omega)\mapsto 819\frac{1}{\ell} \sum_{H\subset E[\ell], \#H=\ell}(E/H,\pi_{*}\omega) 820=\ell^{k-1}\sum_{H\subset E[\ell]} (E/H,\hat{\pi}^{*}(\omega)).$$
821
822% Lecture 4, 1/24/96
823
824We consider modular forms $f$ on $\Gamma_1(1)=\sl2z$, that
825is, holomorphic functions on $\sH\cup\{\infty\}$ which satisfy
826$$f(\tau)=f(\frac{a\tau+b}{c\tau+d})(c\tau+d)^{-k}$$
827for all $\abcd\in\sl2z$. Using a Fourier expansion we write
828$$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$
829and say $f$ is a cusp form if $a_0=0$.
830There is a correspondence between modular forms $f$ and
831lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$
832given by $F(\bZ\tau+\bZ)=f(\tau)$.
833
834\section{Explicit Description of Sublattices}
835The $n$th Hecke operator $T_n$ of weight $k$ is defined by
836$$T_n(L)=n^{k-1}\sum_{\substack{L'\subset L\\(L:L')=n}} L'.$$
837What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and
838$$L'/nL\subset L/nL=(\bZ/n\bZ)^2.$$
839Write $L=\bZ\omega_1+\bZ\omega_2$, let $Y_2$ be the cyclic subgroup
840of $L/L'$ generated by $\omega_2$ and let $d=\#Y_2$. Let
841$Y_1=(L/L')/Y_2$, then $Y_1$ is generated by the image
842of $\omega_1$ so it is a cyclic group of order $a=n/d$.
843We want to exhibit a basis of $L'$. Let
844$\omega_2'=d\omega_2\in L'$ and use the fact that $Y_1$ is
845generated by $\omega_1$ to write $a\omega_1=\omega_1'+b\omega_2$
846for some integer $b$ and some $\omega_1'\in L'$. Since $b$ is only
847well-defined modulo $d$ we may assume $0\leq b\leq d-1$.
848Thus
849$$850\Bigl(\begin{matrix}\omega_1'\\ \omega_2'\end{matrix}\Bigr) 851= 852\Bigl(\begin{matrix}a&b\\0&d\end{matrix}\Bigr) 853\Bigl(\begin{matrix}\omega_1\\ \omega_2\end{matrix}\Bigr) 854$$
855and the change of basis matrix has determinent $ad=n$.
856Since
857$$\bZ\omega_1'+\bZ\omega_2'\subset L' \subset L=\bZ\omega_1+\bZ\omega_2$$
858and $(L:\bZ\omega_1'+\bZ\omega_2')=n$ (since the change of basis matrix has
859determinent $n$) and $(L:L')=n$ we see that $L'=\bZ\omega_1'+\bZ\omega_2'$.
860
861Thus there is a one-to-one correspondence between sublattices $L'\subset L$
862of index $n$ and matrices
863$\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr)$
864with $ad=n$ and $0\leq b\leq d-1$.
865In particular, when $n=p$ is prime there $p+1$ of these. In general, the
866number of such sublattices equals the sum of the positive divisors
867of $n$.
868
869\section{Action of Hecke Operators on Modular Forms}
870Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular
871form with corresponding lattice function $F$. How can we describe the
872action of the Hecke operator $T_n$ on $f(\tau)$? We have
873\begin{align*}
874T_nF(\bZ\tau+\bZ) & =  n^{k-1}\sum_{\substack{a,b,d\\ ab=n\\ 0\leq b<d}}
875F((a\tau+b)\bZ + d\bZ)\\
876& = n^{k-1}\sum d^{-k} F(\frac{a\tau+b}{d}\bZ+\bZ)\\
877& = n^{k-1}\sum d^{-k} f(\frac{a\tau+b}{d})\\
878& = n^{k-1}\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\\
879& = n^{k-1}\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}}
880\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\\
881& = n^{k-1}\sum_{\substack{ad=n\\m'\geq 0}}d^{1-k} c_{dm'}e^{2\pi i a m' \tau}\\
882& = \sum_{\substack{ad=n\\m'\geq 0}} a^{k-1} c_{dm'}q^{am'}.
883\end{align*}
884In the second to the last expression we
885 let $m=dm'$, $m'\geq 0$, then used the fact that the
886 sum
887$\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$
888is only nonzero if $d|m$.
889
890Thus
891$$T_nf(q)=\sum_{\substack{ad=n\\m\geq 0}} a^{k-1}c_{dm} q^{am}$$
892and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is
893$$\sum_{\substack{a|n\\ a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$
894
895\begin{remark}
896When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong
897to the $\bZ$-module generated by the $c_m$.
898\end{remark}
899
900\begin{remark}
901Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is
902$$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$
903Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic
904since its original definition is as a finite sum of holomorphic
905functions.)
906\end{remark}
907
908\begin{remark}
909Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is
910$\sum_{a|1}1^{k-1}c_n=c_n$. As an immediate corollary we have the
911following important result.
912\end{remark}
913
914\begin{cor}
915Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient
916of $q$ for all $n\geq 1$, then $f=0$.
917\end{cor}
918
919\begin{remark}
920When $n=p$ is prime we get an interesting formula for the
921action of $T_p$ on the $q$-expansion of $f$.
922One has
923$$T_p f = \sum_{\mu\geq 0} \sum_{\substack{a|n\\a|\mu}}a^{k-1} 924 c_{\frac{n\mu}{a^2}} q^{\mu}.$$
925Since $n=p$ is prime either $a=1$ or $a=p$. When
926$a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$
927and when $a=p$, we can write $\mu=p\lambda$ and we get
928terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$.
929Thus
930$$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+ 931 p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$
932\end{remark}
933
934%% Lecture 5, 1/26/96
935
936\chapter{Embedding Hecke Operators in the Dual}
937
938\section{The Space of Modular Forms}
939Let $\Gamma=\Gamma_1(1)=\sl2z$ and for $k\geq 0$
940let
941\begin{align*}
942M_k&=\{f=\sum_{n=0}^{\infty}a_n q^n : \text{$f$ is a modular form
943for $\Gamma$}\}\\
944&\subset S_k=\{f=\sum_{n=1}^{\infty}a_n q^n\}\end{align*}
945These are finite dimensional $\bC$-vector spaces whose dimensions
946are easily computed. Furthermore, they are generated by familiar elements
947(see Serre \cite{serre2} or Lang \cite{lang1}.)
948\index{Serre}
949The main tool is the formula
950$$\sum_{p\in D\union\{\infty\}} \frac{1}{e(p)}\ord_p(f) = \frac{k}{12}$$
951where $D$ is the fundamental domain for $\Gamma$ and
952$$e(p)=\begin{cases} 1&\text{otherwise}\\ 953 2&\text{if p=i}\\ 954 3&\text{if p=\rho} 955 \end{cases}$$
956One can alternatively define $e(p)$
957as follows. If $p=\tau$ and $E=\bC/(\bZ\tau+\bZ)$
958then $e(p)=\frac{1}{2}\#\aut(E)$.
959
960For $k\geq 4$ we define the {\em Eisenstein series}\index{Eisenstein series} $G_k$ by
961$$G_k(q)=\frac{1}{2}\zeta(1-k)+\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n,$$
962then the map
963$$\tau\mapsto\sum_{\substack{(m,n)\neq(0,0)\\m,n\in\bZ}} 964\frac{1}{(m\tau+n)^k}$$
965differs from $G_k$ by a constant (no proof).
966Also, $\zeta(1-k)\in\bQ$ and one may say, {\em symbolically} at least,
967$\displaystyle \zeta(1-k)=\sum_{d=1}^{\infty} d^{k-1} = 968\sigma_{k-1}(0)$.''
969The {\em $n$th Bernoulli number $B_n$} is defined by the equation
970$$\frac{x}{e^x-1}=\sum_{n=0}^{\infty} \frac{B_nx^n}{n!}.$$
971One can show that $\zeta(1-k)=-\frac{B_k}{k}$ so the constant
972%% How?? Reference???
973coefficient of $G_k$ is $-\frac{B_k}{2k}$ which is rational.
974
975\section{Inner Product}\index{inner product}
976In what follows we assume $k\geq 2$ to avoid trivialities..
977The Hecke operators $T_n$ acts on the space $M_k$. Fix a
978subspace $V\subset M_k$ which is stable under the action
979of the $T_n$. Let $\bT(V)$ be the $\bC$-algebra generated by
980the endomorphism $T_n$ acting on $V$ and note that $\bT(V)$
981is actually a finite dimensional $\bC$-vector space since it
982is a subspace of $End(V)$ and $V$ is finite dimensional.  Recall
983that $\bT$ is commutative.
