Sharedwww / talks / ants6 / disc.texOpen in CoCalc
Author: William A. Stein
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64%opening
65\title{\Large\bf\dblue Modular Degrees of Elliptic Curves\\and\\ Discriminants
66of Hecke Algebras}
67\author{\rd{\Large William Stein\footnote{This is joint work
68with F.~Calegari.\hspace{2ex}\photo{1}{frank}}}\\
69{\tt http://modular.fas.harvard.edu}}
70\date{ANTS VI, June 18, 2004\vspace{1ex}
71}
72
73\begin{document}
74
75\maketitle
76
77
78\begin{slide}
79\h{\rd{Goal}\hspace{2in}}
80
81Let $p$ be a prime.
82My goal is to explain and justify
83the following \underline{Calegari-Stein conjectures}
84(note: 3 implies 2 implies 1):
85\vspace{-1.5ex}
86
87{\Large{\dblue\bf Conjecture 1:} }{\large If $E/\Q$ is an elliptic
88  curve of conductor $p$, then the modular degree $m_E$ of $E$ is not
89  divisible by $p$.}
90\vspace{-1.5ex}
91
92{\Large{\dblue\bf Conjecture 2:} } {\large If
93$\T_2(p)$ is the Hecke algebra associated to
94$S_2(p)$, then $p$ does not divide the
95index of $\T_2(p)$ in its normalization.}
96\vspace{-1.5ex}
97
98{\Large{\dblue\bf Conjecture 3:} } {\large
99If $p\geq k-1$, then there is an explicit
100formula for the $p$-part of the index of $\T_k(p)$
101in its normalization.}
102\vspace{-1.5ex}
103\end{slide}
104
105
106\begin{slide}
107%\h{Conjecture 1}
108\htwo{Conj 1: If $E$
109of conductor $p_E$,
110then $p_E\nmid m_E$.}
111
112
113\rd{\bf A Motivation:} \photo{1.3}{flach} Conjecture 1 looks like
114Vandiver's conjecture, which asserts that $p\nmid h_p^-$.  Flach
115proved the modular degree annihilates $\Sha({\rm Sym}^2(E))$,
116which is an analogue of a class group.
117\end{slide}
118
119\begin{slide}
120\htwo{Conj 1: If $E$
121of conductor $p_E$,
122then $p_E\nmid m_E$.}
123
124\rd{\bf Watkins Data:} \photo{1.2}{watkins}  For $p_E<10^7$ there
125are 52878 curves of
126prime conductor whose modular degree Watkins computed.
127No counterexamples to Conjecture 1 in the data.
128There are $23$ curves
129such that $m_E$ is divisible by a prime $\ell> p_E$.  For example the
130curve $y^2 + xy = x^3 - x^2 -391648x -94241311$ of prime conductor
131$p_E=4\,847\,093$ has modular degree $2\cdot 21\,695\,761$.
132Smallest $p_E$ with some $\ell>p_E$ is $p_E=1\,194\,923$.
133\end{slide}
134
135\begin{slide}
136\h{More Data}
137\begin{itemize}
138\item The \rd{maximum} known ratio $\ds\frac{m_E}{p_E}$ is $\sim 23.2$, attained for
139$p_E=7\,944\,197$.
140\item
141\rd{First} curve with $\ds \frac{m_E}{p_E}>1$ has $p_E=13723$ and
142$m_E=16176=2^4\cdot 3\cdot 337$.
143\item \rd{Smallest} known $\ds\frac{m_E}{p_E}>1$ is
144$1.0004067\ldots$ for $p_E=1\,757\,963$ where $m_E=p_E+715$.
145\end{itemize}
146%Conjecture is consistent with $m_E\gg p_E^{7/6-\varepsilon}$.
