CoCalc Public Fileswww / tables / Notes / discriminants / disctalk.texOpen with one click!
Author: William A. Stein
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\documentclass[11pt]{article}
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\title{Discriminants of Hecke Algebras at Prime Level}
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\author{William A. Stein\footnote{This will eventually be a joint paper with Frank Calegari.}}
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\date{September 24, 2002}
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\DeclareMathOperator{\charpoly}{charpoly}
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\DeclareMathOperator{\splittemp}{split}
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\renewcommand{\split}{\splittemp}
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\DeclareMathOperator{\deriv}{deriv}
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\begin{document}
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\maketitle
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\begin{abstract}
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We study $p$-divisibility of discriminant of Hecke algebras associated
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to spaces of cusp forms of prime level. By considering cusp forms of
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weight bigger than~$2$, we are are led to make a conjecture about
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indexes of Hecke algebras in their normalization which, if true,
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implies that there are no mod~$p$ congruences between
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non-conjugate newforms in $S_2(\Gamma_0(p))$.
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\end{abstract}
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\section{Introduction}
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I started working in modular forms when Ken Ribet asked about discriminants of
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Hecke algebras at prime level. I've recently revisited this question and, with
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the help of Frank Calegari, have made some interesting discoveries.
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\section{Discriminants of Hecke Algebras}
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Let~$R$ be a ring and let~$A$ be an~$R$ algebra that
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is free as an~$R$ module.
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The trace of an element of~$A$ is the trace, in the
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sense of linear algebra, of left
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multiplication by that element on~$A$.
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\begin{definition}[Discriminant]
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Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis
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for~$A$. Then the {\em discriminant} of~$A$, denoted $\disc(A)$,
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is the determinant of the $n\times n$ matrix
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$(\tr(\omega_i\omega_j))$, which is well defined
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modulo squares of units in~$A$.
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\end{definition}
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When $R=\Z$ the discriminant is well defined, since
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the only units are $\pm 1$.
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\begin{proposition}\label{prop:separable}
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Suppose~$R$ is a field. Then~$A$ has discriminant~$0$ if and only
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if~$A$ is separable over~$R$, i.e., for every extension
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$R'$ of $R$, the ring $A\tensor R'$ contains no nilpotents.
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\end{proposition}
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The following proof is summarized from Section~26 of Matsumura.
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If~$A$ contains a nilpotent then that nilpotent is in the kernel of
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the trace pairing. If~$A$ is separable then we may assume that~$R$ is
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algebraically closed. Then~$A$ is an Artinian reduced ring, hence
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isomorphic as a ring to a finite product of copies of~$R$, since~$R$
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is algebraically closed. Thus the trace form on~$A$ is
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nondegenerate.
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\subsection{The Discriminant Valuation}
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Let $\Gamma$ be a congruence subgroup of $\SL_2(\Z)$, e.g.,
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$\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$.
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For any integer $k\geq 1$, let $S_k(\Gamma)$ denote
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the space of holomorphic weight-$k$ cusp forms for $\Gamma$. Let
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$$
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\T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma))
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$$
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be the associated Hecke algebra.
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Then~$\T$ is a commutative ring that is free and of finite rank
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as a $\Z$-module. Also of interest is the image $\T^{\new}$
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of~$\T$ in $\End(S_k(\Gamma)^{\new})$.
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\begin{example}
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Let $\Gamma=\Gamma_0(243)$, which is illustrated on
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my T-shirt. Since $243=3^5$, experts will immediately
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deduce that $\disc(\T) = 0$. A computation shows that
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$$
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\disc(\T^{\new}) = 2^{13} \cdot 3^{40},
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$$
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which reflects the mod-$2$ and mod-$3$ intersections all
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over my shirt.
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\end{example}
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\begin{definition}[Discriminant Valuation]
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Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
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$\Gamma_1(p)$.
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The {\em discriminant valuation} is
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$$
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d_k(\Gamma) = \ord_p(\text{the discriminant of $\T$}).
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$$
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%When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to
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%be~$+\infty$.
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\end{definition}
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\section{Motivation and Applications}
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Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
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$\Gamma_1(p)$.
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The quantity $d_k(\Gamma)$ is of interest because it measures mod~$p$
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congruences between eigenforms in $S_k(\Gamma)$.
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\begin{proposition}
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Suppose that $d_k(\Gamma)$ is finite.
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Then the discriminant valuation $d_k(\Gamma)$ is nonzero
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if and only if there is a mod-$p$ congruence
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between two Hecke eigenforms in $S_k(\Gamma)$
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(note that the two congruent eigenforms might
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be Galois conjugate).