984
985There is a bilinear form
986\begin{align*}
987\bT\times V &\into \bC \\
988\langle T,f\rangle & \mapsto a_1(f|T)
989\end{align*}
990where $f|T=\sum_{n=0}^{\infty}a_n(f|T)q^n$.
991We thus get maps
992\begin{align*}
993V\into\Hom(\bT,\bC)=\bT^{*}\\
994\bT\into \Hom(V,\bC)=V^{*}.
995\end{align*}
996
997\begin{thm}
998The above maps are isomorphisms.
999\end{thm}
1000\begin{proof}
1001It just remains to show each map is injective.
1002Then since a finite dimensional
1003vector space and its dual have the same dimension the result follows.
1004First suppose $f\mapsto 0\in\Hom(\bT,\bC)$, then
1005$a_1(f|T)=0$ for all $T\in\bT$ so, in particular,
1006$a_n=a_1(f|T_n)=0$ for all $n\geq 1$. Thus $f$ is a constant,
1007but since $k\geq 2$ this implies $f=0$ (otherwise $f$ wouldn't
1008transform correctly with respect to the action of the modular group).
1009
1010Next suppose $T\mapsto 0\in \Hom(V,\bC)$, then
1011$a_1(f|T)=0$ for all $f\in V$. Substiting $f|T_n$ for
1012$f$ and using the commutativity of $\bT$ we have
1013\begin{align*}
1014a_1((f|T_n)|T)&=0 && \text{for all $f$, $n\geq 1$}\\
1015a_1((f|T)|T_n)&=0 && \text{by commutativity}\\
1016a_n(f|T)&=0 && \text{$n\geq 1$}\\
1017f|T&=0 && \text{since $k\geq 2$, as above}
1018\end{align*}
1019Thus $T=0$ which completes the proof.
1020\end{proof}
1021
1022\begin{remark}
1023The above isomorphisms are {\em $\bT$-equivariant}.
1024$\Hom(\bT,\bC)$ is a $\bT$-module if we let $T\in\bT$ act
1025on $\varphi\in\Hom(\bT,\bC)$ by
1026$(T\cdot\varphi)(T')=\varphi(TT')$. If $\alpha:V\into\Hom(\bT,\bC)$
1027is the above isomorphism
1028(so $\alpha:f\mapsto\varphi_f:=(T'\mapsto a_1(f|T'))$)
1029then equivariance is the statement that $\alpha(Tf)=T\alpha(f).$
1030This follows since
1031\begin{align*}
1032\alpha(Tf)(T')&=\varphi_{Tf}(T')=a_1(Tf|T')=a_1(f|T'T)\\
1033&=\varphi_{f}(T'T)=T\varphi(T')=T\alpha(f)(T').
1034\end{align*}
1035\end{remark}
1036
1037\section{Eigenforms}
1038\index{eigenforms}
1039We continue to assume that $k\geq 2$.
1040\begin{defn}
1041A modular form $f\in M_k$ is an {\em eigenform for $\bT$} if
1042$f|T_n=\lambda_n f$ for all $n\geq 1$ and some complex numbers $\lambda_n$.
1043\end{defn}
1044Let $f$ be an eigenform, then
1045$a_n(f)=a_1(f|T_n)=\lambda_n a_1(f)$
1046so if $a_1(f)=0$ then $a_n(f)=0$ for all $n\geq 1$ so
1047since $k\geq 2$ this would imply $f=0$. Thus $a_1(f)\neq 0$
1048and we may as well divide through by $a_1(f)$ to obtain
1049the {\em normalized eigenform} $\frac{1}{a_1(f)}f$. We thus
1050assume that $a_1(f)=1$, then the formula becomes $a_n(f)=\lambda_n$
1051and so $f|T_n = a_n(f) f$, for all $n\geq 1$.
1052
1053\begin{thm}
1054Let $f\in V$ and let $\psi$ be the image of
1055$f$ in $\Hom(\bT,\bC)$, thus $\psi(T)=a_1(f|T)$.
1056Then $f$ is a normalized eigenform iff $\psi$ is a
1057
1058ring homomorphism.
1059\end{thm}
1060
1061\begin{proof}
1062First suppose $f$ is a normalized eigenform so $f|T_n=a_n(f)f$.
1063Then
1064\begin{align*}
1065\psi(T_nT_m) &=a_1(f|T_nT_m)=a_m(f|T_n)\\
1066             &=a_m(a_n(f)f)=a_m(f)a_n(f)\\
1067             &=\psi(T_n)\psi(T_m),
1068\end{align*}
1069so $\psi$ is a homomorphism.
1070
1071Conversely, assume $\psi$ is a homomorphism. Then
1072$f|T_n=\sum a_m(f|T_n)q^m$, so to show that $f|T_n=a_n(f)f$
1073we must show that $a_m(f|T_n)=a_n(f)a_m(f)$. Recall that %% remark 3.3
1074$\psi(T_n)=a_1(f|T_n)=a_n$, thus
1075\begin{align*}
1076a_n(f)a_m(f) &= a_1(f|T_n)a_1(f|T_m) = \psi(T_n)\psi(T_m) \\
1077             & = \psi(T_n T_m) = a_1(f|T_n|T_m) \\
1078             & = a_m(f| T_n)
1079\end{align*}
1080as desired.
1081
1082\end{proof}
1083
1084%% Lecture 6, 1/29/96
1085
1086\chapter{Rationality and Integrality Questions}
1087
1088\section{Review}
1089In the previous lecture we looked at subspaces
1090$V \subset M_k \subset \bC[[q]]$, $(k\geq 4)$, and considered the
1091space $\bT=\bT(V)=\bC[\ldots,T_n,\ldots]\subset\End_{\bsC}V$
1092of Hecke operators on $V$. We defined a pairing
1093$\bT\cross V\into \bC$ by $(T,f)\mapsto a_1(f|T)$ and
1094showed this pairing is nondegenerate and that
1095it induces isomorphisms
1096$\bT\isom\Hom(V,\bC)$ and $V\isom\Hom(\bT,\bC)$.
1097
1098\section{Integrality}
1099Fix $k\geq 4$ and let $S=S_k$ be the space of weight $k$
1100cusp forms with respect to the action of $\sl2z$. Let
1101\begin{align*}
1102S(\bQ)& =S_k\intersect \bQ[[q]]\\
1103S(\bZ)& = S_k\intersect \bZ[[q]].
1104\end{align*}
1105
1106\begin{thm}
1107There is a $\bC$-basis of $M_k$ consisting of forms
1108with integral coefficients.
1109\end{thm}
1110\begin{proof}
1111This is seen by
1112exhibiting a basis. Recall that for all $k\geq 4$
1113$$G_k=-\frac{b_k}{2k}+\sum_{k=1}^{\infty}\sum_{d|k}d^{k-1}q^n$$
1114is the $k$th Eisenstein series\index{Eisenstein series} which is a modular form of weight $k$
1115and
1116$$E_k=-\frac{2k}{b_k}\cdot G_k=1+\cdots$$
1117is its normalization. Since the Bernoulli numbers $b_2,\ldots,b_8$
1118have $1$ as numerator (this isn't always the case,
1119$b_{10}=\frac{5}{66}$) we see that $E_4$ and $E_6$ have coefficients
1120in $\bZ$ and constant term $1$. Furthermore one shows by dimension and
1121independence arguments that the forms
1122$$\{E_4^aE_6^b|4a+6b=k\}$$
1123form a basis for $M_k$.
1124\end{proof}
1125
1126\section{Victor Miller's Thesis}
1127Let $d=\dim_{\bsC}S_k$, then Victor Miller showed in his thesis (see
1128\cite{lang2}, ch. X, theorem 4.4) that there exists
1129$$f_1,\ldots,f_d\in S_k(\bZ) \quad\text{such that}\quad a_i(f_j)=\delta_{ij}$$
1130for $1\leq i,j\leq d$. The $f_i$ clearly form a basis.
1131\begin{prop}
1132Let $R=\bZ[\ldots,T_n,\ldots]\subset End(S_k)$, then
1133$R=\bigoplus_{i=1}^{d} \bZ T_i$.
