\documentclass{article} \bibliographystyle{amsalpha} %\voffset=-0.2\textheight \textheight=1.4\textheight %\hoffset=-0.2\textwidth \textwidth=1.4\textwidth %\voffset=-0.05\textheight \textheight=1.1\textheight %\hoffset=-0.05\textwidth \textwidth=1.1\textwidth \usepackage{amsmath} \usepackage{amsfonts} \usepackage[all]{xy} \newcommand{\vphi}{\varphi} \newcommand{\eps}{\varepsilon} \newcommand{\CR}[1]{{\bf #1}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\an}{{\rm an}} \newcommand{\ncisom}{\approx} % noncanonical isomorphism \newcommand{\isom}{\cong} \newcommand{\tensor}{\otimes} \newcommand{\h}{\mathfrak{h}} \renewcommand{\P}{\mathbb{P}} \newcommand{\smallmtwo}[4]{\left( \begin{smallmatrix}#1\\#3 \end{smallmatrix}\right)} \DeclareMathOperator{\Div}{Div} \DeclareMathOperator{\Stab}{Stab} \DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\Jac}{Jac} \DeclareMathOperator{\Sym}{Sym} \DeclareMathOperator{\tor}{tor} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\PSL}{PSL} \DeclareMathOperator{\Imm}{Im} \renewcommand{\Im}{\Imm} %%%% Theoremstyles \usepackage{amsthm} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{conjecture}[theorem]{Conjecture} \usepackage{hyperref} \title{Numerical Computation of Chow-Heegner Points} \author{William Stein} \begin{document} \maketitle \begin{abstract} In this note, we consider a special case of Chow-Heegner points that has a simple concrete description due to Shouwu Zhang. Given a pair $E$, $F$ of nonisogenous elliptic curves, and a fixed choice of surjective morphisms $\vphi_E: X_0(N) \to E$ and $\vphi_F: X_0(N) \to F$ of curves over $\Q$, we associate a rational point $P\in E(\Q)$. We describe a relatively elementary numerical approach to computing $P$, state some motivating results of Zhang et al. about the height of $P$, and present a table of data. \end{abstract} \section{Introduction: Zhang's Construction} Consider a pair $E,F$ of nonisogenous elliptic curves over $\Q$ and fix surjective morphisms from $X_0(N)$ to each curve. We do {\em not} assume that $N$ is the conductor of either $E$ or $F$, though $N$ is necessarily a multiple of the conductor. $$ \xymatrix{ & {X_0(N)}\ar[ld]_{\vphi_E}\ar[rd]^{\vphi_F} & \\ E & & F} $$ Let $(\vphi_E)_*:\Div(X_0(N))\to \Div(E)$ and $\vphi_F^*:\Div(F) \to \Div(X_0(N))$ be the pushforward and pullback maps on divisors on algebraic curves. Let $Q\in F(\C)$ be any point, and let $$ P_{\vphi_E, \vphi_F,Q} = \sum (\vphi_E)_* \vphi_F^*(Q) \in E(\C), $$ where $\sum$ means the sum of the points in the divisor using the group law on $E$, i.e., given a divisor $D = \sum n_i P_i \in \Div(E)$, we have $(\sum D) - \infty \sim D - \deg(D)\infty$, which uniquely determines $\sum D$. \begin{proposition} The point $P_{\vphi_E, \vphi_F, Q}$ does not depend on the choice of $Q$. \end{proposition} \begin{proof} The composition $(\vphi_E)_* \circ \vphi_F^*$ induces a homomorphism of elliptic curves $$ \psi: \Pic^0(F)=\Jac(F) \to \Jac(E)=\Pic^0(E). $$ Our hypothesis that $E$ and $F$ are nonisogenous implies that $\psi=0$. We denote by $[D]$ the linear equivalence class of a divisor in the Picard group. If $Q'\in F(\C)$ is another point, then under the above composition of maps, $$ [Q - Q']\mapsto \left[(\vphi_E)_* \vphi_F^*(Q) - (\vphi_E)_* \vphi_F^*(Q')\right] = [P_Q - P_{Q'}]. $$ Thus the divisor $P_Q - P_{Q'}$ is linearly equivalent to $0$. But $F$ has genus $1$, so there is no rational function on $F$ of degree 1, hence $P_Q = P_{Q'}$, as claimed. \end{proof} We let $P_{\vphi_E, \vphi_F} = P_{\vphi_E, \vphi_F,Q} \in E(\C)$, for any choice of $Q$. When there is a canonical choice of $\vphi_E$, $\vphi_F$, e.g., when $E$ and $F$ are both are optimal curves of the same conductor $N$, then we write $P_{E,F} = P_{\vphi_E, \vphi_F}$. %(we fix the sign of the modular parameterization %map by taking the one that sends the cusp $\infty$ to the 0 point %$\mathcal{O}$ on the curve), \begin{corollary} We have $P_{\vphi_E, \vphi_F} \in E(\Q)$. \end{corollary} \begin{proof} Taking $Q=\mathcal{O} \in F(\Q)$, we see that the divisor $(\vphi_E)_* \circ \vphi_F^*(Q)$ is rational, so its sum is also rational. \end{proof} %It is natural at this point to speculate: {\bf surely $P_{E,F}$ % is always $0$?} \subsection{Outline} In Section~\ref{sec:37} we discuss in detail a point $P_{E,F}$, where $E$ is the rank 1 curve of conductor $37$. Section~\ref{sec:yzz} is about a recent Gross-Zagier style formula of Yuan-Zhang-Zhang for the height of $P_{E,F}$ in terms of the derivative of an $L$-function, in some cases. In Section~\ref{sec:integrals} we discuss the connection between the present paper and \cite{darmon_daub_lichtenstein_rotger:chow_heegner}, which is about computing Chow-Heegner points using iterated integrals. The heart of the present paper is Section~\ref{sec:numerical}, which describes our numerical approach to approximating $P_{E,F}$. Finally, Section~\ref{sec:data} presents extensive tables of points $P_{E,F}$. {\bf Acknowledgement: } We would like to thank Ralph Greenberg, Ken Ribet, Barry Mazur, Karl Rubin, Shouwu Zhang, and Jon Bober for helpful conversations related to this paper, and especially thank Xinyi Yuan for introducing us to this topic, Henri Darmon for encouraging us to flesh out the details and write this up for publication, and Victor Rotger for correcting a serious mistake in an early draft. \section{Example: $N=37$}\label{sec:37} The smallest conductor curve of rank $1$ is the curve $E$ with Cremona label \CR{37a} (see \cite{cremona:onlinetables}). The beautiful paper \cite{mazur-swinnerton-dyer:arithmetic} discusses the modular curve $X_0(37)$ in detail. In particular, it presents the affine equation $y^2=-x^6-9x^4-11x^2 +37$ for $X_0(37)$, and describes explicitly how $X_0(37)$ is equipped with three independent involutions $w$, $T$, and $S$. The quotient of $X_0(37)$ by $w$ is $E$, the quotient by $T$ is an elliptic curve $F$ with $F(\Q)\ncisom \Z/3\Z$ and Cremona label \CR{37b}, and the quotient by $S$ is the projective line $\P^1$. $$ \xymatrix{ & {X_0(37)}\ar[ld]_{\vphi_E}\ar[d]^{\vphi_F}\ar[rd] & \\ E=X/w & F=X/T & {\P^1}=X/S }$$ The maps $\vphi_E$ and $\vphi_F$ both have degree $2$, since they are induced by quotienting out by an involution. As explained in \cite{mazur-swinnerton-dyer:arithmetic}, the fiber over $Q=0\in F(\Q)$ contains 2 points: \begin{enumerate}\setlength{\itemsep}{-.5ex} \item the cusp $[\infty] \in X_0(37)(\Q)$, and \item the noncuspidal affine rational point $(-1,-4) = T(\infty) \in X_0(37)(\Q)$. \end{enumerate} We have $\vphi_E([\infty])=0\in E(\Q)$ and \cite[Prop.~3, pg.~30]{mazur-swinnerton-dyer:arithmetic} explains that $$\vphi_F((-1,-4)) = (6,14) = -6(0,-1),$$ where $(0,-1)$ generates $E(\Q)$. We conclude that $$ P_{E,F} = (6,14)\qquad\text{and}\qquad[E(\Q):\Z P_{E,F}] = 6. $$ On \cite[pg.