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\end{proposition}
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\begin{proof}
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It follows from Proposition~\ref{prop:separable} that
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$d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not
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separable. The Artinian ring $\T\tensor\Fpbar$ is
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not separable if and only if the number of ring
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homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is
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less than
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$$
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\dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma).
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$$
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Since $d_k(\Gamma)$ is finite, the number of ring
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homomorphisms $\T\tensor\Qpbar \ra \Qpbar$ equals
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$\dim_\C S_k(\Gamma)$. Using the standard bijection between
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congruences and normalized eigenforms, we see that
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$\T\tensor\Fpbar$ is not separable if and only
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if there is a mod-$p$ congruence between two eigenforms.
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\end{proof}
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\begin{example}
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If $\Gamma=\Gamma_0(389)$ and $k=2$,
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then $\dim_\C S_2(\Gamma) = 32$.
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Let~$f$ be the characteristic polynomial of $T_2$.
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One can check that~$f$ is square free and $389$ exactly
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divides the discriminant of~$f$, so $T_2$ generated
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$\T\tensor \Z_{389}$ as a ring. (If it generated a subring of $\T\tensor\Z_{389}$
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of finite index, then the discriminant of~$f$ would be divisible
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by $389^2$.)
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Modulo~$389$ the polynomial~$f$ is congruent to
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$$\begin{array}{l}
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(x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\
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(x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\
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113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\
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213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101)
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\end{array}
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$$
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The factor $(x+175)^2$ indicates that
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$\T\tensor \Fbar_{389}$ is not separable
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since the image of $T_2+175$ is nilpotent
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(its square is~$0$). There are $32$ eigenforms over~$\Q_2$
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but only $31$ mod-$389$ eigenforms, so there must be a congruence.
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Let~$F$ be the $389$-adic newform whose $a_2$ term is a root of
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$$
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x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 + 43\cdot 389 +
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19\cdot 389^2 + \cdots).
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$$
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Then the congruence is between~$F$
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and its $\Gal(\Qbar_{389}/\Q_{389})$-conjugate.
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\end{example}
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\begin{example}
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The discriminant of the Hecke algebra $\T$ associated
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to $S_2(\Gamma_0(389))$ is
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$$
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2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\! 37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\! 477439237737571441
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$$
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I computed this using the following algorithm, which was suggested
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by Hendrik Lenstra. Using the Sturm bound I found a~$b$
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such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module. I then
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found a subset~$B$ of the $T_i$ that form a $\Q$-basis for
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$\T\tensor_\Z\Q$.
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Next, viewing $\T$ as a ring of matrices acting on $\Q^{32}$,
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I found a random vector $v\in\Q^{32}$ such that the set of
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vectors $C=\{T(v) : T \in B\}$ is linearly independent. Then
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I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations
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of the elements of~$C$. Next I found a $\Z$-basis~$D$ for the
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$\Z$-span of these $\Q$-linear combinations of elements of~$C$.
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Tracing everything back, I find the trace pairing on
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the elements of~$D$, and deduce the discriminant by computing
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the determinant of the trace pairing matrix. The most difficult
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step is computing~$D$ from $T_1(v),\ldots,T_b(v)$ expressed
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in terms of~$C$, and this explains why we embed $\T$ in $\Q^{32}$
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instead of viewing the elements of $\T$ as vectors in $\Q^{32^2}$.
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This whole computation takes one
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second on an Athlon 2000 processor.
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\end{example}
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\subsection{Literature}
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I've seen a version of Theorem~\ref{thm:disc} referred to in the
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following papers:
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\begin{enumerate}
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\item Ribet: {\em Torsion points on $J_0(N)$ and Galois representations}
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\item Lo\"\i{}c Merel and William Stein: {\em The field generated by the points of small prime
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order on an elliptic curve}
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\item Ken Ono and William McGraw:
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{\em Modular form Congruences and Selmer groups}
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(McGraw will speak about this next week in this seminar!)
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\item Momose and Ozawa: {\em Rational points of
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modular curves $X_{\split}(p)$}
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\end{enumerate}
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\section{Data About Discriminant Valuations}
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\subsection{Weight Two}
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\begin{theorem}\label{thm:disc}
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The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$.
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(Except possibly $50923$ and $51437$, which I
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haven't finished checking yet.)