1134\end{prop}
1135
1136\begin{proof}
1137To see that $T_1,\cdots,T_d\in \bT=\bT(S_k)$
1138are linearly independent over $\bC$  suppose
1139$\sum_{i=1}^{d} c_i T_i = 0$, then
1140$$0=a_1(f_j|\sum c_i T_i)=\sum_{i}c_i a_i(f_j) = 1141\sum_{i} c_i \delta_{ij} = c_j.$$ From the isomorphism
1142$\bT\isom\Hom(S_k,\bC)$ we know that $\dim_{\bsC}\bT=d$,
1143so we can write any $T_n$ as a $\bC$-linear combination
1144$$T_n=\sum_{i=1}^{d}c_{n_i}T_i,\quad c_{n_i}\in\bC.$$
1145But
1146$$\bZ\ni a_n(f_j)=a_1(f_j|T_n)=\sum_{i=1}^{d}c_{n_i}a_1(f_j|T_i) 1147=\sum_{i=1}^{d}c_{n_i}a_i(f_j) = c_{n_j}$$
1148so the $c_{n_i}$ all lie in $\bZ$ which completes the proof.
1149\end{proof}
1150
1151Thus $R$ is an integral Hecke algebra of finite rank $d$ over $\bZ$.
1152We have a map
1153\begin{align*}
1154S(\bZ)\cross R & \into \bZ \\
1155(f,T) & \mapsto a_1(f|T)
1156\end{align*}
1157which induces an embedding
1158$$S(\bZ)\hookrightarrow\Hom(R,\bZ)\isom \bZ^d.$$
1159
1160\begin{exercise}
1161Prove that the map $S(\bZ)\hookrightarrow\Hom(R,\bZ)$ is
1162in fact an isomorphism of $\bT$-modules.
1163[Hint: Show the cokernel is torsion free.]
1164\end{exercise}
1165
1167
1168The main theorem is
1169\begin{thm}
1170The $T_n\in\bT(S_k)$ are all diagonalizable over $\bC$.
1171\end{thm}
1172
1173To prove this we note that $S_k$ supports a non-degenerate positive definite
1174Hermitean inner product (the Petersson inner product)
1175$$(f,g)\mapsto\langle f,g\rangle\in\bC$$
1176such that $\langle f|T_n,g\rangle =\langle f,g|T_n\rangle$.
1177We need some background facts.
1178
1179\begin{defn}
1180An operator $T$ is {\em normal} if it commutes with its adjoint, thus
1181$TT^{*}=T^{*}T$.
1182\end{defn}
1183$T_n$ is clearly normal since $T_n^{*}=T_n$,
1184\begin{thm}
1185A normal operator is diagonalizable.
1186\end{thm}
1187Thus each $T_n$ is diagonalizable.
1188\begin{thm}
1189A commuting family of semisimple (=diagonalizable) operators
1190can be simultaneously diagonalized.
1191\end{thm}
1192Since the $T_n$ commute this implies $S_k$ has a basis consisting
1193of normalized eigenforms $f$. Their eigenvalues are real since
1194\begin{align*}
1195a_n(f)\langle f,f\rangle &=\langle a_n(f)f,f\rangle =\langle f|T_n,f\rangle\\
1196           &=\langle f,a_n(f)f\rangle =\overline{a_n(f)}\langle f,f\rangle.
1197\end{align*}
1198\begin{exercise}
1199The coefficients $a_n$ of the eigenforms are totally real algebraic integers.
1200[Hint: The space $S_k$ is stable under the action of $\aut(\bC)$ on
1201coefficients: if $f=\sum_{n=1}^{\infty}c_n q^n\in S_k$ and
1202$\sigma\in\aut(\bC)$ then $\sigma(f)=\sum_{n=1}^{\infty}\sigma(c_n)q^n$
1203is again in $S_k$ (check this by writing $f$ in terms of a basis
1204$f_1,\ldots,f_d\in S(\bZ)$). Next use the fact that $f$ is an eigenform
1205iff $\sigma(f)$ is an eigenform.]
1206%%% I have absolutely no idea how to do this!!!
1207\end{exercise}
1208
1209Let
1210$$\sH=\{x+iy : x, y\in \bR, \text{ and } y>0\}$$
1211be the upper half plane. Then the volume form
1212$\frac{dx\wedge dy}{y^2}$ is invariant under the action of
1213$$\gl_2^{+}(\bR)=\{M\in\gl_2(\bR) | \det(M)>0\}.$$
1214If $\alpha=\abcd\in\gl_2^{+}(\bR)$ then $\abcd$ acts on $\sH$ by
1215$$\bigabcd:\quad z\mapsto\frac{az+b}{cz+d}$$
1216and one has
1217$$\imag(\frac{az+b}{cz+d})=\frac{\det(\alpha)}{|cz+d|^2}y.$$
1218Differentiating $\frac{az+b}{cz+d}$ gives
1219\begin{align*}
1220d(\frac{az+b}{cz+d})&=
1221\frac{a(cz+d)dz-c(az+b)dz}{(cz+d)^2}\\
1223&= \frac{det(\alpha)}{(cz+d)^2}dz
1224\end{align*}
1225Thus, under the action of $\alpha$, $dz\wedge d{\overline z}$
1226takes on a factor of
1227$$\frac{\det(\alpha)^2}{(cz+d)^2(c\overline{z}+d)^2} 1228=\Bigl(\frac{\det(\alpha)}{|cz+d|^2}\Bigr)^2.$$
1229
1230\begin{defn}
1231The {\em Petersson inner product} of forms $f,g\in S_k$ is defined by
1232$$<f,g>=\int_{\Gamma\backslash\sH}(f(z)\overline{g(z)}y^k) 1233\frac{dx\wedge dy}{y^2},$$
1234where $\Gamma=\sl2z$.
1235\end{defn}
1236
1237Integrating over $\Gamma\backslash\sH$ can be taken to mean integrating
1238over a fundamental domain for the action of $\sH$. Showing that the
1239operators $T_n$ are self-adjoint with respect to the Petersson inner
1240product is a harder computation than Serre \cite{serre2}
1241\index{Serre}
1242might lead one to believe --- it takes a bit of thought.
1243
1244%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1245%% Lecture 7, 1/31/96
1246
1247\chapter{Modular Curves}
1248\index{modular curves}
1249% define various gammas
1250\section{Cusp Forms}
1251Recall that if $N$ is a positive integer we define the congruence
1252subgroups
1253$\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by
1254\begin{align*}
1255\Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\
1256\Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\
1257\Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv
1258             \bigl(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\bigr) \pmod{N}\}.
1259\end{align*}
1260
1261Let $\Gamma$ be one of the above subgroups.
1262One can give a construction of the space $S_k(\Gamma)$ of cusp forms
1263of weight $k$ for the action of $\Gamma$ using the language of
1264algebraic geometry.
1265Let $X_{\Gamma}=\overline{\Gamma\backslash\H^{*}}$
1266be the compactifaction of the upper half plane (union the cusps)
1267modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure
1268of Riemann surface and
1269$S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where
1270$\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$.
1271This works since an element of $H^0(X_{\Gamma},\Omega^1)$
1272is a differential form $f(z)dz$, holomorphic on $\H$ and
1273the cusps, which is invariant with respect to the action
1274of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then
1275$$d(\gamma(z))/dz=(cz+d)^{-2}$$
1276so
1277$$f(\gamma(z))d(\gamma(z))=f(z)dz$$
1278iff $f$ satisfies the modular condition
1279$$f(\gamma(z))=(cz+d)^{2}f(z).$$
1280
1281There is a similar construction of $S_k$ for $k>2$.
1282
1283\section{Modular Curves}\index{modular curves}
1284One knows that $\sl2z\backslash\sH$ parameterizes isomorphism
1285classes of elliptic curves. The other congruence subgroups also
1286give rise to similar parameterizations.
1287Thus $\Gamma_0(N)\backslash\sH$ parameterizes pairs $(E,C)$ where
1288$E$ is an elliptic curve and $C$ is a cyclic subgroup of order
1289$N$, and $\Gamma_1(N)\backslash\H$ parameterizes pairs $(E,P)$ where
1290$E$ is an elliptic curve and $P$ is a point of exact order $N$.
1291Note that one can also give a point of exact order $N$ by giving
1292an injection $\bZ/N\bZ\hookrightarrow E[N]$
1293or equivalently an injection $\Mu_N\hookrightarrow E[N]$
1294where $\Mu_N$ denotes the $N$th roots of unity.
1295$\Gamma(N)\backslash\sH$ parameterizes pairs $(E,\{\alpha,\beta\})$
1296where $\{\alpha,\beta\}$ is a basis for
1297$E[N]\isom(\bZ/N\bZ)^2$.