~31]{mazur-swinnerton-dyer:arithmetic}, they remark: ``It would be of the utmost interest to link this index [the index of 6 above] to something else in the theory.'' This remark motivates our desire to compute many more examples. Unfortunately, it is very difficult to generalize the above approach directly, since it involves computations with $X_0(37)$ and its quotients that rely heavily on having simple explicit defining equations. Just as there are multiple complementary approaches to computing Heegner points, there are several approaches to computing $P_{E,F}$: \begin{itemize}\setlength{\itemsep}{-.5ex} \item Section~\ref{sec:yzz}: a Gross-Zagier style formula for the height of $P_{E,F}$, \item Section~\ref{sec:integrals}: explicit evaluation of iterated integrals, and \item Section~\ref{sec:numerical}: numerical approximation of the fiber in the upper half plane over a point on an elliptic curve using a polynomial approximation to $\vphi_F$. \end{itemize} The present paper is mainly about the last approach listed above. \section{The Formula of Yuan-Zhang-Zhang}\label{sec:yzz} Consider the triple product $L$-function of \cite{gross-kudla} \begin{equation}\label{eqn:tprod} L(E,F,F,s) = L(E,s) \cdot L(E,\Sym^{2}(F),s), \end{equation} where $E$ and $F$ are elliptic curves of the same conductor $N$. For simplicity, in this section all $L$-functions are normalized so that $1/2$ is the center of the critical strip. The following theorem is proved in \cite{yuan-zhang-zhang:triple}: \begin{theorem}[Yuan-Zhang-Zhang]\label{thm:yzz} Assume that the local root number of $L(E,F,F,s)$ at every prime of bad reduction is $+1$ and that the root number at infinity is $-1$. Then $ \hat{h}(P_{E,F}) = (*) \cdot L'(E,F,F,\frac{1}{2}), $ where $(*)$ is nonzero. \end{theorem} \begin{remark} The above formula resembles the Gross-Zagier formula $$ \hat{h}(P_K) = (*) \cdot (L(E/\Q,s) \cdot L(E^K/\Q,s))'|_{s=\frac{1}{2}}, $$ where $K$ is a quadratic imaginary field satisfying certain hypotheses. \end{remark} Unfortunately, at present it appears that nobody has implemented a computer program to evaluate the formula of Theorem~\ref{thm:yzz} numerically in any interesting cases yet. %One problem is that if $E$ %and $F$ have conductor $N$, then $L(E,F,F,s)$ has the enormous %conductor $N^{10}$ (though the factor $L(E,\Sym^2 F,s)$ has smaller %conductor). If one could evaluate $L'(E,F,F,\frac{1}{2})$, along with the factor $(*)$ in the theorem, this would yield an algorithm to compute $\pm P_{E,F}\pmod{E(\Q)_{\tor}}$ in the special case when $N$ is squarefree and all roots numbers for $E$ are $+1$. We hope to carry out this approach (using \cite{dokchitser:lfun}) in future work. We have the following proposition that we can apply in specific examples. This follows from \cite[\S1]{gross-kudla}, which implies that the local root number of $L(E,F,F,s)$ at $p$ is the same as the local root number of $E$ at $p$ when the level is square free (computing the local root number when the level is not square free is more complicated). \begin{proposition}\label{prop:esymf} Assume that $E$ and $F$ have the same conductor $N$, that $N$ is square free, that the local root numbers of $E$ at primes $p\mid N$ are all $+1$ (equivalently, that we have $a_p(E)=-1$) and that $r_{\an}(E/\Q)=1$. Then $L(E, \Sym^2 F, \frac{1}{2})\neq 0$ if and only if $\hat{h}(P_{E,F})\neq 0$. \end{proposition} \begin{proof} By hypothesis, we have $L(E,\frac{1}{2})=0$ and $L'(E,\frac{1}{2})\neq 0$. %First suppose $\hat{h}(P_{E,F})\neq 0$. Theorem~\ref{thm:yzz} and the factorization \eqref{eqn:tprod} then imply that $$ \hat{h}(P_{E,F}) = (*) \cdot L'(E,\frac{1}{2}) \cdot L(E, \Sym^2 F, \frac{1}{2}), $$ from which the result follows. \end{proof} Section~\ref{sec:data} contains numerous examples in which $E$ has rank $1$, $F$ has rank $0$, and yet $P_{E,F}$ is a torsion point. The first example is when $E$ is \CR{91b} and $F$ is \CR{91a}. Then $P_{E,F}=(1,0)$ is a torsion point (of order 3). In this case, we cannot apply Proposition~\ref{prop:esymf} since $\eps_7=\eps_{13}=-1$ for $E$. Another example is when $E$ is \CR{99a} and $F$ is \CR{99c}, where we have $P_{E,F}=0$, and $\eps_3=\eps_{11}=+1$, but Proposition~\ref{prop:esymf} does not apply since the level is not square free. Fortunately, there is an example with squarefree level $158=2\cdot 79$: here $E$ is \CR{158b}, $F$ is \CR{158d}, we have $P_{E,F}=0$ and $\eps_{2}=\eps_{79}=+1$, so Proposition~\ref{prop:esymf} implies that $L(E,\Sym^2 F, \frac{1}{2})=0$. %{\em Does this mean anything in terms of the arithmetic of $E$ and $F$?