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