1298
1299The above quotients spaces are called {\em moduli spaces} for the
1300{\em moduli problem} of determining equivalence classes of
1301pairs ($E +$ extra structure). \index{$\Gamma(N)$-structures}
1302
1303\section{Classifying $\Gamma(N)$-structures}
1304\begin{defn}
1305Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve}
1306$E/S$ is a proper smooth curve
1307$$\begin{CD} E \\ @VfVV \\ S\end{CD}$$
1308with geometrically connected fibers all of genus one, give with a
1309section 0''.
1310\end{defn}
1311
1312Loosely speaking, proper is a generalization of projective
1313and smooth generalizes nonsingularity. See
1314Hartshorne \cite{hartshorne}, chapter III, section 10,
1315for the precise definitions.
1316
1317\begin{defn}
1318Let $S$ be any scheme and $E/S$ an elliptic curve.
1319A {\bfseries $\Gamma(N)$-structure} on $E/S$ is
1320a group homomorphism
1321$$\varphi:(\bZ/N\bZ)^2\into E[N](S)$$
1322whose image generates'' $E[N](S)$.
1323\end{defn}
1324
1325A good reference is chapter 3 of Katz and Mazur \cite{katzmazur}.
1326
1327Define a functor from the category of $\Q$-schemes to the
1328category of sets by sending a scheme $S$ to the
1329set of isomorphism classes of pairs
1330$$(E, \Gamma(N)\text{-structure})$$\index{$\Gamma(N)$-structures}
1331where $E$ is an elliptic curve defined over $S$ and
1332isomorphisms (preserving the $\Gamma(N)$-structure) are taken
1333over $S$. An isomorphism preserves the $\Gamma(N)$-structure
1334if it takes the two distinguished generators to the two
1335distinguished generators in the image (in the correct order).
1336
1337\begin{thm}
1338For $N\geq 4$ the functor defined above is representable and
1339the object representing it is the modular curve $X$ corresponding
1340to $\Gamma(N)$.
1341\end{thm}
1342
1343What this means is that given a $\Q$-scheme $S$, the
1344set $X(S)=\Mor_{\Q\text{-schemes}}(S,X)$ is isomorphic to
1345the image of the functor's value on $S$.
1346
1347There is a natural way to map a pair $(E,\Gamma(N)\text{-structure})$
1348\index{$\Gamma(N)$-structures}
1349to an $N$th root of unity.
1350If $P,Q$ are the distinguished basis of $E[N]$ we send
1351the pair $(E,\Gamma(N)\text{-structure})$ to
1352$$e_N(P,Q)\in\Mu_N$$
1353where $e_N:E[N]\cross E[N]\into \Mu_N$ is the Weil pairing. For
1354the definition of this pairing see chapter III, section 8 of
1355Silverman \cite{silverman1}. The Weil pairing
1356is bilinear, alternating, non-degenerate, Galois invariant, and
1357maps surjectively onto $\Mu_N$.
1358
1359%% 2/2/96
1360\section{More on Integral Hecke Operators}
1361
1362We are considering the algebra of integral Hecke operators $\T=\T_{\Z}$
1363on the space of cusp forms $S_k(\C)$ with respect to the action
1364of the full modular group $\sl2z$. Our goal is to see why
1365$\T\isom\Z^d$ where $d=\dim_{\bsC}S_k(\C)$.
1366
1367Suppose $A\subset\C$ is any {\em subring} of $\C$ and recall that
1368$$\T_A=A[\ldots,T_n,\ldots]\subset \End_{\C}S_k.$$
1369We have a natural map
1370$$\T_A\tensor_A\C\into\T_C$$
1371but we do not yet know that it is
1372an isomorphism.
1373
1374\section{Complex Conjugation}
1375We have a conjugate linear map on functions
1376$$f(\tau)\mapsto \overline{f(-\overline{\tau})}.$$
1377Since $\overline{(e^{-2\pi i\overline{\tau}})}=e^{2\pi i \tau}$,
1378it follows that
1379$$\sum_{n=1}^{\infty} a_n q^n \mapsto 1380 \sum_{n=1}^{\infty}\overline{a_n}q^n$$
1381so it is reasonable to call this map complex conjugation''. Furthermore,
1382if we know that
1383$$S_k(\C)=\C\tensor_{\bsQ}S_k(\Q)$$
1384then it follows that complex conjugation
1385takes $S_k(\C)$ into $S_k(\C)$. To see this note that if
1386we have the above equality then every element of
1387$S_k(\C)$ is a $\C$-linear combination of elements of
1388$S_k(\Q)$ and conversely, and it is clear that the set of such
1389$\C$-linear combinations is invariant under the action
1390of complex conjugation.
1391
1392\section{Isomorphism in the Real Case}
1393
1394\begin{prop} $\T_{\R}\tensor_{\bsR}\C \isom \T_{\C}$, as $\C$-vector spaces.
1395\end{prop}
1396\begin{proof}
1397Since $S_k(\R)=S_k(\C)\intersect\R[[q]]$ and since
1398theorem 5.1 assures us that there is a $\C$-basis of
1399$S_k(\C)$ consisting of forms with integral coefficients,
1400we see that $S_k(\R)\isom \R^d$ where $d=\dim_{\bsC}S_k(\C)$.
1401(Any element of $S_k(\R)$ is a $\C$-linear combination
1402of the integral basis, hence equating real and imaginary
1403parts, an $\R$-linear combination of the integral basis,
1404and the integral basis stays independent over $\R$.)
1405By considering the explicit formula for the action
1406of the Hecke operators $T_n$ on $S_k$ (see section 3)
1407we see that $\T_{\R}$ leaves $S_k(\R)$ invariant, thus
1408$$\T_{\R}=\R[\ldots,T_d,\ldots]\subset \End_{\R}S_k(\R).$$
1409In section 4 we defined a perfect'' pairing
1410$$\T_{\C}\times S_k(\C)\into \C$$
1411which allowed us to show that
1412$\T_{\C}\isom S_k(\C).$ By restricting
1413to $\R$ we again get a perfect pairing so we see that
1414$\T_{\R}\isom S_k(\R) \isom {\R}^d$ which
1415implies that
1416$$\T_R\tensor_{\bsR}\C \xrightarrow{\sim}\T_{\C}.$$
1417\end{proof}
1418
1419This also shows that $S_k(\C)\isom \C\tensor_{\bsR} S_k(\R)$
1420so we have complex conjugation over $\R$.
1421%% For some reason I feel something circular is going on here, but
1422%% i can't put my finger on it... :(
1423
1424\section{The Eichler-Shimura Isomorphism}
1425\index{Eichler-Shimura}
1426
1427Our goal in this section is to outline a homological interpretation
1428of $S_k$. For details see chapter 6 of Lang \cite{lang2},
1429the original paper of Shimura \cite{shimura2},
1430or chapter VIII of Shimura \cite{shimura1}.
1431
1432How is $S_k(\C)$ sort of isomorphic to $H^1(X_{\Gamma},\R)$?
1433Suppose $k=2$ and $\Gamma\subset\sl2z$ is a congruence subgroup,
1434let $X_{\Gamma}=\overline{\Gamma\backslash\H}$ be the
1435Riemann surface obtained by compactifying the upper half plane modulo
1436the action of $\Gamma$. Then $S_k(\C)=H^0(X_{\Gamma},\Omega^{1})$ so
1437we have a pairing
1438$$H_1(X_{\Gamma},\Z)\cross S_k(\C)\into \C$$
1439given by integration
1440$$(\gamma,\omega)\mapsto \int_{\gamma}\omega.$$
1441This gives an embedding
1442$$\Z^{2d}\isom{}H_1(X_{\Gamma},\Z)\hookrightarrow 1443 \Hom_{\C}(S_k(\C),\C)\isom {\C}^d$$
1444of a lattice'' in $\C^d$. (We say lattice'' since there
1445were some comments by Ribet that $Z^{2d}$ isn't a lattice because the rank
1446might be too small since a subring of $\C^d$ having $\Z$-rank $2d$
1447might not spans $\C^d$ over $\C$).