147\end{slide}
148
149\begin{slide}
150\htwo{Modular Forms\hspace{1in}\photo{1.3}{hecke}}
151\vspace{-2ex}
152
153\rd{Congruence Subgroup:}
154$$155\Gamma_0(N) = \left\lbrace 156\mtwo{a}{b}{c}{d} \in \SL_2(\Z) \text{ such that } N \mid c 157\right\rbrace. 158$$
159\vspace{-2ex}
160
161\rd{Cusp Forms:}
162$\displaystyle 163 S_k(N) = \Bigl\{f : \mathfrak{h}\to\C \text{ such that } 164$
165\vspace{-2ex}
166$$167 \qquad\qquad\qquad f(\gamma(z)) = (cz+d)^{-k}f(z) \text{ all } 168 \gamma\in\Gamma_0(N), 169\vspace{-1ex}$$
170$$\qquad \qquad \qquad \text{ and f is holomorphic at the cusps}\Bigr\} 171$$
172\vspace{-1ex}
173\rd{Fourier Expansion:}
174$$175 f = \sum_{n\geq 1} a_n e^{2\pi i z n} = \sum_{n\geq 1} a_n q^n \in \C[[q]]. 176$$
177\end{slide}
178
179\begin{slide}
180\htwo{\photo{1}{merel} Computing Modular Forms \photo{1}{cremona}}
181$S_k(N)=0$ if $k$ is odd, so we will not consider odd $k$ further.
182
183For $k\geq 2$, a basis of $S_k(N)$ can be computed to any given
184precision using \rd{modular symbols}.  Appears
185that no formal analysis of complexity has been done.
186Certainly polynomial time in $N$ and required precision.
187Is polynomial factorization over $\Z$ the theoretical
188bottleneck?
189
190
191%GUESS: Maybe $O(N^6)$ to
192%compute setup data, and computing $q$-expansions to precision $n$ is
193%maybe $O(N\cdot n^3)$?  (TODO)
194
195\end{slide}
196\begin{slide}
197\h{Implemented in MAGMA \photo{2}{magma}}
198\begin{verbatim}
199> S := CuspForms(37,2);
200> Basis(S);
201    q + q^3 - 2*q^4 - q^7 + O(q^8),
202    q^2 + 2*q^3 - 2*q^4 + q^5 - 3*q^6 + O(q^8)
203\end{verbatim}
205\begin{center}
206\photo{3}{mfd}
207\end{center}
208\end{slide}
209
210\begin{slide}
211\rd{Basis for $S_{14}(11)$:}
212\begin{verbatim}
213> S := CuspForms(11,14); SetPrecision(S,17);
214> Basis(S);
215    q   - 74*q^13 - 38*q^14 + 441*q^15 + 140*q^16 + O(q^17),
216    q^2 - 2*q^13 + 78*q^14 + 24*q^15 - 338*q^16 + O(q^17),
217    q^3 + 18*q^13 - 72*q^14 + 89*q^15 + 492*q^16 + O(q^17),
218    q^4 + 12*q^13 + 31*q^14 - 18*q^15 - 193*q^16 + O(q^17),
219    q^5 - 10*q^13 + 46*q^14 - 63*q^15 - 52*q^16 + O(q^17),
220    q^6 + 11*q^13 - 18*q^14 - 74*q^15 - 4*q^16 + O(q^17),
221    q^7 - 7*q^13 - 16*q^14 + 42*q^15 - 84*q^16 + O(q^17),
222    q^8 - q^13 - 16*q^14 - 18*q^15 - 34*q^16 + O(q^17),
223    q^9 - 8*q^13 - 2*q^14 - 3*q^15 + 16*q^16 + O(q^17),
224    q^10 - 5*q^13 - 2*q^14 - 6*q^15 + 14*q^16 + O(q^17),
225    q^11 + 12*q^13 + 12*q^14 + 12*q^15 + 12*q^16 + O(q^17),
226    q^12 - 2*q^13 - q^14 + 2*q^15 + q^16 + O(q^17)
227\end{verbatim}
228\end{slide}
229
230\begin{slide}
231\h{Hecke Algebras \photo{1}{hecke}}
232
233\rd{Hecke Operators:}
234Let $p$ be a prime.