} \section{Iterated Complex Path Integrals}\label{sec:integrals} The paper \cite{darmon_daub_lichtenstein_rotger:chow_heegner} contains a general approach using iterated path integrals to compute certain Chow-Heegner points, of which $P_{E,F}$ is a specific instance. Comparing our data (Section~\ref{sec:data}) with theirs, we find that if $E$ and $F$ are optimal elliptic curves over $\Q$ of the same conductor $N\leq 100$, if $e, f\in S_2(\Gamma_0(N))$ are the corresponding newforms, and if $P_{f,e,1} \in E(\Q)\tensor_{\Q}\Q$ the associated Chow-Heegner point in the sense of \cite{darmon_daub_lichtenstein_rotger:chow_heegner}, then $2P_{E,F} = P_{f,e,1}$. This is (presumably) a consequence of \cite{darmon_rotger_sols:iterated}. %[[TODO: Actually, I've only checked this up to level 77.]] % The algorithm of \cite{darmon_daub_lichtenstein_rotger:chow_heegner} % has been implemented using Sage \cite{sage}, and run for the $17$ rank % $1$ curves $E$ with $N<100$, and various newforms $f$, and one obtains % the table of data below. We have simplified the data to only include % points $P_{g,f,1}$ with $f$ corresponding to an elliptic curve (and % not a higher degree newform), and also labeled $f$ and $g$ using % standard Cremona labels, which makes comparison of our data % (Section~\ref{sec:data}) with theirs much easier. In the column % labeled $P_{g,f,1}$, we give the element $P_{g,f,1} \in % E(\Q)\tensor\Q$ as an integral multiple of $P$. % \begin{center} % \begin{tabular}{|c|c|c|c|}\hline % $g$& $f$& $P$ & $P_{g,f,1}$ \\\hline % \CR{37a} & \CR{37b} & $(0,-1)$ & $-12P$\\\hline % \CR{57a} & \CR{57c} & $(0,-1)$ & $16P$\\\hline % \CR{57a} & \CR{57b} & $(2,1)$ & $-16P$\\\hline % \CR{57a} & \CR{19a} & $(2,1)$ & $-8P$\\\hline % \CR{58a} & \CR{58b} & $(0,-1)$ & $16P$\\\hline % \CR{77a} & \CR{77b} & $(2,3)$ & $48P$\\\hline % \CR{77a} & \CR{77c} & $(2,3)$ & $-8P$\\\hline % TODO & .. & .. & TODO \\\hline % \CR{91b} & \CR{91a} & ... & $0$\\\hline % \CR{92b} & \CR{92a} & ... & $0$\\\hline % \end{tabular} % \end{center} \section{A Numerical Approach to Computing $P_{E,F}$}\label{sec:numerical} The numerical approach to computing $P$ that we describe in this section is easy to explain and implement and uses little abstract theory. It is inspired by work of Delaunay (see \cite{delaunay:thesis}) on computing the fiber of the map $X_0(389)\to E$, over rational points on the rank $2$ curve $E$ of conductor $389$. We make no guarantees about how many digits of our approximation to $P_{E,F}$ are correct, instead viewing this as an algorithm to produce something useful for experimental mathematics. % \subsection{Modular Correspondence Degree} % Define the {\em modular correspondence degree} $m_{E,F}$ of two % nonisogenous elliptic curves $E$ and $F$ to be the degree of the % divisor $((\vphi_E)_* \circ \vphi_F^*)(Q)$, for any $Q\in F(\C)$, i.e., % the degree of the correspondence $(\vphi_E)_* \circ \vphi_F^*$. % We do not know an easy way to compute $m_{E,F}$. However, we at least % have an upper bound. Let $W$ be the group of Atkin-Lehner involutions % of $X_0(N)$, which is a group of order a power of $2$. Because $E$ % and $F$ are elliptic curves corresponding to newforms, there is an % induced action of $W$ on $E$ and $F$. % \begin{proposition}\label{prop:jointmoddeg} % Let $G$ be the subgroup of $W$ of elements that fix both $E$ and $F$. % If ..., % Then $$m_{E,F} \leq \frac{m_F}{\#G}.