1448Passing to the quotient (and compactifying) gives a complex torus
1449called the Jacobian\index{Jacobian} of $X_{\Gamma}.$
1450Again using the above pairing we get an embedding
1451$${\C}^d\isom S_k(\C)\hookrightarrow 1452 \Hom(H_1(X_{\Gamma},\Z),\C)\isom\C^{2d}$$
1453which, upon taking the real part, gives
1454$$1455S_k(\C) \into \Hom(H_1(X_{\Gamma},\Z),\R) 1456 \isom H^1(X_{\Gamma},\R) \isom H^1_p(\Gamma,\R) 1457$$
1458where $H^1_p(\Gamma,\R)$ denotes the {\em parabolic} group cohomology
1459of $\Gamma$ with respect to the trivial action. It is this result, that we
1460may view $S_k(\C)$ as the cohomology group $H^1_p(\Gamma,\R)$, that was
1461alluded to above.
1462
1463Shimura\index{Shimura} generalized this for arbitrary $k\geq 2$ so that
1464$$S_k(\C)\isom{}H_p^1(\Gamma,V_k)$$
1465where $V_k$ is a $k-1$ dimensional $\R$-vector space.
1466The isomorphism is (approximately) the following:
1467$f\in S_k(\C)$ is sent to the map
1468$$\gamma\mapsto \real\int_{\tau_0}^{\gamma\tau_0}f(\tau){\tau}^{i}d\tau,\quad 1469i=0,\ldots,k-2.$$
1470Let $W=\R\oplus\R$, then $\Gamma$ acts on $W$ by
1471$$\begin{pmatrix}a&b\\c&d\end{pmatrix}: 1472 \begin{pmatrix}x\\y\end{pmatrix}\mapsto 1473 \begin{pmatrix}ax+by\\cx+dy\end{pmatrix} 1474$$
1475so $\Gamma$ acts on
1476$$V_k=\Sym^{k-2}W=W^{\otimes k-2}/S_{k-2}$$
1477where $S_{k-2}$ is the symmetric group on $k-2$ symbols
1478(note that $\dim V_k = k-1$).
1479Let $$L=H_p^1(\Gamma,\Sym^{k-2}(\Z\oplus\Z))$$
1480then under the isomorphism $$S_k(\C)\isom H^1_p(\Gamma,\R)$$
1481$L$ is a sublattice of $S_k(\C)$ of $\Z$-rank 2 which
1482is $T_n$-stable for all $n$. Thus we have an embedding
1483$$\T_{\Z} = \T \hookrightarrow \End L$$
1484and so
1485$\T_{\R}\subset\End_{\bsR}(L\tensor\R)$
1486and $\T_{\Z}\tensor_{\bsZ}\R\isom \T_{\R}$ which has rank $d$.
1487
1488%% What is the point of this last bit??
1489
1490
1491\section{The Petterson Inner Product is Hecke Compatible}\index{inner product}
1492
1493\begin{thm}
1494Let $\Gamma=\sl2z$, let $f,g\in S_k(\C)$,  and let
1495$$\langle f,g\rangle=\int_{\Gamma\backslash\H} f(\tau)\overline{g(\tau)}y^{k} 1496 \frac{dx dy}{y^2}.$$
1497Then this integral is well-defined and Hecke compatible, that is,
1498$\langle f|T_n,g\rangle=\langle f,g|T_n\rangle$ for all $n$.
1499\end{thm}
1500\begin{proof}
1501See Chapter 3 of Lang \cite{lang2}.
1502\end{proof}
1503
1504%% 2/5/96
1505
1506\chapter{Higher Weight Modular Forms}
1507
1508We are considering the spaces $S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$
1509which all have rank $d$. Each is acted upon by the Hecke algebra $\T$.
1510We defined a Hecke compatible inner product (the Peterrson product)
1511and used it to show that $$S_k(\Z)\isom\Hom_{\bsZ}(\T,\Z).$$
1512
1513\section{Definitions of $\T$}
1514
1515\begin{quote}
1516I may be asking you to explain something we have already discussed,
1517but have we intrinsically defined the Hecke operators yet?''
1518
1519-- Saul Schliemer
1520\end{quote}
1521
1522The $T_n$ are defined as operators on $S_k(\C)$ by defining their action
1523on modular forms and noting from explicit formulas that $S_k(\C)$ is
1524preserved. But the $T_n$ can be thought of in other ways, for example,
1525since $$S_k(\C)\isom H_p^1(\Gamma,\Sym^{k-2}(\R\oplus\R))$$
1526one may give an {\em explicit} description of the action of $T_n$ on
1527$H_p^1$.
1528
1529\section{Double Cosets}
1530Let $p$ be a prime and
1531$$\sM_p=\{\alpha\in M_2(\Z) : \det(\alpha)=p\}.$$
1532Let $F:\sL\into\C$ be a function on the free abelian group of lattices
1533and recall that $T_p$ acts on $F$ by
1534$$(T_{p}F)(L)=p^{k-1}\sum_{\substack{L'\subset L\\(L:L')=p}}F(L').$$
1535One can write $\sM_p$ as a disjoint union of left cosets,
1536$$\sM_p= 1537\sl2z \begin{pmatrix}p&0\\0&1\end{pmatrix} \sl2z 1538= \bigcup_{\substack{ad=p\\0\leq b<d}} 1539\begin{pmatrix}a&b\\0&d\end{pmatrix} \sl2z.$$
1540Then $\sum_{(L:L')=p}L'$ may be thought of as the sum of the images
1541of $L$ under the action of the left cosets of $\sM_p$.
1542For a complete exposition, in greater generality, see Shimura\index{Shimura}
1543\cite{shimura1}, especially chapter 3.
1544
1545\section{More General Congruence Subgroups}
1546\begin{defn}
1547A {\bfseries Dirichlet character} mod $N$ is a homomorphism
1548$$\varepsilon:(\Z/n\Z)^{*}\into\C^{*}$$
1549extended to $\Z/N\Z$ by putting $\varepsilon(m)=0$ if
1550$(m,N)\neq 1$.
1551\end{defn}
1552
1553Fix integers $k\geq 0$ and $N\geq 1$. In this section we consider the spaces
1554$$S_k(\Gamma_1(N),\epsilon)$$
1555for Dirichlet characters $\varepsilon$ mod $N$
1556and explicitly describe the action of the Hecke operators
1557$$\begin{cases} 1558 T_n,&n\geq{}1\\ 1559 \dbd{d},&d\in(\Z/N\Z)^{*} 1560\end{cases} 1561$$
1562on these spaces.
1563
1564\begin{remark}
1565Let $n$ be a positive integer. If $(n,N)=1$, then the $T_n$ behave
1566like they do for $\sl2z$. In fact, the $T_n$ and $\dbd{d}$ commute and
1567$$(f|T_n,g)=(f,g|\dbd{n}^{-1}T_n)$$
1568$$(f|\dbd{d},g)=(f,g|\dbd{d}^{-1})$$
1569so the $T_n$ (for $n$ prime to $N$) and $\dbd{d}$ are
1570simultaneously diagonalizable. But if
1571$(n,N)\neq{}1$ then $T_n$ may not be diagonalizable.
1572\end{remark}
1573
1574\begin{defn}
1575Let
1576$$S_k(\Gamma_1(N)) = \{ f : f(\gamma \tau) = (c\tau+d)^{k}f(\tau) \text{ all } 1577 \gamma \in \Gamma_1(N) \}$$
1578where the $f$ are assumed holomorphic on $\sH\union\{\text{cusps}\}$.
1579For each Dirichlet character $\varepsilon$ mod $N$ let
1580$$S_k(\Gamma_1(N),\varepsilon)=\{ f : 1581 f(\gamma\tau)(c\tau+d)^{-k} = \varepsilon(d) f 1582 \text{ all } \gamma=\abcd \in \Gamma_0(N) \}.$$
1583\end{defn}
1584When $\varepsilon\neq 0$ and $f\in{}S_k(\Gamma_1(N),\varepsilon)$
1585one calls $\varepsilon$ the {\bfseries nebentypus} of $f$.
1586
1587Let $d\in(\Z/N\Z)^{*}$ and let $f\in S_k(\Gamma_1(N))$. Let
1588$\gamma=\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr) 1589 \in\Gamma_1(N)$ be a matrix
1590whose lower right entry is congruent to $d$ mod $N$. Then we define
1591$$f(\tau)|\dbd{d} = f(\gamma\tau)(c\tau+d)^{-k}.$$
1592
1593Since $f|\dbd{d}=\varepsilon(d)f$, $S_k(\Gamma_1(N),\varepsilon)$ is
1594the $\varepsilon(d)$ eigenspace of $\dbd{d}$ and $\dbd{d}$ is diagonalizable
1595so one has a direct sum decomposition
1596$$S_k(\Gamma_1(N))=\bigoplus_{\varepsilon:(\bsZ/N\bsZ)^{*}\into\bsC^{*}} 1597 S_k(\Gamma_1(N),\varepsilon).$$
1598If $f\in{}S_k(\Gamma_1(N),\varepsilon)$ then
1599 $$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\in\Gamma_0(N)$$
1600so $$f(-\tau)(-1)^{-k}=\varepsilon(-1)f(\tau)$$
1601so that $S_k(\Gamma_1(N),\varepsilon)=0$ unless $\varepsilon(-1)=(-1)^{k}$.