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\end{theorem}
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\begin{proof}
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This is the result of a large computer computation, and perhaps
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couldn't be verified any other way, since I know of no general
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theorems about $d_2(\Gamma_0(p))$. The rest of this proof describes
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how I did the computation, so you can be convinced that there is valid
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mathematics behind my computation, and that you could verify the
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computation given sufficient time. The computation described below
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took about one week using $12$ Athlon 2000MP processors. In 1999 I
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had checked the result stated above but only for $p<14000$ using a
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completely different implementation of the algorithm and a 200Mhz
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Pentium computer. These computations are nontrivial; we compute
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spaces of modular symbols, supersingular points, and Hecke operators
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on spaces of dimensions up to~$5000$.
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The aim is to determine whether or not~$p$ divides the discriminant of
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the Hecke algegra of level~$p$ for each $p < 60000$. If~$T$ is an
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operator with integral characteristic polynomial, we write $\disc(T)$
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for $\disc(\charpoly(T))$, which also equals $\disc(\Z[T])$. We will
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often use that
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$$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$
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Most levels~$p<60000$ were ruled out by
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computing characteristic polynomials of Hecke operators using an
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algorithm that David Kohel and I implemented in MAGMA, which is based
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on the Mestre-Oesterle method of graphs (our implementation is ``The Modular of
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Supersingular Points'' package that comes with MAGMA). I computed
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$\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most
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cases found a~$q$ such that this discriminant is nonzero. The
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following table summarizes how often we used each prime~$q$ (note
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that there are $6057$ primes up to $60000$):
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\begin{center}
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\begin{tabular}{|l|l|}\hline
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$q$ & number of $p< 60000$ where~$q$ smallest
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s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline
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2& 5809 times\\
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3& 161 (largest: 59471)\\
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5& 43 (largest: 57793)\\
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7& 15 (largest: 58699)\\
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11& 15 (the smallest is 307; the largest 50971)\\
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13& 2 (they are 577 and 5417)\\
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17& 3 (they are 17209, 24533, and 47387)\\
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19& 1 (it is 15661 )\\\hline
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\end{tabular}
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\end{center}
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The numbers in the right column sum to 6049, so 8 levels are
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missing. These are
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$$
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389,487,2341,7057,15641,28279, 50923, \text{ and } 51437.
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$$
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(The last two are still being processed. $51437$ has the property
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that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.)
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We determined the situation with the remaining 6 levels
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using Hecke operators $T_n$ with~$n$ composite.
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\begin{center}
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\begin{tabular}{|l|l|}\hline
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$p$ & How we rule level~$p$ out, if possible\\\hline
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389& $p$ does divide discriminant\\
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487& using charpoly($T_{12}$)\\
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2341& using charpoly($T_6$)\\
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7057& using charpoly($T_{18}$)\\
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15641& using charpoly($T_6$)\\
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28279& using charpoly($T_{34}$)\\\hline
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\end{tabular}
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\end{center}
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Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is
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large, so it is important to choose the right $T_n$ quickly.
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For $p=28279$, here is the trick I used to quickly find an~$n$ such
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that $\disc(T_n)$ is not divisible by~$p$. This trick might be used
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to speed up the computation for some other levels. The key idea is to
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efficiently discover which $T_n$ to compute. Though computing $T_n$
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on the full space of modular symbols is quite hard, it turns out that
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there is an algorithm that quickly computes $T_n$ on subspaces of
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modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis).
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Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$,
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and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$. Let~$V$ be the
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kernel of $f(T_2)$ (this takes 7 minutes to compute). If $V=0$, we
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would be done, since then $\disc(T_2)\neq 0\in\F_p$. In fact,~$V$ has
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dimension~$7$. We find the first few integers~$n$ so that the
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charpoly of $T_n$ on $V_1$ has distinct roots, and they are
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$n=34$, $47$, $53$, and $89$. I then computed
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$\charpoly(T_{34})$ directly on the whole space and found that it has
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distinct roots modulo~$p$.
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\end{proof}
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\subsection{Higher Weight Data}
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\begin{enumerate}
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\item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of the discriminant
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of the Hecke algebras associated to $S_4(\Gamma_0(p))$ for $p<500$.
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\hspace{-4em}\shadowbox{\begin{minipage}[b]{1.15\textwidth}
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\begin{tabular}{|c|ccccccccccccccccc|}\hline
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$p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
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$d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
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$p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131& 137& 139\\
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$d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline
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$p$ & 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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$d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
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$p$ & 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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$d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
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$p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
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$d$ & 56& 58& 58& 58& 60&62& 62& 62& 65 &66& 66& 68& 68& 70& 70& 72& 72\\\hline
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$p$ & 443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
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$d$ & 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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\comment{\item
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For each prime~$p$, let
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$$
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\delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)).
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$$
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Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$.
