\documentclass[landscape]{slides} \usepackage{fancybox} \usepackage{amsmath} \usepackage{amssymb} \usepackage{pstricks} \usepackage{graphicx} \usepackage[all]{xy} \newcommand{\mtwo}[4]{\left( \begin{matrix}\hfill #1&\hfill #2\\\hfill #3&\hfill #4 \end{matrix}\right)} \DeclareMathOperator{\moddeg}{moddeg} \DeclareMathOperator{\HH}{H} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\Jac}{Jac} \DeclareMathOperator{\Disc}{Disc} \DeclareMathOperator{\Det}{Det} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\res}{res} \renewcommand{\H}{\HH} \newcommand{\ds}{\displaystyle} \DeclareMathOperator{\Vis}{Vis} \DeclareMathOperator{\Coker}{Coker} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\Reg}{Reg} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\op}{odd} \DeclareMathOperator{\tor}{tor} \DeclareFontEncoding{OT2}{}{} % to enable usage of cyrillic fonts \newcommand{\textcyr}[1]{% {\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}% \selectfont #1}} \newcommand{\Sha}{{\mbox{\textcyr{Sh}}}} \newcommand{\first}[1]{\hline#1} \newcommand{\add}[1]{\\} \newcommand{\sha}[3]{{\bf #1}, dim #2, $\ell=#3$} \newcommand{\higher}[3]{$\mathbf{p=#1}$: dim #2, rank #3} \newrgbcolor{dblue}{0 0 0.8} \newcommand{\h}[1]{\begin{center}\LARGE\bf\dblue #1\end{center}} \newcommand{\htwo}[1]{\begin{center}\Large\bf\dblue #1\end{center}} \newrgbcolor{dred}{0.7 0 0} \newrgbcolor{dgreen}{0 0.3 0} \newcommand{\rd}[1]{{\bf \dred #1}} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\charpoly}{charpoly} \newcommand{\Q}{\mathbf{Q}} \newcommand{\F}{\mathbf{F}} \newcommand{\C}{\mathbf{C}} \newcommand{\T}{\mathbf{T}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\Qbar}{\mathbf{\overline{Q}}} \newcommand{\con}{\equiv} \newcommand{\hra}{\hookrightarrow} \newcommand{\isom}{\cong} \newcommand{\ce}{\stackrel{\mbox{\tiny conj.}}{=}} \newcommand{\nT}{\tilde{\T}} %\newcommand{\photo}[2]{{\small [Photo: #1 #2]}} \newcommand{\photo}[2]{\includegraphics[width=#1in]{graphics/#2.eps}} %opening \title{\Large\bf\dblue Modular Degrees of Elliptic Curves\\and\\ Discriminants of Hecke Algebras} \author{\rd{\Large William Stein\footnote{This is joint work with F.~Calegari.\hspace{2ex}\photo{1}{frank}}}\\ {\tt http://modular.fas.harvard.edu}} \date{ANTS VI, June 18, 2004\vspace{1ex} } \begin{document} \maketitle \begin{slide} \h{\rd{Goal}\hspace{2in}} Let $p$ be a prime. My goal is to explain and justify the following \underline{Calegari-Stein conjectures} (note: 3 implies 2 implies 1): \vspace{-1.5ex} {\Large{\dblue\bf Conjecture 1:} }{\large If $E/\Q$ is an elliptic curve of conductor $p$, then the modular degree $m_E$ of $E$ is not divisible by $p$.} \vspace{-1.5ex} {\Large{\dblue\bf Conjecture 2:} } {\large If $\T_2(p)$ is the Hecke algebra associated to $S_2(p)$, then $p$ does not divide the index of $\T_2(p)$ in its normalization.} \vspace{-1.5ex} {\Large{\dblue\bf Conjecture 3:} } {\large If $p\geq k-1$, then there is an explicit formula for the $p$-part of the index of $\T_k(p)$ in its normalization.