1602Thus about half of the direct summands vanish.
1603
1604\section{Explicit Formulas}
1605Let $$f=\sum_{n=1}^{\infty}a_n q^n \in S_k(\Gamma_1(N),\varepsilon)$$
1606and let $p$ be a prime, then
1607$$f|T_p = \begin{cases} 1608\displaystyle 1609\sum_{n=1}^{\infty} a_{np}q^n + p^{k-1}\varepsilon(p) 1610 \sum_{n=1}^{\infty} a_n{}q^{pn}, &p\nd N\\ 1611\displaystyle 1612\sum_{n=1}^{\infty} a_{np}q^n + 0, &p|N 1613\end{cases} 1614$$
1615When $p|N$, $T_p$ is often denoted $U_p$ and called an Atkin-Lehner
1616operator.
1617
1618We have the relations
1619\begin{align*}
1621T_{p^k}&=\begin{cases}
1622   (T_p)^k, & p|N\\
1623   ?, & p\nd N
1624\end{cases}
1625\end{align*}
1626
1627\section{Old and New Forms}
1628\index{newform}
1629{\bfseries Warning:} $T_p$ is not necessarily diagonalizable if $p|N$.
1630There is an example due to Shimura\index{Shimura}, to present it we must first
1631introduce old and new forms.
1632
1633Let $M$ and $N$ be positive integers such that $M|N$ and let $d|\frac{N}{M}$.
1634If $f(\tau)\in S_k(\Gamma_1(M))$ then $f(d\tau)\in{}S_k(\Gamma_1(N))$.
1635We thus have a map $S_k(\Gamma_1(M))\into{}S_k(\Gamma_1(N))$ for each
1636$d|\frac{N}{M}$. Combining these gives a map
1637$$\varphi_M:\bigoplus_{d|\frac{N}{M}}S_k(\Gamma_1(d))\into S_k(\Gamma_1(N)).$$
1638
1639\begin{defn} The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the subspace
1640generated by the images of the $\varphi_M$ for $M|N$, $M\neq N$.
1641\end{defn}
1642
1643We remark that the {\bfseries new part} of $S_k(\Gamma_1(N))$ is the
1644orthogonal complement of the old part with respect to the Petersson inner
1645product.
1646
1647%% NOte; he goes on to map S_k(\Gamma_1(N))^2\into S_k(\Gamma_1(p))
1648%% which makes no sense to me in this framework -- see the written notes
1649%% if it makes sense wednesday.
1650
1651%% 2/7/96
1652
1653\chapter{New Forms}
1654
1655Today we discuss how the Hecke operators $T_n$ on $S_k(\Gamma_1(N))$
1656can fail to be diagonalizable. Let $N$ be a positive integer and
1657$M$ a divisor of $N$. For each $d|\frac{N}{M}$ we define a map
1658$$\alpha_{d}:S_k(\Gamma_1(M))\into S_k(\Gamma_1(N)):\quad{} 1659 f(\tau)\mapsto{}f(d\tau).$$
1660Note that when $T_p$ acts on the image space $S_k(\Gamma_1(N))$ we
1661will often denote it by $U_p$.
1662We must check that $f(d\tau)\in{}S_k(\Gamma_1(N))$. Define for
1663$\gamma=\abcd$,
1664$$(f|[\gamma]_k)(\tau)=\det(\gamma)^{k-1}(cz+d)^{-k}f(\gamma(\tau)).$$
1665Thus $f\in S_k(\Gamma_1(N))$ iff $f|[\gamma]_k(\tau)=f(\tau)$ (and
1666$f$ is holomorphic). Now let $f(\tau)\in\Gamma_1(M)$ and let
1667$\iota_d=\bigl(\begin{smallmatrix}d&0\\0&1\end{smallmatrix}\bigr)$. Then
1668$f|[\iota_d]_k(\tau)=d^{k-1}f(d\tau)$ is a modular form on
1669$\Gamma_1(N)$ since $\iota_d^{-1}\Gamma_1(M)\iota_d$ contains
1670$\Gamma_1(N)$ (check this directly by conjugating an element
1671of $\Gamma_1(N)$ by $\iota_d$).
1672Moreover if $f$ is a cusp form then so is $f|[\iota_d]_k$.
1673If $f\in S_k(\Gamma_1(M))$ is nonzero, then as $d$ varies over divisors
1674of $\frac{N}{M}$, the various $f(d\tau)$ are linearly
1675independent.
1676
1677Suppose $f\in S_k(\Gamma_1(M))$ is a normalized eigenform for
1678all of the Hecke operators $T_n$ and $\dbd{n}$, and $p$ is a prime
1679not dividiing $M$. Then
1680$$f|T_p=a f \quad \text{and} \quad f|\dbd{p}=\varepsilon(p)f.$$
1681Assume $N=p^{r}M$ where $r$ is an integer $\geq 1$.
1682Let $$f_i(\tau)=f(p^i\tau),$$
1683so $f_0,\ldots,f_r$ are the images of $f$ under the maps
1684defined above and $f=f_0$. Consider the action of $U_p$ on the $f_i$.
1685From previous work we have
1686\begin{align*}
1687f|T_p & = \sum_{n\geq 1} a_{np}q^n+\varepsilon(p)p^{k-1}\sum a_n{}q^{pn}\\
1688      & = f_0|U_p + \varepsilon(p)p^{k-1} f_1
1689\end{align*}
1690so
1691$$f_0|U_p = f|T_p - \varepsilon(p)p^{k-}f_1 1692 = af_0 - \varepsilon(p)p^{k-1}f_1.$$
1693Also
1694$$f_1|U_p = (\sum a_n q^{pn}) | U_p = \sum a_n q^n = f_0.$$
1695More generally one can show that $f_i|U_p = f_{i-1}$.
1696
1697$U_p$ preserves the two dimensional vector space spanned by
1698$f_0$ and $f_1$. The matrix of $U_p$ is
1699$$A=\Bigl(\begin{matrix}a&1\\-\varepsilon(p)p^{k-1}&0\end{matrix}\Bigr)$$
1700which has characteristic polynomial
1701$$\chi_A(X)=X^2 - aX + p^{k-1}\varepsilon(p).$$
1702
1703\section{Connection With Galois Representations}
1704\index{Galois representations}
1705This leads to a striking connection with Galois representations.
1706Let $f$ be a modular form and $E$ be the field generated over $\Q$
1707by the coefficients of $f$. Let $\ell$ be a prime and $\lambda$
1708a prime lying over $\ell$. Then one constructs a representation
1709$$\rho_{\lambda}:\gal(\overline{\Q}/\Q)\into\GL(2,E_{\lambda}).$$
1710If $p\nd N\ell$, then $\rho_{\lambda}$ is unramified at $p$,
1711so there is a Frobenious element $\frob_p\in\gal(\overline{\Q}/\Q)$.
1712One can show that
1713\begin{align*}
1714\det(\rho_{\lambda}(\frob_p)) &= p^{k-1}\varepsilon(p) \\
1715\tr(\rho_{\lambda}(\frob_p)) & = a_p = a,
1716\end{align*}
1717so the characteristic polynomial of $\rho_{\lambda}(\frob_p)\in\GL_2(E_{\lambda})$
1718is $$X^2-a_p X + p^{k-1}\varepsilon(p).$$
1719
1720\section{Semisimplicity of $U_p$}
1721\begin{question}
1722Is $U_p$ semisimple on the span of $f_0$ and $f_1$?
1723\end{question}
1724
1725If the eigenvalues are distinct the answer is clearly yes.
1726If the eigenvalues are the same, then $X^2-aX+p^{k-1}\varepsilon(p)$
1727has discriminant zero, that is, $4\varepsilon(p)p^{k-1}=a^2$ so
1728$$a=2p^{\frac{k-1}{2}}\sqrt{\varepsilon(p)}.$$
1729Is this possible? The answer is still {\em unknown}, although it is
1730a curious fact that the Ramanujan conjectures (proved by Delign in 1973)
1731imply that $|a|\leq 2p^{\frac{k-1}{2}}$, so the above equality remains
1732taunting.
