Sharedwww / tables / Notes / artintalk.texOpen in CoCalc
Author: William A. Stein
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\title{\bf\Huge New examples of modular icosahedral
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Artin representations\vspace{5ex}}
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\author{\LARGE William A. Stein\vspace{2ex}\\
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\Large (joint work with Kevin Buzzard)\vspace{2ex}\\}
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\date{\Large \today \vspace{2ex}\\}
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\pagestyle{myheadings}
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\markboth{}{W.\thinspace{}A. Stein}
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\newcommand{\Page}[1]{\newpage\centerline{\LARGE \bf \stepcounter{Pagecount}
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\S\thePagecount. #1}}
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\theoremstyle{plain}
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\newtheorem{thm}{Theorem}
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\newtheorem{prp}[thm]{Proposition}
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\newtheorem{conj}[thm]{Conjecture}
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\begin{document}
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\Large
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\maketitle
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\pagenumbering{Roman}
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\setcounter{page}{0}
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% 7 minutes per page
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\Page{Artin's conjecture}
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Consider irreducible
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$$\rho : \GQ=\Gal(\Qbar/\Q) \ra \GL_n(\C),\qquad n\geq 2.$$
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\begin{thm}[Artin, 1924]
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Some power of $L(\rho,s)$ is meromorphic.
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\end{thm}
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\begin{conj}[Artin, 1924]
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$L(\rho,s)$ is entire.
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\end{conj}
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\begin{conj}
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Assume $n=2$ and $\rho$ is odd. Then~$\rho$ arises
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from a classical weight one modular form.
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\end{conj}
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By Hecke, this conj. implies $L(\rho\tensor\eps,s)$ entire for all~$\eps$.
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{\bf What we know about Conjecture 3:}\\
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Image of $\proj \rho : \GQ\ra \PGL_2(\C)$
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solvable or $\ncisom A_5$.\\
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{\em Solvable case:} Langlands, Tunnel, Hecke prove $\rho = \rho_f$
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for~$f$ classical weight one eigenform.\\
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{\em Icosahedral case:}
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\begin{bulletlist}
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\item Buhler (1970): the first example; conductor $800$.
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\item Frey, Kiming, Wang (1994): 7 more, by systematizing Buhler's work.
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Use Deligne-Serre + Langlands-Weil. Compute the number of weight
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1 newforms and the number of rep's $\rho$ and get the same number.
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\item Buzzard, Dickinson, Shepherd-Barron, Taylor (1999): infinitely
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many examples.
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Use a frightening amount of deep mathematics, including Shimura-Taniyama.
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\end{bulletlist}
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\Page{Our strategy}
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%$\xymatrix{
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% *++{\lambda}\[email protected]{^(->}[r]& \O_L\[email protected]{-}[d]\\
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% &{\Zl}}$
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%$\O_L/\Zl$, max ideal $\lambda$.
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\begin{thm}[Buzzard-Taylor]
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Suppose $p\geq 5$, and $\rho:\GQ\ra \GL_2(\O_L)$. If
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$\rho$ satisfies the following, then $\rho$ is modular:
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\begin{bulletlist}
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\item finitely many ramified primes
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\item $\rhobar$ is modular and abs. irred.
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\item $\rho$ is unramified at $p$ and $\rho(\Frob_p)$
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has \nobreak{distinct} eigenvalues mod $\lambda$.
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\end{bulletlist}
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\end{thm}
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{\bf Strategy:}
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\begin{numlist}
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\item Choose an $A_5$-extension $K/\Q$ such that
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\begin{bulletlist}
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\item $5$ is unramified, and
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\item the order of $\Frob_5\in\Gal(K/\Q)$ is not $5$.
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\end{bulletlist}
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\item Lift to $\rho:\GQ\ra \GL_2(\C)$ of conductor $N$.
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\item Pretend $\rho$ comes from a weight one form $f_?\in S_1(N,\eps;\C)$.
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\item Reduce to get $\overline{f}_?\in S_1(N,\eps;\Fbar_5)$.
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\item Multiply $\overline{f}_?$ by $E_4 \con 1 \in S_4(N;\Fbar_5)$.
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\item Search for
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$f=\overline{f}_?\cdot E_4 \in S_5(N,\eps;\Fbar_5)$.
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\item Twist to get a form $g=f\tensor \eps'\in
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S_5(N\cdot M,\eps{\eps'}^2;\F_5)$.
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\item Then $\proj \rho_g:\GQ\ra \PSL(2,\F_5)\ncisom A_5$.
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\item Prove that $\rho_g$ is unramified at $5$.
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\item Prove that image of $\proj\rho_g$ is $A_5$.
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\item Prove that the fixed field $L$
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of $\proj\rho_g$ has root field of small discriminant $\leq 2083^2$;
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hence $L$ is in Kiming's table, so has no choice but to be $K$.
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\item Apply Buzzard-Taylor to a lift of $\rhobar=\rho_g$;
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conclude that $\rho$ is modular.