} \vspace{-1.5ex} \end{slide} \begin{slide} %\h{Conjecture 1} \htwo{Conj 1: If $E$ of conductor $p_E$, then $p_E\nmid m_E$.} \rd{\bf A Motivation:} \photo{1.3}{flach} Conjecture 1 looks like Vandiver's conjecture, which asserts that $p\nmid h_p^-$. Flach proved the modular degree annihilates $\Sha({\rm Sym}^2(E))$, which is an analogue of a class group. \end{slide} \begin{slide} \htwo{Conj 1: If $E$ of conductor $p_E$, then $p_E\nmid m_E$.} \rd{\bf Watkins Data:} \photo{1.2}{watkins} For $p_E<10^7$ there are 52878 curves of prime conductor whose modular degree Watkins computed. No counterexamples to Conjecture 1 in the data. There are $23$ curves such that $m_E$ is divisible by a prime $\ell> p_E$. For example the curve $y^2 + xy = x^3 - x^2 -391648x -94241311$ of prime conductor $p_E=4\,847\,093$ has modular degree $2\cdot 21\,695\,761$. Smallest $p_E$ with some $\ell>p_E$ is $p_E=1\,194\,923$. \end{slide} \begin{slide} \h{More Data} \begin{itemize} \item The \rd{maximum} known ratio $\ds\frac{m_E}{p_E}$ is $\sim 23.2$, attained for $p_E=7\,944\,197$. \item \rd{First} curve with $\ds \frac{m_E}{p_E}>1$ has $p_E=13723$ and $m_E=16176=2^4\cdot 3\cdot 337$. \item \rd{Smallest} known $\ds\frac{m_E}{p_E}>1$ is $1.0004067\ldots$ for $p_E=1\,757\,963$ where $m_E=p_E+715$. \end{itemize} %Conjecture is consistent with $m_E\gg p_E^{7/6-\varepsilon}$. \end{slide} \begin{slide} \htwo{Modular Forms\hspace{1in}\photo{1.3}{hecke}} \vspace{-2ex} \rd{Congruence Subgroup:} $$ \Gamma_0(N) = \left\lbrace \mtwo{a}{b}{c}{d} \in \SL_2(\Z) \text{ such that } N \mid c \right\rbrace. $$ \vspace{-2ex} \rd{Cusp Forms:} $\displaystyle S_k(N) = \Bigl\{f : \mathfrak{h}\to\C \text{ such that } $ \vspace{-2ex} $$ \qquad\qquad\qquad f(\gamma(z)) = (cz+d)^{-k}f(z) \text{ all } \gamma\in\Gamma_0(N), \vspace{-1ex}$$ $$\qquad \qquad \qquad \text{ and $f$ is holomorphic at the cusps}\Bigr\} $$ \vspace{-1ex} \rd{Fourier Expansion:} $$ f = \sum_{n\geq 1} a_n e^{2\pi i z n} = \sum_{n\geq 1} a_n q^n \in \C[[q]]. $$ \end{slide} \begin{slide} \htwo{\photo{1}{merel} Computing Modular Forms \photo{1}{cremona}} $S_k(N)=0$ if $k$ is odd, so we will not consider odd $k$ further. For $k\geq 2$, a basis of $S_k(N)$ can be computed to any given precision using \rd{modular symbols}. Appears that no formal analysis of complexity has been done. Certainly polynomial time in $N$ and required precision. Is polynomial factorization over $\Z$ the theoretical bottleneck? %GUESS: Maybe $O(N^6)$ to %compute setup data, and computing $q$-expansions to precision $n$ is %maybe $O(N\cdot n^3)$? (TODO) \end{slide} \begin{slide} \h{Implemented in MAGMA \photo{2}{magma}} \begin{verbatim} > S := CuspForms(37,2); > Basis(S); q + q^3 - 2*q^4 - q^7 + O(q^8), q^2 + 2*q^3 - 2*q^4 + q^5 - 3*q^6 + O(q^8) \end{verbatim} See also {\tt http://modular.fas.harvard.edu/mfd} \begin{center} \photo{3}{mfd} \end{center} \end{slide} \begin{slide} \rd{Basis for $S_{14}(11)$:} \begin{verbatim} > S := CuspForms(11,14); SetPrecision(S,17); > Basis(S); q - 74*q^13 - 38*q^14 + 441*q^15 + 140*q^16 + O(q^17), q^2 - 2*q^13 + 78*q^14 + 24*q^15 - 338*q^16 + O(q^17), q^3 + 18*q^13 - 72*q^14 + 89*q^15 + 492*q^16 + O(q^17), q^4 + 12*q^13 + 31*q^14 - 18*q^15 - 193*q^16 + O(q^17), q^5 - 10*q^13 + 46*q^14 - 63*q^15 - 52*q^16 + O(q^17), q^6 + 11*q^13 - 18*q^14 - 74*q^15 - 4*q^16 + O(q^17), q^7 - 7*q^13 - 16*q^14 + 42*q^15 - 84*q^16 + O(q^17), q^8 - q^13 - 16*q^14 - 18*q^15 - 34*q^16 + O(q^17), q^9 - 8*q^13 - 2*q^14 - 3*q^15 + 16*q^16 + O(q^17), q^10 - 5*q^13 - 2*q^14 - 6*q^15 + 14*q^16 + O(q^17), q^11 + 12*q^13 + 12*q^14 + 12*q^15 + 12*q^16 + O(q^17), q^12 - 2*q^13 - q^14 + 2*q^15 + q^16 + O(q^17) \end{verbatim} \end{slide} \begin{slide} \h{Hecke Algebras \photo{1}{hecke}} \rd{Hecke Operators:} Let $p$ be a prime. $$T_p\left(\sum_{n\geq 1} a_n\cdot q^n\right) = \sum_{n\geq 1} a_{np}\cdot q^n + p^{k-1} \sum_{n\geq 1} a_n \cdot q^{np}$$ (If $p\mid N$, drop the second summand.) This preserves $S_k(N)$, so defines a linear map $$ T_p : S_k(N) \to S_k(N). $$ Similar definition of $T_n$ for any integer~$n$. \rd{Hecke Algebra:} A {\em commutative ring}: $$\T_k(N) = \Z[T_1, T_2,T_3,T_4,T_5,\ldots] \subset \End_\C(S_k(N))$$ \end{slide} \begin{slide} \h{Computing Hecke Algebras} \rd{Fact:} $\T_k(N) = \Z[T_1, T_2,T_3,T_4,T_5,\ldots]$ is free as a $\Z$-\rd{module} of rank equal to $\dim S_k(N)$. \rd{Sturm Bound:} $\T_k(N)$ is generated as a $\Z$-module by $T_1, T_2, \ldots, T_b$, where $$b = \Bigl\lceil \frac{k}{12}\cdot N \cdot \prod_{p\mid N} \left(1+\frac{1}{p}\right)\Bigr\rceil.$$ \rd{Example:} For $N=37$ and $k=2$, the bound is $7$. In fact, $\T_2(37)$ has $\Z$-basis $T_1=\mtwo{1}{0}{0}{1}$ and $T_2=\mtwo{-2}{0}{1}{0}$. There are several other $\T_k(N)$-modules isomorphic to $S_2(N)$, and I use these instead to compute $\T_k(N)$ as a ring. \end{slide} \begin{slide} \h{Discriminants} The discriminant of $\T_k(N)$ is an integer. It measures ramification, or what's the same, congruences between simultaneous eigenvectors for $\T_k(N)$, hence is related to the modular degree. \rd{Discriminant:} $$ \Disc(\T_k(N)) = \Det(\Tr(t_i \cdot t_j)), $$ where $t_1,\ldots, t_n$ are a basis for $\T_k(N)$ as a free $\Z$-module. \rd{Examples:}\\ $\quad\Disc(\T_2(37)) = \Det\mtwo{ 2}{-2}{-2}{ 4}=4$\\ $\quad\Disc(\T_{14}(11)) = $ {\small $2^{46}\cdot 3^{14}\cdot 5^2\cdot 11^{42} \cdot 79 \cdot 241\cdot 1163 \cdot 40163 \cdot 901181111 \cdot \quad{}\qquad{} \qquad{} \qquad{} \qquad{} \qquad{} 47552569849 \cdot 124180041087631 \cdot 205629726345973.$ } \end{slide} \begin{slide} \h{Ribet's Question \hspace{2ex}\photo{1.6}{ribet}} I became interested in computing with modular forms when I was a grad student and Ken Ribet started asking: \vspace{-1ex} \rd{Question:} (Ribet, 1997) Is there a prime $p$ so that $p\mid \Disc(\T_2(p))$? \vspace{-2ex} Ribet proved a theorem about $X_0(p) \cap J_0(p)_{\tor}$ under the hypothesis that $p\nmid \Disc(\T_2(p))$, and wanted to know how restrictive his hypothesis was. Note: When $k>2$, usually $p\mid \Disc(\T_k(p))$. \end{slide} \begin{slide} \h{Computations\hspace{2ex}\photo{1.5}{mestre}} Using a \photo{0.9}{pari} script of Joe Wetherell, I set up a computation on