235$$T_p\left(\sum_{n\geq 1} a_n\cdot q^n\right) 236 = \sum_{n\geq 1} a_{np}\cdot q^n + p^{k-1} \sum_{n\geq 1} a_n \cdot q^{np}$$
237(If $p\mid N$, drop the second summand.)
238This preserves $S_k(N)$, so defines a linear map
239$$240T_p : S_k(N) \to S_k(N). 241$$
242  Similar definition of $T_n$ for any
243integer~$n$.
244
245\rd{Hecke Algebra:}  A {\em commutative ring}:
246$$\T_k(N) = \Z[T_1, T_2,T_3,T_4,T_5,\ldots] \subset \End_\C(S_k(N))$$
247
248
249\end{slide}
250
251\begin{slide}
252\h{Computing Hecke Algebras}
253
254\rd{Fact:} $\T_k(N) = \Z[T_1, T_2,T_3,T_4,T_5,\ldots]$ is free as a
255$\Z$-\rd{module} of rank equal to $\dim S_k(N)$.
256
257\rd{Sturm Bound:}
258$\T_k(N)$ is generated as a $\Z$-module by
259$T_1, T_2, \ldots, T_b$, where
260$$b = \Bigl\lceil \frac{k}{12}\cdot N \cdot \prod_{p\mid N} \left(1+\frac{1}{p}\right)\Bigr\rceil.$$
261
262\rd{Example:} For $N=37$ and $k=2$, the bound is $7$.  In fact, $\T_2(37)$
263has $\Z$-basis $T_1=\mtwo{1}{0}{0}{1}$ and $T_2=\mtwo{-2}{0}{1}{0}$.
264
265There are several other $\T_k(N)$-modules isomorphic to $S_2(N)$, and
266I use these instead to compute $\T_k(N)$ as a ring.
267\end{slide}
268
269\begin{slide}
270  \h{Discriminants}
271
272  The discriminant of $\T_k(N)$ is an integer.  It measures
273  ramification, or what's the same, congruences between simultaneous
274  eigenvectors for $\T_k(N)$, hence is related to the modular degree.
275
276\rd{Discriminant:}
277$$278 \Disc(\T_k(N)) = \Det(\Tr(t_i \cdot t_j)), 279$$
280where $t_1,\ldots, t_n$ are a basis for $\T_k(N)$
281as a free $\Z$-module.
282
283\rd{Examples:}\\
284$\quad\Disc(\T_2(37)) = \Det\mtwo{ 2}{-2}{-2}{ 4}=4$\\
285$\quad\Disc(\T_{14}(11)) =$ {\small $2^{46}\cdot 3^{14}\cdot 5^2\cdot 11^{42} 286\cdot 79 \cdot 241\cdot 1163 \cdot 40163 \cdot 901181111 \cdot 287\quad{}\qquad{} \qquad{} \qquad{} \qquad{} \qquad{} 47552569849 \cdot 124180041087631 \cdot 205629726345973.$
288}
289\end{slide}
290
291\begin{slide}
292\h{Ribet's Question \hspace{2ex}\photo{1.6}{ribet}}
293
294I became interested in computing with modular forms when I was
296
297\vspace{-1ex}
298\rd{Question:} (Ribet, 1997) Is there a prime $p$ so that
299$p\mid \Disc(\T_2(p))$?
300\vspace{-2ex}
301
302Ribet proved a theorem about $X_0(p) \cap J_0(p)_{\tor}$ under the
303hypothesis that $p\nmid \Disc(\T_2(p))$, and wanted to know how restrictive his
304hypothesis was.   Note: When $k>2$, usually $p\mid \Disc(\T_k(p))$.
305\end{slide}
306
307\begin{slide}
308\h{Computations\hspace{2ex}\photo{1.5}{mestre}}
309Using a \photo{0.9}{pari} script of Joe Wetherell, I set up a computation on my
310laptop and found exactly one example in which
311$p\mid \Disc(\T_2(p))$.  It was $p=389$, now my favorite number.