1733
1734When $k=2$ Weil showed that $\rho_{\lambda}(\frob_p)$ is semisimple
1735so if the eigenvalues of $U_p$ are equal then $\rho_{\lambda}(\frob_p)$
1736is a scalar. But Edixhoven and Coleman \cite{edixcole}
1737show that it is not a scalar by looking at the abelian
1738variety attached to $f$.
1739
1740\section{Shimura's Example of Nonsemisimple $U_p$}
1741\index{Shimura}
1742
1743Let $W$ be the space spanned by $f_0, f_1$ and let
1744$V$ be the space spanned by $f_0, f_1, f_2, f_3$.
1745$U_p$ acts on $V/W$ by $\overline{f_2}\mapsto 0$
1746and $\overline{f_3}\mapsto \overline{f_2}$. Thus the matrix of the
1747action of $U_p$ on $V/W$ is
1748$\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$
1749which is nonzero and nilpotent hence not semisimple.
1750Since $W$ is invariant under $U_p$ this shows that
1751$U_p$ is not semisimple on $V$.
1752%% I CAN'T SEE THIS! It seems like basic representation theory...
1753
1754\section{An Interesting Duality}
1755
1756Now suppose $N=1$ thus $\Gamma_1(N)=\sl2z$. Because of the
1757Petersson product all the $T_n$ are diagonalizable, so
1758$S_k=S_k(\Gamma_1(1))$ has a basis
1759$$f_1,\ldots,f_d$$
1760of normalized eigenforms where $d=\dim S_k$. Let $\T=\T_{\C}$, then
1761there is a {\em canonical} map
1762$$\T_{\C}\hookrightarrow{}\C^d: \quad T\mapsto(\lambda_1,\ldots,\lambda_d)$$
1763where $f_i|T=\lambda_{i}f_i$. This map is clearly injective and we
1764know by previous arguments that $\dim\T_{\C}=d$ so the map is an isomorphism
1765of $\C$-vector spaces.
1766
1767The form
1768$$v=f_1+\cdots+f_n$$ generates $S_k$ as a $\T$-module.
1769Since $v$ corresponds to the vector $(1,\ldots,1)$ and
1770$\T\isom\C^d$ acts on $S_k\isom\C^d$ componentwise this
1771is just the statement that $\C^d$ is generated by
1772$(1,\ldots,1)$ as a $\C^d$-module -- which is clear.
1773Thus we have simultaneously:
1774
17751) $S_k$ is free of rank 1 over $\T$, and
1776
17772) $S_k=\Hom_{\C}(\T,\C)$ as $\T$-modules, thus
1778
1779$$\T\isom\Hom_{\C}(\T,\C).$$ The isomorphism sends
1780an element of $T\in \T$ to $Tv\in S_k$. Since
1781the identification $S_k=\Hom_{\C}(\T,\C)$ was
1782constructed using the Petersson product it is canonical
1783and since the choice of a normalized eigenbasis $f_1,\ldots,f_d$
1784is canonical we see that the isomorphism $T\isom\Hom_{\C}(\T,\C)$
1785is canonical.
1786
1787\begin{prop}
1788$v\in S_k(\Q)$
1789\end{prop}
1790\begin{proof}
1791Let $\sigma\in\gal(\overline{\Q}/\Q)$, then if $f_i$ is
1792a normalized eigenform so is $\sigma(f_i)$ (from the explicit
1793formula). Thus $\sigma(f_1+\cdots+f_n)=f_1+\cdots+f_n$ for
1794all $\sigma$ as desired.
1795\end{proof}
1796
1797Now we consider the case for general $N$.
1798Recall that we have defined maps
1799$$S_k(\Gamma_1(M))\into S_k(\Gamma_1(N))$$
1800for all $M$ dividing $N$ and all divisors $d$ of $\frac{N}{M}$.
1801
1802\begin{defn}
1803The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the space
1804generated by all images of these maps with $M|N$ but $M\neq N$.
1805The {\bfseries new part} is the orthogonal complement of the
1806old part with respect to the Petersson product.
1807\end{defn}
1808
1809There is an algebraic definition of the new part. One defines
1810certain trace maps
1811$$S_k(\Gamma_1(N))\into S_k(\Gamma_1(M))$$
1812for all $M<N$, $M|N$
1814Then $f$ is in the new part of $S_k(\Gamma_1(N))$ iff $f$
1815is killed by all of these maps.
1816
1817It can be shown that the $T_n$ act semisimply on
1818$S_k(\Gamma_1(M))_{\text{new}}$ for all $M\geq 1$.
1819Thus $S_k(\Gamma_1(M))_{\text{new}}$ has a basis of eigenforms.
1820We have a natural map
1821$$\bigoplus_{M|N} S_k(\Gamma_1(M))_{\text{new}}\hookrightarrow 1822 S_k(\Gamma_1(N)).$$
1823The image in $S_k(\Gamma_1(N))$ of an eigenform $f$ for some
1824$S_k(\Gamma_1(M))_{\text{new}}$ is called a {\bfseries newform}
1825of level $M_f=M$. Note that a newform is not necessarily
1826an eigenform for the Hecke operators acting on $S_k(\Gamma_1(N))$.
1827Let
1828$$v=\sum_{f} f(q^{\frac{N}{M_f}})\in S_k(\Gamma_1(N))$$
1829where the sum is taken over all newforms $f$ of weight
1830$k$ and some level $M|N$.
1831This generalizes the $v$ constructed above when $N=1$
1832and has many of the same good properties. For example,
1833$S_k(\Gamma_1(N))$ is free of rank $1$ over $\T$ with
1834basis element $v$. The coefficients of $v$ lie in $\Q$,
1835but to show this we need to know the new part of $S_k(\Gamma_1(N))$
1836is stable under the action of the Galois group of $\Q$.
1837This is not easy since the new part is defined in terms of
1838the Petersson product which is an analytic construction.
1839Serre\index{Serre} circumvents this problem by giving an alternative
1840definition in terms of trace maps going the other way.
1841
1842
1843%% 2/9/96
1844\section{Observations on $T_n$}
1845Let $\T_{\Q}=\Q[\cdots,T_n,\cdots]$ and
1846$\Gamma=\Gamma(1)=\modgp$. Let $f_1,\ldots,f_d$ be a basis of $\Gamma$
1847consisting of normalized eigenforms.
1848\begin{prop}
1849The coefficients of the $f_i$ are totally real algebraic integers.
1850\end{prop}
1851\begin{proof}
1852$\gal(\C/\Q)$ acts on $f_i$ by acting on the coefficients
1853of its $q$-expansion. From the explicit formula in section 3.2
1854one sees that the set $\{f_1,\ldots,f_d\}$ is
1855stable under the action of $\gal(\C/\Q)$.
1856For any $i$, $a_n(f_i)$ is an eigenvalue of $T_n$ since
1857$f_i|T_n = a_n(f_i)f_i$, and $T_n$ is self-adjoint so
1858$a_n(f_i)$ must be real. Thus all conjugates of $a_n(f_i)$
1859are real and there are only finitely many since a conjugate
1860of $a_n(f_i)$ must be $a_n(f_j)$ for some $j$, $1\leq j\leq d$.
1861\end{proof}
1862
1863\begin{prop}
1864The operators $\dbd{d}$ on $S_k(\Gamma_1(N))$
1865lie in $\Z[\ldots,T_n,\ldots]$.
1866\end{prop}
1867\begin{proof}
1868It is enough to show $\dbd{p}\in\Z[\ldots,T_n,\ldots]$ for
1869There is a formula relating $\dbd{p}$ and $T_p$,
1870$$p^{k-1}\dbd{p}={T_p}^2-T_{p^2}.$$
1871By Dirichlet's theorem on prime's in arithmetic progression,
1872see VIII.4 of Lang \cite{lang1},
1873there is another prime $q$ congruent to $p$ mod $N$.
1874Since $p^{k-1}$ and $q^{k-1}$ are relatively prime there
1875exist integers $a$ and $b$ so that
1876$a p^{k-1} + b q^{k-1} = 1$. Then
1877$$1878\dbd{p}=\dbd{p}(a p^{k-1} + b q^{k-1}) 1879 = a({T_p}^2-T_{p^2}) + b({T_q}^2-T_{q^2}). 1880$$
1881\end{proof}
1882
1883Let $\Sigma$ be a set of representatives of
1884$\{f_1,\ldots,f_d\} \backslash \gal(\C/\Q)$. It
1885is unknown whether or not $\#\Sigma$ can be
1886larger than one, that is, whether the eigenforms
1887are all conjugate under the action of Galois.