my laptop and found exactly one example in which $p\mid \Disc(\T_2(p))$. It was $p=389$, now my favorite number. Last year I checked that for $p<50000$ there are no other examples in which $p\mid \Disc(\T_2(p))$. For this I used the Mestre method of graphs, which involves computing with the free abelian group on the supersingular $j$-invariants in $\F_{p^2}$ of elliptic curves. \end{slide} \begin{slide} \h{Index in the Normalization} Let $\nT_k(p)$ be the \rd{normalization} of $\T_k(p)$. Since $\T_k(p)$ is an order in a product of number fields, $\nT_k(p)$ is the product of the rings of integers of those number fields. It turned out that Ribet could prove his theorem under the weaker hypothesis that $p\nmid [\nT_2(p):\T_2(p)]$. I was unable to find a counterexample to this divisibility. (Note: Matt Baker's Ph.D. was a complete proof of the result Ribet was trying to prove, but used different methods.) [[Picture on blackboard of ${\rm Spec}(\T_k(p))$]] \end{slide} \begin{slide} \h{\photo{0.5}{questions} \hspace{1in} Conjecture 2 \hspace{1in}\photo{0.5}{questions}} {\Large{\bf Conjecture 2. (--).} } {\large If $\T_2(p)$ is the Hecke algebra associated to $S_2(\Gamma_0(p))$, then $p$ does not divide the index of $\T_2(p)$ in its normalization.} The primes that divide $[\nT_2(p):\T_2(p)]$ are called \rd{congruence primes}. They are the primes of congruence between non-$\Gal(\Qbar/\Q)$-conjugate eigenvectors for $\T_2(p)$. Using this observation and another theorem of Ribet (and Wiles's theorem), we see that Conjecture 2 implies that $p$ does not divide the modular degree of any elliptic curve of conductor~$p$. This is why Conjecture 2 implies Conjecture 1. But is there any reason to believe Conjecture 2, beyond knowing that it is true for $p<50000$? \end{slide} \comment{\begin{slide} \h{Example of Weight $k=14$} Totally stuck on Conjecture 2, so look at weight $k>2$. We have $\quad\Disc(\T_{14}(11)) = $ {\small $2^{46}\cdot 3^{14}\cdot 5^2\cdot \mathbf{11}^{42} \cdot 79 \cdot 241\cdot 1163 \cdot 40163 \cdot 901181111 \cdot \quad{}\qquad{} \qquad{} \qquad{} \qquad{} \qquad{} 47552569849 \cdot 124180041087631 \cdot 205629726345973.$} Notice the large power of $11$. Upon computing the $p$-maximal order in $\T_{14}(11)\otimes_\Z\Q$, we find that $11\nmid \Disc(\nT_{14}(11))$, so all the $11$ is in the index of $\T_{14}(11)$ in $\nT_{14}(11)$. Thus $$ \ord_{11}([\nT_{14}(11) : \T_{14}(11)]) = 21. $$ \end{slide}} \begin{slide} \h{Data for $k=4$} For inspiration, consider weight $>2$. Each row contains pairs $p$ and $\ord_p(\Disc(\T_{4}(p)))$. {\tiny \hspace{-3em}\begin{minipage}[b]{\textwidth} \begin{tabular}{|ccccccccccccccccc|}\hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\ 0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131& 137& 139\\ & & & & & & & & & & & & & & & & \vspace{-2ex}\\ 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22& {\bf 24}\\\hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199& 211& 223& 227& 229& 233\\ & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 239& 241& 251& 257& 