312
313Last year I checked that for $p<50000$ there are no other
314examples in which $p\mid \Disc(\T_2(p))$.  For this
315I used the Mestre method of graphs, which involves
316computing with the free abelian group on the supersingular
317$j$-invariants in $\F_{p^2}$ of elliptic curves.
318\end{slide}
319
320\begin{slide}
321  \h{Index in the Normalization}
322
323Let $\nT_k(p)$ be the \rd{normalization} of $\T_k(p)$.  Since
324$\T_k(p)$ is an order in a product of number fields, $\nT_k(p)$ is the
325product of the rings of integers of those number fields.
326
327It turned out that Ribet could prove his theorem under
328the weaker hypothesis that $p\nmid [\nT_2(p):\T_2(p)]$.
329I was unable to find a counterexample to this divisibility.
330(Note: Matt Baker's Ph.D. was a complete proof of the result Ribet
331was trying to prove, but used different methods.)
332
333[[Picture on blackboard of ${\rm Spec}(\T_k(p))$]]
334\end{slide}
335
336\begin{slide}
337\h{\photo{0.5}{questions} \hspace{1in} Conjecture 2   \hspace{1in}\photo{0.5}{questions}}
338
339{\Large{\bf Conjecture 2. (--).} } {\large If
340$\T_2(p)$ is the Hecke algebra associated to
341$S_2(\Gamma_0(p))$, then $p$ does not divide the
342index of $\T_2(p)$ in its normalization.}
343
344The primes that divide $[\nT_2(p):\T_2(p)]$ are called \rd{congruence
345  primes}.  They are the primes of congruence between
346non-$\Gal(\Qbar/\Q)$-conjugate eigenvectors for $\T_2(p)$.  Using this
347observation and another theorem of Ribet (and Wiles's theorem), we
348see that Conjecture 2 implies that $p$ does not divide the modular
349degree of any elliptic curve of conductor~$p$.  This is why Conjecture
3502 implies Conjecture 1.
351
352But is there any reason to believe Conjecture 2, beyond knowing that
353it is true for $p<50000$?
354\end{slide}
355
356\comment{\begin{slide}
357\h{Example of Weight $k=14$}
358Totally stuck on Conjecture 2, so look at weight $k>2$.
359We have
360
361$\quad\Disc(\T_{14}(11)) =$ {\small $2^{46}\cdot 3^{14}\cdot 5^2\cdot \mathbf{11}^{42} 362\cdot 79 \cdot 241\cdot 1163 \cdot 40163 \cdot 901181111 \cdot 363\quad{}\qquad{} \qquad{} \qquad{} \qquad{} \qquad{} 47552569849 \cdot 124180041087631 \cdot 205629726345973.$}
364
365Notice the large power of $11$.  Upon computing the $p$-maximal order
366in $\T_{14}(11)\otimes_\Z\Q$, we find that $11\nmid \Disc(\nT_{14}(11))$,
367so all the $11$ is in the index of $\T_{14}(11)$ in $\nT_{14}(11)$. Thus
368$$369\ord_{11}([\nT_{14}(11) : \T_{14}(11)]) = 21. 370$$
371\end{slide}}
372
373\begin{slide}
374\h{Data for $k=4$}
375For inspiration, consider weight $>2$.
376
377Each row contains pairs $p$ and $\ord_p(\Disc(\T_{4}(p)))$.
378
379{\tiny \hspace{-3em}\begin{minipage}[b]{\textwidth}
380\begin{tabular}{|ccccccccccccccccc|}\hline
381 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
3822& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
3830& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
384 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
38561& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131&  137& 139\\
386 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-2ex}\\
38710& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&
388{\bf 24}\\\hline
389 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
390 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199&
391  211& 223& 227& 229& 233\\
392 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
393 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
394 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
395 239& 241& 251& 257& 263& 269& 271& 277&
396  281& 283& 293& 307& 311& 313& 317& 331& 337\\
397 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
398 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
399 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
400 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
401 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-2ex}\\
402 56& 58& 58& 58& 60&62& 62& 62& {\bf 65}  &66& 66& 68& 68& 70& 70& 72& 72\\\hline
403 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
404  443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
405 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \vspace{-1ex}\\
406  72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline
407\end{tabular}
408\end{minipage}}
409\end{slide}
410
411\begin{slide}
412\h{A Pattern? \hspace{1in}\photo{2.5}{frank2}}
413\rd{F.~ Calegari} (during a talk I gave):
414There is \rd{almost} a pattern!!!