$$ % \end{proposition} % \begin{proof} % \end{proof} % The upper bound of the proposition is not in general sharp. For % example, we have the following table, which we compute using % Section~\ref{sec:method}, so the values of $m_{E,F}$ are only highly % likely to be correct. % [[todo: insert table]] % \begin{proposition} % If ..., then % We have $\displaystyle \frac{m_{F}}{m_{E,F}} \mid [E(\Q)/_{\tor}:\Z P_{E,F}]$. % \end{proposition} % \begin{proof} % \end{proof} % \begin{corollary} % If ..., then $2\mid [E(\Q)/_{\tor}:\Z P_{E,F}]$. % \end{corollary} % \subsection{The Numerical Method}\label{sec:method} Let $\h$ be the upper half plane, and let $Y_0(N) = \Gamma_0(N)\backslash \h \subset X_0(N)$ be the affine modular curve. Let $E$ and $F$ be nonisogenous optimal elliptic curve quotients of $X_0(N)$, with modular parametrization maps $\vphi_E$ and $\vphi_F$, and assume both Manin constants are $1$. Let $\Lambda_E$ and $\Lambda_F$ be the period lattices of $E$ and $F$, so $E\isom \C/\Lambda_E$ and $F\isom \C/\Lambda_F$. Viewed as a map $[\tau] \mapsto \C/\Lambda_E$, we have (using square brackets to denote equivalence classes), $$ \vphi_E([\tau]) = \left[\sum_{n=1}^{\infty} \frac{a_n}{n} e^{2\pi i n \tau}\right], $$ and similarly for $\vphi_F$. This is explained in \cite[\S2.10]{cremona:algs}, which uses the oppositive convention for the sign. Here $a_n = a_n(E)$ are the $L$-series coefficients of $E$, so for good primes $p$, we have $a_p = p+1-\#E(\F_p)$. For any positive integer $B$, define the polynomial $$\vphi_{E,B}=\sum_{n=1}^B \frac{a_n}{n} T^n \in \Q[T],$$ and similarly for $\vphi_{F,B}$. To compute $P_{E,F}$, we proceed as follows. First we make some choices, and after making these choices we run the algorithm, which will either find a ``probable'' numerical approximation to $P_{E,F}$ or fail. \begin{itemize}\setlength{\itemsep}{-.5ex} \item $y\in \R_{>0}$ -- minimum imaginary part of points in fiber in upper half plane. \item $d\in\Z_{>0}$ -- degree of the first approximation to $\vphi_F$ in Step~\ref{step:lowprecroots}. \item $r \in \R_{\neq 0}$ -- real number specified to $b$ bits of precision that defines $Q \in \C/\Lambda$. \item $b'$ -- bits of precision when dividing points into $\Gamma_0(N)$ orbits. \item $n$ -- number of trials before we give up and output FAIL. \end{itemize} We compute $P_{\vphi_E, \vphi_F,Q}$ using an approach that will always fail if $Q$ is a ramification point. Our algorithm will also fail if any points in the fiber over $Q$ are cusps. This is why we do not allow $r=0$. It is possible to modify the algorithm to usually work when $Q$ is a torsion point, by using modular symbols and keeping track of images of cusps. To increase our confidence that we have computed the right point $P_{E,F}$, we often carry out the complete computation with more than one choice of $r$. %Note that if we choose an $r$ that is not in %the $\Q$ span of $\Lambda$, then it is highly likely that $Q \not \in %E(\overline{\Q})$, in which case $Q$ cannot be a ramification point. \begin{enumerate} \item\label{step:lowprecroots} {\bf Low precision roots:} Compute all complex double precision roots of the polynomial $\vphi_{F,d} - r = 0$. One way to do this is to use ``balanced QR reduction of the companion matrix'', as implemented in GSL.