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Moreover, for every $p\neq 139$ we have that $\delta(p)\geq
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d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but
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$d_4(\Gamma_0(p))=24$.
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}
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\end{enumerate}
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\section{The Conjecture}
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\newcommand{\tT}{\tilde{\T}}
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Let~$k=2m$ be an even integer and~$p$ a prime.
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Let $\T$ be the Hecke algebra associated to $S_k(\Gamma_0(p))$ and
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let $\tT$ be the normalization of $\tT$ in $\T\tensor\Q$.
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\begin{conjecture}\label{conj:big}
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$$
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\ord_p([\tT : \T])
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= \left\lfloor\frac{p}{12}\right\rfloor\cdot \binom{m}{2} + a(p,m),
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$$
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where
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$$
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a(p,m) =
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\begin{cases}
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0 & \text{if $p\con 1\pmod{12}$,}\\
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3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if $p\con 5\pmod{12}$,}\\
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2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if $p\con 7\pmod{12}$,}\\
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a(5,m)+a(7,m) & \text{if $p\con 11\pmod{12}$.}
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\end{cases}
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$$
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In particular, when $k=2$ we conjecture that
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$[\tT:\T]$ is not divisible by~$p$.
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\end{conjecture}
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Here $\binom{x}{y}$ is the binomial coefficient ``$x$ choose $y$'',
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and floor and ceiling are as usual. We have checked this conjecture
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against significant numerical data. (Will describe here.)
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\section{Conjectures}
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\begin{conjecture}
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Suppose~$p$ is a prime and $k\geq 4$ is an even integer.
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If
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\begin{align*}
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(p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\
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&(3,4),(3,6), (3,8),\\
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&(5,4), (5,6), (7,4), (11,4)\}
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\end{align*}
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then $d_k(\Gamma_0(p))>0$.
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\end{conjecture}
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Frank Calegari outlined a possible strategy for proving this conjecture.
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\begin{conjecture}
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Suppose $p>2$ is a prime and $k\geq 3$ is an integer.
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If
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\begin{align*}
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(p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\
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&(5,3),(5,4), (5,5), (5,6), (5,7)\\
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&(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\
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&(13,3), (17,3), (19,3)\}
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\end{align*}
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then $d_k(\Gamma_1(p))>0$.
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\end{conjecture}
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\end{document}
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\begin{conjecture}
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For any even $k\geq 4$,
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$$\lim_{p\ra \infty} d_k(\Gamma_0(p)) = \infty.$$
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\end{conjecture}
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\begin{conjecture}
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For any integer $k\geq 3$,
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$$\lim_{p\ra \infty} d_k(\Gamma_1(p)) = \infty.$$
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\end{conjecture}
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The sequence of the limit need not be nonincreasing.
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\begin{conjecture}
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For any even integer $k\geq 4$, let
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$$
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\delta_k(p) = \dim S_k(\Gamma_0(p)) - \dim S_{k+p-1}(\Gamma_0(1)).
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$$
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Then
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$$
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\lim_{p\ra \infty}
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\frac{\delta_k(p)}{d_k(\Gamma_0(p))} = 1.
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$$
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\end{conjecture}
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\comment{By Theorem~\ref{thm:disc} the only~$p<50923$ such that
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$d_2(\Gamma_0(p))>0$ is $p=389$. We have
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$d_{2}(\Gamma_0(389))=1$ and $d_4(\Gamma_0(389))=65$.
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This is mysterious, because there's no (obvious) ring-theoretic
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relationship between weight-$2$ and weight-$4$ Hecke algebras.
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\begin{question}
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Is $d_{k}(\Gamma_0(389))$ odd for all even integers $k\geq 2$?
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\end{question}
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\begin{question}\label{ques:divdown}
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If $d_k(\Gamma_0(389))$ is odd for some~$k$,
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does it necessarily follow that $d_2(\Gamma_0(p))$
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is odd? Note that $d_4(\Gamma_0(p))$ is even for all $p<500$,
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except $p=389$.
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\end{question}
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The answer to Question~\ref{ques:divdown}
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is ``no'' if $\Gamma_0(p)$ is replaced
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by $\Gamma_1(p)$. For example,
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$d_3(\Gamma_1(23))=1$,
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but $d_2(\Gamma_1(23))=0$.
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There is a newform in $S_3(\Gamma_1(23))$
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with quadratic character that
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is defined over
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$K=\Q(\alpha)$ where $\alpha^3-12\alpha+7=0$, and
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the discriminant of~$K$ is $3^3\cdot 23$.
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We have $d_4(\Gamma_1(23)) = 4$.% and $d_5(\Gamma_1(23))=???$.
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}
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