415Frank, Romyar Sharifi and I computed $2\cdot 416[\nT_4(p):\T_4(p)]$ and obtained the numbers as in the table, except for
417$p=389$ (which gives $64$) and $139$ (which gives $22$).
418We also considered many other examples... and found a pattern!
419\end{slide}
420
421
422\begin{slide}
423\h{Conjecture 3}
424
425In all cases, we found the following \rd{amazing} pattern:
426
427\rd{Conjecture 3.} Suppose $p\geq k-1$.  Then
428  $$429 \ord_p([\nT_k(p) : \T_k(p)]) = \left\lfloor\frac{p}{12}\right\rfloor\cdot 430 \binom{k/2}{2} + a(p,k), 431$$
432  where
433  $$434 a(p,k) = 435\begin{cases} 436 0 & \text{if p\con 1\pmod{12},}\\ 437\vspace{-2ex}\\ 438 3\cdot\ds\binom{\lceil \frac{k}{6}\rceil}{2} & \text{if p\con 5\pmod{12},}\\ 439\vspace{-1ex}\\ 440 2\cdot\ds\binom{\lceil \frac{k}{4}\rceil}{2} & \text{if p\con 7\pmod{12},}\\ 441\vspace{-1ex}\\ 442 a(5,k)+a(7,k) & \text{if p\con 11\pmod{12}.} 443\end{cases} 444$$
445\end{slide}
446\begin{slide}
447\h{Warning}
448
449The conjecture is false without the constraint that $p\geq k-1$.
450
451%It works for our running example $p=11$,
452%$k=14$, where the formula yields $0 + 3\cdot \binom{3}{2} + 453%2\cdot\binom{4}{2} = 9 + 12 = 21$, which is correct.
454
455For example, if $p=5$ and $k=12$, then the conjecture predicts
456that the index is $0 + 3\cdot 1 = 3$,
457but in fact
458$\ord_p([\nT_k(p) : \T_k(p)]) = 5$.
459
460In our data when $k> p+1$, then the conjectural $\ord_p$ is
461often less than the actual $\ord_p$.
462\end{slide}
463
464\begin{slide}
465  \h{Summary}
466
467  For many years I had no idea whether there
468  should or shouldn't be mod~$p$ congruence between nonconjugate
469  eigenforms.  (I.e., whether~$p$ divides modular degrees at
470  prime level.)  By considering weight $k\geq 4$, and computing examples,
471  a simple conjectural formula emerged.  When specialized to weight $2$
472  this formula is the conjecture that there are no mod $p$ congruences.
473
474  \rd{Future Direction.} Explain why there are so many mod~$p$
475  congruences at level $p$, when $k\geq 4$.  See paper for a strategy.
476
477%  \rd{Computational Question.} Push computation of
478%  $\ord_p(\Disc(\T_2(p)))$ to $100000$ using sparse
479%  charpoly algorithm (e.g., Wiedemann).
480
481  \rd{Connection with Vandiver's Conjecture?} Investigate the
482  connection between Conjecture~1 and Flach's results on modular
483  degrees annihilating Selmer groups.
484
485\end{slide}
486
487\begin{slide}
488\htwo{This Concludes ANTS VI: THANKS!}
489\begin{center}
490\includegraphics[width=6in]{graphics/ants6.eps}
491\end{center}
492Many thanks to the organizers (Sands, Kelly, Buell):
493\begin{center}
494       \photo{1}{sands}, \photo{1}{kelly}, and Duncan Buell
495     \end{center}
496\end{slide}
497
498\end{document}
499
500