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\end{numlist}
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{\bf Motivation:}
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Maybe find a theorem: ``the existence of a mod~$5$ representation
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that plausibly looks like it is $\rhobar$ must be $\rhobar$.''
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%Modularity of reps $\rho:\GQ\ra\GL(2,\F_5)$
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%with $\det(\rho)$ the cyclotomic character is {\em known}.
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\Page{A new example of conductor $1376$}
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The splitting field $K$ of
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$$h=x^5 + 2x^4+6x^3+8x^2+10x+8$$
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has group $A_5$ and discriminant $2^6\cdot 43^2$.
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Since
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$$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5},$$
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the order of $\Frob_5$ is $2$.
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A minimal lift $\rho:\GQ\ra \GL_2(\C)$ has conductor
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$$N=1376=2^5\cdot 43.$$
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\begin{thm}[-, Buzzard]
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$\rho$ is modular.
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\end{thm}
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Let
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$$\eps:(\Z/2^5\Z)^*\cross(\Z/43\Z)^*\ra\C^*$$
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be
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one of the two characters of order~$3$ at~$43$ and
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conductor $4$ at $2$. Then $\rho$ conjecturally comes from some
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$f_?\in S_1(1376,\eps;\C)$.
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{\bf Search for $f=f_?\cdot E_4$:}
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If $\Frob_p$ has order~$2$ then $a_p=0$. The first
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few such $p$ are: $19, 31, 97, 107, 127$.
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Search for $f$ by finding
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$\ker(T_{19})\intersect \ker(T_{31})
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\intersect \ker(T_{97})$, etc.
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\Page{Step 1: Find~$f$ using modular symbols}
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Using modular symbols, we compute a space
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$\sS_5(1376,\eps;\Fbar_5)$ isomorphic to
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a lattice in $S_5(1376,\eps;\Q[\eps])$ tensor
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$\Fbar_5$; plus possibly some spurious torsion.
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Use the Hijikata trace formula to rule out the torsion.
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Using modular symbols, we
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compute, in under~$4$ minutes on a 450Mhz Intel processor,
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$\sS_5(1376,\eps;\Fbar_{25})$ and
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$$V := \ker(T_{19})\intersect \ker(T_{31}) \intersect \ker(T_{97}).$$
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Find $\dim V=8$.\\
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$T_2|_V=0$.\\
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Let $\alp^2+4\alp+2 = 0\in\F_{25}$.\\
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$T_3|_V$ satisfies
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$(T_3|_V + \alp^4)^4(T_3|_V+\alp^{16})^4.$\\
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$V_1 := \ker(T_3+\alp^4)$ has dim $2$.\\
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$T_5|_{V_1}$ satisfies
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$(T_5|_{V_1} + \alp^{10})(T_5|_{V_1}+\alp^{22}).$
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The newform in~$V_1$ with $a_5+\alp^{10}=0$ is
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$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
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+ \alp^{14}q^9 + 4q^{11}+\cdots.$$
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{\em We check:} the $p\leq 997$ such that $a_p=0$ are
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the same as those for which $\Frob_p$ has order $2$.
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Morally we have found $f_?\cdot E_4$.
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\Page{Step 2: Twist down to $\GL_2(\F_5)$}
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No way to show
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$$\rho_f : \GQ\ra \GL_2(\F_{25})$$
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has projective image $A_5$!
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Write $\eps=\eps_2\cdot \eps_{43}$.
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$$g := f\tensor\eps_{43}$$
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\begin{prp}
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For $p\nmid 5N$, $a_p(g)\in\F_5$.
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\end{prp}
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\begin{proof}[Proof sketch]
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Lift $f$ to $\tilde{f}\in S_5(1376,\eps;\Qbar)$. \\
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The corresponding automorphic representation $\pi_f$ is
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ramified principal series because $43||1376$.
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Thus
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$$\rho_{\tilde{f},\lambda}|D_{43}\sim\mtwo{\Psi_1}{0}{0}{\Psi_2},$$
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with $\Psi_2$ unramified.
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Since
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$\det(\rho_{\tilde{f},\lambda})=\chi^{k-1}\eps$
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and $\chi$ only ramified at $5$, get
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$\Psi_1|_{I_{43}}=\eps_{43}|_{I_{43}}$:
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$$\rho_{\tilde{f},\lambda}|I_{43}\sim\mtwo{\eps_{43}}{0}{0}{1}$$
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Thus
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$$\rho_{{\tilde{f}\tensor\eps_{43}^{-1}},\lambda}
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\sim \mtwo{1}{0}{0}{\eps_{43}^{-1}}$$
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has the same conductor at $43$ as $\tilde{f}$.