263& 269& 271& 277& 281& 283& 293& 307& 311& 313& 317& 331& 337\\ & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\ & & & & & & & & & & & & & & & & \vspace{-2ex}\\ 56& 58& 58& 58& 60&62& 62& 62& {\bf 65} &66& 66& 68& 68& 70& 70& 72& 72\\\hline & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\ & & & & & & & & & & & & & & & & \vspace{-1ex}\\ 72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline \end{tabular} \end{minipage}} \end{slide} \begin{slide} \h{A Pattern? \hspace{1in}\photo{2.5}{frank2}} \rd{F.~ Calegari} (during a talk I gave): There is \rd{almost} a pattern!!! Frank, Romyar Sharifi and I computed $2\cdot [\nT_4(p):\T_4(p)]$ and obtained the numbers as in the table, except for $p=389$ (which gives $64$) and $139$ (which gives $22$). We also considered many other examples... and found a pattern! \end{slide} \begin{slide} \h{Conjecture 3} In all cases, we found the following \rd{amazing} pattern: \rd{Conjecture 3.} Suppose $p\geq k-1$. Then $$ \ord_p([\nT_k(p) : \T_k(p)]) = \left\lfloor\frac{p}{12}\right\rfloor\cdot \binom{k/2}{2} + a(p,k), $$ where $$ a(p,k) = \begin{cases} 0 & \text{if $p\con 1\pmod{12}$,}\\ \vspace{-2ex}\\ 3\cdot\ds\binom{\lceil \frac{k}{6}\rceil}{2} & \text{if $p\con 5\pmod{12}$,}\\ \vspace{-1ex}\\ 2\cdot\ds\binom{\lceil \frac{k}{4}\rceil}{2} & \text{if $p\con 7\pmod{12}$,}\\ \vspace{-1ex}\\ a(5,k)+a(7,k) & \text{if $p\con 11\pmod{12}$.} \end{cases} $$ \end{slide} \begin{slide} \h{Warning} The conjecture is false without the constraint that $p\geq k-1$. %It works for our running example $p=11$, %$k=14$, where the formula yields $0 + 3\cdot \binom{3}{2} + %2\cdot\binom{4}{2} = 9 + 12 = 21$, which is correct. For example, if $p=5$ and $k=12$, then the conjecture predicts that the index is $0 + 3\cdot 1 = 3$, but in fact $\ord_p([\nT_k(p) : \T_k(p)]) = 5$. In our data when $k> p+1$, then the conjectural $\ord_p$ is often less than the actual $\ord_p$. \end{slide} \begin{slide} \h{Summary} For many years I had no idea whether there should or shouldn't be mod~$p$ congruence between nonconjugate eigenforms. (I.e., whether~$p$ divides modular degrees at prime level.) By considering weight $k\geq 4$, and computing examples, a simple conjectural formula emerged. When specialized to weight $2$ this formula is the conjecture that there are no mod $p$ congruences. \rd{Future Direction.} Explain why there are so many mod~$p$ congruences at level $p$, when $k\geq 4$. See paper for a strategy. % \rd{Computational Question.} Push computation of % $\ord_p(\Disc(\T_2(p)))$ to $100000$ using sparse % charpoly algorithm (e.g., Wiedemann). \rd{Connection with Vandiver's Conjecture?} Investigate the connection between Conjecture~1 and Flach's results on modular degrees annihilating Selmer groups. \end{slide} \begin{slide} \htwo{This Concludes ANTS VI: THANKS!} \begin{center} \includegraphics[width=6in]{graphics/ants6.eps} \end{center} Many thanks to the organizers (Sands, Kelly, Buell): \begin{center} \photo{1}{sands}, \photo{1}{kelly}, and Duncan Buell \end{center} \end{slide} \end{document}