1888Let $K_f = \Q(\ldots,a_n(f),\ldots)$ and defined a homomorphism
1889of $\Q$-algebras
1890$$T_{\bsQ}\into K_f : T_n\mapsto \lambda 1891 \text{ where }T_n f = \lambda f$$
1892Taking the product over a set of representatives of the $f_i$
1893yields a map
1894$$\T_{\bsQ}\xrightarrow{\sim}\prod_{f\in\Sigma}K_f$$
1895which one can show is an isomorphism of $\Q$-algebras.
1896
1897\begin{example}
1898Consider $S_2(\Gamma_0(N))$ with $N$ prime, then
1899$$\T_{\bsQ}\isom E_1\times\cdots E_t$$
1900with the $E_i$ totally real fields. When $N=37$,
1901that $\T_{\bsQ}\isom \Q\cross\Q$.
1902\end{example}
1903
1904\chapter{Some Explicit Genus Computations}
1905\section{Computing the Dimension of $S_k(\Gamma)$}
1906Let $k=2$ unless otherwise noted, and let
1907$\Gamma\subset\modgp$ be a congruence subgroup.
1908Then $$S_2(\Gamma)=H^{0}(X_{\Gamma},\Omega^1)$$
1909where $$X_{\Gamma}=(\Gamma\backslash\H)\union 1910 (\Gamma\backslash\bP^1(\Q)).$$
1911By definition $\dim H^{0}(X_{\Gamma},\Omega^1)$
1912is the genus of $X_{\Gamma}$.
1913\begin{exercise}
1914Prove that when $\Gamma=\modgp$ then $\Gamma\backslash\bP^1(\Q)$
1915is a point.
1916\end{exercise}
1917
1918Since $\Gamma\subset\Gamma(1)$ there is a covering
1919$$\begin{CD} 1920\Gamma\backslash\H @>>> X_{\Gamma} \\ 1921@VVV @VVV \\ 1922\Gamma(1)\backslash\H @>>> X_{\Gamma(1)} @>j>> \bP^1(\C) 1923\end{CD}$$
1924which is only ramified at points above $0, 1728, \infty$
1925($0$ corresponds to $i$ and $\rho$ to $1728$ under $j$).
1926
1927\begin{example} Suppose $\Gamma=\Gamma_0(N)$, then the degree
1928of the covering is the index $(\modgp/\{\pm 1\}:\Gamma_0(N)/\{\pm 1\})$.
1929A point on $Y_{\Gamma(1)}$ corresponds to an elliptic curve, whereas
1930a points on $Y_{0}(N)$ correspond to a pair consisting of an
1931elliptic curve and a subgroup of order $N$.
1932\end{example}
1933
1934\section{Application of Riemann-Hurwitz}
1935Now we compute the genus of $X_{\Gamma}$ by applying the
1936Riemann-Hurwitz formula.
1937Intuitively the Euler charcteristic
1938should be totally additive, that is, if $A$ and $B$ are
1939disjoint spaces then
1940$$\chi(A\union B)=\chi(A)+\chi(B).$$
1941Let $X$ be a compact Riemann surface of genus $g$, then
1942$\chi(X)=2-2g$. Since $\chi(\{\text{point}\})=1$ we should
1943have that
1944$$\chi(X-\{p_1,\ldots,p_n\})=\chi(X)-n\chi(1)=(2-2g)-n.$$
1945If we have an umramified covering $X\into Y$
1946of degree $d$ then $\chi(X)=d\cdot\chi(Y)$.
1947Consider the covering
1948$$\begin{CD}X_{\Gamma}-\{\text{points over 0,1728,\infty}\}\\ 1949 @VVV 1950 \\ X_{\Gamma(1)}-\{0,1728,\infty\} 1951\end{CD}$$
1952Since $X_{\Gamma(1)}$ has genus $0$, $X_{\Gamma(1)}-\{0,1728,\infty\}$
1953has Euler characteristic $2-3=-1$. If we let $g=\chi(X_{\Gamma})$ then
1954$\chi(X_{\Gamma}-\{\text{points over$0,1728,\infty$}\} 1955 = 2-2g -n_{0} - n_{1728} - n_{\infty}$,
1956where $n_p$ denotes the number of points lying over $p$.
1957Thus $-d=2-2g-n_0-n_{1728}-n_{\infty}$ whence
1958$$2g-2 = d - n_0 -n_{1728}-n_{\infty}.$$
1959
1960Suppose $\Gamma=\Gamma_0(N)$ with $N>3$, then $n_0=d/3$ and $n_{1728}=d/2$
1961(I'm not sure why).
1962The degree $d$ of the covering is equal to the number of unordered
1963ordered basis of $E[N]$, thus
1964$$d=\#\SL_2(\Z/N\Z)/2.$$
1965We still need to compute $n_{\infty}$. $\modgp$ acts on $\bP^1(\Q)$
1966if we view $\bP^1(\Q)$ as all pairs $(a,b)$ of relatively prime integers
1967and suppose $\infty$ corresponds to $(1,0)$. The stabilizer of $(1,0)$ is
1968the sugroup $\{\abcd \in \modgp : c=0 \}$ of upper triangular matrices.
1969Since the points lying over $\infty$ are all conjugate by the Galois
1970group of the covering (which is $\SL_2(\Z/N\Z)/\{\pm 1\}$),
1971$$\text{number of cusps}=\frac{\text{order of \SL_2(\Z/N\Z)/\{\pm 1\}}} 1972 {\text{order of stabilizer of \infty}}.$$
1973We thus have
1974$$2g(X(N))-2=\frac{d}{6}-\frac{d}{N}$$
1975where $\frac{d}{N}$ is the number of cusps.
1976
1977%% 2/12/96
1978\section{Explicit Genus Computations}
1979
1980Let $N>3$ and consider the modular curve $X=X(N)$.
1981There is a natural covering map $X\into{}X(1)\xrightarrow{j}\C$.
1982Let $d$ be the degree, then
1983$$2g-2=d-m_0-m_{1728}-m_{\infty}$$
1984where $g$ is the genus of $X$ and
1985$m_x$ is the number of points lying over $x$.
1986Since $m_0$ is approximately $\frac{d}{3}$ and $m_{1728}$
1987is approximately $\frac{d}{2}$,
1988$$2g-2=\frac{d}{6}-m_{\infty}\pm\text{ small correction factor}.$$
1989
1990\section{The Genus of $X(N)$}
1991Now we count the number of cusps of $X(N)$, that is, the
1992size of $\Gamma(N)\backslash\bP^1(\Q)$. There is a surjective map
1993from $\modgp$ to $\bP^1(\Q)$ given by
1994$$\bigabcd\mapsto \bigabcd\Bigl(\begin{matrix}1\\0\end{matrix}\Bigr).$$
1995Let $U$ be the kernel, thus $U$ is the stabilizer of
1996$\infty=\bigl(\begin{smallmatrix}1\\0\end{smallmatrix}\bigr)$,
1997so $U=\{\pm\bigl(\begin{smallmatrix}1&a\\0&1\end{smallmatrix}\bigr):a\in\Z\}$.
1998Then the cusps of $X(N)$ are the elements of
1999$$\Gamma(N)\backslash(\modgp/U)=(\Gamma(N)\backslash\modgp)/U=\SL_2(\Z/N\Z)/U$$
2000which has order $$\frac{\#\SL_2(\Z/N\Z)}{2N}=\frac{d}{N}.$$
2001
2002Substituting this into the above formula gives
2003$$2g-2=\frac{d}{6}-\frac{d}{N}=\frac{d}{6N}(N-6)$$
2004so $$g=1+\frac{d}{12N}(N-6).$$
2005When $N$ is prime
2006$$d=\frac{1}{2}\#\SL_2(\Z/N\Z)=\frac{1}{2}\cdot\frac{(N^2-1)(N^2-N)}{N-1}.$$
2007Thus when $N=5$, $d=60$ so $g=0$, and when $N=7$, $d=168$ so $g=3$.
2008
2009\section{The Genus of $X_0(N)$}
2010Suppose $N>3$ and $N$ is prime. The covering map
2011$X_0(N)\into X(1)$ is of degree $N+1$ since a point of $X_0(N)$
2012corresponds to an elliptic curve along with a subgroup of order $N$
2013and there are $N+1$ such subgroups because $N$ is prime.
2014\begin{exercise}
2015$X_0(N)$ has two cusps; they are the orbit of $\infty$ which
2016is unramified and $0$ which is ramified of order $N$.