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By level lowering,
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$$f\tensor\eps_{43}^{-1}\in S_5(1376,\eps_2\eps_{43}^2;\F_{25}).$$
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{\em Claim:} $f\tensor\eps_{43}^{-1} = \sigma(f)$,
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where $\sigma$ is the $5$th power map $\F_{25}\ra\F_{25}$.\\
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{\sc Proof.} We have $a_p(f\tensor\eps_{43}^{-1})=0$ for
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$p=19, 31,97$.
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Also, $\eps_{43}(3)=\alp^8$, $\eps_{43}(5) =\alp^8$, so
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$$a_3(f\tensor\eps_{43}^{-1})=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
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$$a_5(f\tensor\eps_{43}^{-1})=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5$$
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$f\tensor\eps_{43}^{-1}$ and $\sigma(f)$ both lie in
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$S_5(1376,\eps_2\eps_{43}^2;\F_{25})$ and have same $a_p$ for
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$p=3,5,19,31,97$, so they are equal by how we found $f$.
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So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
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Thus $a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
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$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10}
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= a_p(f) \eps_{43} = a_p(g)$.
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\end{proof}
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We have
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$$\proj \rho_g:\GQ\ra \PSL(2,\F_5)\ncisom A_5.$$
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\Page{Step 3: Prove $\rho_g$ unramified at $5$}
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\begin{prp}
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Prove that $\rho_g$ is unramified at $5$.
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\end{prp}
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\begin{proof}
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$V_1$ is spanned by the eigenforms
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$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
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+ \alp^{14}q^9 + 4q^{11}+\cdots$$
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and
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$$f_1=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7
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+ \alp^{14}q^9 + 4q^{11}+\cdots$$
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It looks like $a_p(f_1)=a_p(f)$ for all $p\neq 5$; finding
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a proof of this stumped us for a while. Suppose this is so
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for now.
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Since $f$ is ordinary, over $\Fbar_5$ Deligne's theorem gives
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$$\rho_f|_{D_5}\sim \mtwo{\alp}{*}{0}{\beta}$$
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Since $\chi_5^{5-1}=1$,
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$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$ and
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$\beta(\Frob_5)=\alp^{22}$.
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(The values of $\alp$ and $\beta$ on $\Frob_5$ are
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part of the theorem, as proved in Gross's paper.)
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Assuming $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
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$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
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with
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$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
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$\beta'(\Frob_5)=\alp^{10}$;
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in particular, $\alp'=\beta$.
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Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so
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$\rho_f|_{D_5}=\alp\oplus\beta$ and $*=0$.
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Remains to check: $a_p(f) = a_p(f_1)$ for $p\neq 5$.\\
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Is the difference
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$f-f_1$ a $q$-expansion involving only powers of $q^5$?\\
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Yes; if $f-f_1$ is in the kernel of $\theta = q \frac{d}{dq}$.
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Using present theorems, to compute
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$\theta : S_5(1376,\eps;\F_{25})\ra S_{11}(1376,\eps;\F_{25})$
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using $q$-expansions requires knowing thousands and
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thousands of terms of the $q$-expansion -- not feasible.\\
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Instead use one of the following:
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\begin{bulletlist}
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\item Define $\theta$ on modular symbols and compute it.
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\item Compute the intersection
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$$\bigcap_{\text{all $p$}} \ker(T_p - p\cdot a_p(f)) \subset S_{11}(1376,\eps;\F_{25}).$$
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The moment the dimension is~$1$, it follows that $\theta(f-f_1)=0$.
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\end{bulletlist}
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We carried out both computations and they worked.
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For the first, $\theta$ is multiplication by
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$X^{\ell}Y - Y^{\ell}X$. For the second, we find that
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$$\dim \bigcap_{p\leq 11}\ker(T_p - p\cdot a_p(f)) = 1.$$
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\end{proof}
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\Page{Step 4: The image of $\proj \rho_g$ is $A_5$}
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List the first few hundred $a_p(g)$.
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Most are nonzero, so the image can't be cyclic or dihedral.
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It can't be $S_4$ since $\#S_4$ does not divide $60$.
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Another trick (suggested by Lenstra) shows that it can't be~$A_4$.
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\Page{Step 5: Bound the ramification}
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This involves considering the order of the image of inertia,
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and then bounding the discriminant.
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\Page{Other examples}
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I have repeated the computation of
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$V$ and $V_1$ for $A_5$ representations of the following conductors
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\begin{eqnarray*}
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2416&=&2^4\cdot 151\\
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3184&=&2^4\cdot 199\\
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3556&=&2^2\cdot 7\cdot 127\\
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3756&=&2^2\cdot 3\cdot 313.
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\end{eqnarray*}
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In each case, the first three primes such that $a_p=0$ were enough
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to obtain an $8$ dimensional space. After that, using a few
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Hecke operators, we find a form $f$ and a form that is (almost
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certainly) a companion of $f$.
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In the $3556$ and $3756$ examples $\Frob_5$ has order~$3$; this
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presents interesting new twists. The dimensions are also large, e.g.,
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$$\dim S_5(3756,\eps;\F_{25}) = 2506.$$
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\end{document}
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