\footnote{GSL is the the GNU scientific library, which is part of Sage \cite{sage}. Rough timings of GSL for this computation: it takes less than a half second for degree 500, about 5 seconds for degree 1000, about 45 seconds for degree 2000, and several minutes for degree 3000.} Record the roots that correspond to $\tau\in \h$ with $\Im(\tau)\geq y$. \item\label{step:highprecroots} {\bf High precision roots:} Compute an integer $B$ such that if $\Im(\tau)\geq y$, then $$\left|\sum_{n=B+1}^{\infty} \frac{a_n(F)}{n}\tau^n\right| < 2^{-b},$$ where $b$ is the number of bits of precision of $r$. Explicitly, by summing the tail end of the series and using that $|a_n|\leq n$ (see \cite[Lem.~2.9]{bsdalg1}), we find that $$ B = \left\lceil \frac{\log(2^{-(b+1)}\cdot (1-e^{-2\pi y_1}))}{-2\pi y} \right\rceil $$ works. Next, compute the polynomial $\vphi_{F,B} \in \Q[T]$, and use Newton iteration to refine all roots saved in Step~\ref{step:lowprecroots} to roots $\alpha$ of $f=\varphi_{F,B} - r \in \R[T]$ to $b$ bits of precision, i.e., so that $|f(\alpha)|<2^{-b}$. Do this by iteratively replacing $\alpha$ by $\alpha - \frac{f(\alpha)}{f'(\alpha)}$, and save those roots that correspond to $\tau\in\h$ with $\Im(\tau)\geq y$. \item\label{step:orbits} {\bf $\Gamma_0(N)$-orbits:} Divide the $\tau$'s from Step~\ref{step:highprecroots} into $\Gamma_0(N)$-equivalence classes, testing equivalence to some bit precision $b'\leq b$. To test $\Gamma_0(N)$ equivalence of two points $\tau_1$ and $\tau_2$, we first decide whether or not they are $\SL_2(\Z)$ equivalent, using the standard fundamental domain, then if they are equivalent, we check whether an explicit transformation matrix is in $\Gamma_0(N)$ -- see Section~\ref{sec:g0nequiv} below for details. It is easy to efficiently compute the modular degree $m_F = \deg(\vphi_F)$ (see \cite{watkins:moddeg}). If we find $m_F$ distinct $\Gamma_0(N)$ classes of points, we suspect that we have found the fiber over $[r]$, so we map each element of the fiber to $E$ using $\vphi_E$ and sum, then apply the elliptic exponential to obtain $P_{E,F}$ to some precision, then output this approximation and terminate. If we find more than $m_F$ distinct classes, that indicates an error in the choices of precision in our computation, so we increase $b$ or possibly decrease $b'$. % \item\label{step:images} {\bf Image points:} Compute an upper bound on % the joint modular degree $m_{E,F}$ using % Proposition~\ref{prop:jointmoddeg}. Map representative $\tau$'s % from Step~\ref{step:orbits} to $E(\C)$, choosing representatives % with largest imaginary part to speed convergence. Use the elliptic % exponential $\C/\Lambda_E \isom E(\C)$ to obtain points on an % algebraic curve. Test equality of these points in $E(\C)$ using % $b_4$ bits of precision. If we find $m_{E,F}$ distinct points, then % we sum them up and multiply by $m_{F}/m_{E,F}$ to obtain a numerical % approximation to $P_{E,F}$ and go to Step~\ref{step:closepoint}. If % we find more than $m_{E,F}$ distinct points, there is a numerical % error due to choice of parameters (maybe $b_4$ is too large). \item\label{step:increase} {\bf Try again:} We did not find enough points in the fiber. Systematically replace $r$ by $r + m\Omega_F$, for $m=1,-1,2,-2,\ldots$ and $\Omega_F$ the least real period of $F$, then try again going to Step~\ref{step:lowprecroots} and including the new points found. If upon trying $n$ choices $r+m\Omega_F$ we find no new points at all, we output FAIL and terminate the algorithm. \end{enumerate} % Regarding complexity, the main difficulty in the above algorithm % arises when $m_F$ is large (e.g., beyond a few hundred). \subsection{Determining $\Gamma_0(N)$ equivalency}\label{sec:g0nequiv} To determine numerically if two points $z_1$ and $z_2$ in the upper half plane $\h$ are equivalent modulo the action of $\Gamma_0(N)$, we first determine whether or not $z_1$ and $z_2$ are equivalent modulo $\SL_2(\Z)$ using the standard fundamental domain, as explained in \cite[\S2.14]{cremona:algs}. If $z_1$ and $z_2$ are not $\SL_2(\Z)$ equivalent, then they are not $\Gamma_0(N)$ equivalent and we are done. If $z_1$ and $z_2$ are $\SL_2(\Z)$ equivalent, then the algorithm mentioned above also produces explicit elements $g_1, g_2\in \SL_2(\Z)$ such that $g_1(z_1) = g_2(z_2)$ is in the standard fundamental domain. Let $g = g_2^{-1} g_1$, so $g(z_1) = z_2$. Suppose $h$ is any other matrix in $\SL_2(\Z)$ such that $h(z_1) = z_2$ as well. Then $$ (h^{-1} g)(z_1) = h^{-1}(g(z_1)) = h^{-1}(z_2) = z_1, $$ so $h^{-1} g$ fixes $z_1$. Assume that $k = h^{-1} g \neq 1$, viewed as elements of $\PSL_2(\Z)$. Then $k$ has a fixed point in the upper half plane. The only elements of $\PSL_2(\Z)$ with a fixed point in the upper half plane are $\Stab(z)$, where \begin{itemize}\setlength{\itemsep}{-.5ex} \item $z = i$, so $\Stab(z)$ is generated by $S=\smallmtwo{0}{-1}{1}{0}$. \item $z = \rho = \exp(2 \pi i/3)$ so $\Stab(z)$ is generated by $ST$, where $T=\smallmtwo{1}{1}{0}{1}$. \item $z = -\overline{\rho} = \exp(\pi i/3)$, so $\Stab(z)$ is generated by $TS$. \end{itemize} Assume that none of the 3 above are the case. Then $g=h$, so there is a matrix in $\Gamma_0(N)$ that sends $z_1$ to $z_2$ if and only if $g\in\Gamma_0(N)$, since $g$ is the unique matrix in $\SL_2(\Z)$ that sends $z_1$ to $z_2$. In the other cases, we check the following: \begin{itemize}\setlength{\itemsep}{-.5ex} \item $z = i$: check that neither of $g$, $g S$ are in $\Gamma_0(N)$ \item $z = \rho$: check that neither of $g$, $g S T$, $g (S T)^2$ are in $\Gamma_0(N)$ \item $z = -\overline{\rho}$: check that neither of $g$, $g T S$, $g (T S)^2$ are in $\Gamma_0(N)$. \end{itemize} \section{Data}\label{sec:data} The columns of the tables in the rest of this section are as follows. The columns labeled $E$ and $F$ contain Cremona labels for elliptic curves, and those labeled $r_E$ and $r_F$ gives the corresponding ranks. The column labeled $E(\Q)$ gives a choice of generators $P_1, P_2, \ldots$ for the Mordell-Weil group as explicit points, with $r_E$ points of infinite order listed first, then 0,1, or 2 torsion points listed with a subscript of their order. The column labeled $P_{E,F}$ contains a near rational point (see below) to the numerically computed Chow-Heegner point, represented in terms of the generators $P_i$ from the column labeled $E(\Q)$, where $P_1$ is the first generator, $P_2$, the second, and so on. The columns labeled $m_E$ and $m_F$ give the modular degrees of $E$ and $F$. The column labeled $\eps$'s contains the local root numbers of $L(E,s)$ at each bad prime. The notes column refers to the notes after the table, which give information about the input parameters needed to compute $P_{E,F}$. We believe that the values of $P_{E,F}$ are ``likely'' to be correct, but we emphasize that {\bf they are not proven correct}. In the table we give an exact point, but the algorithm computes a numerical approximation $\tilde{P}_{E,F}$ to $P_{E,F} \in E(\Q)$. We find the exact point by running through several hundred points in $E(\Q)$ and finding the one closest to $\tilde{P}_{E,F}$. % In our computation, the maximum absolute %difference between $x$ and $y$ coordinates and the exact rational %point is always at most $10^{-??}$. %TODO: what is the ?? The table contains {\em every} pair $E,F$ of nonisogenous optimal elliptic curves of the same conductor $N\leq 184$ with $r_E=1$, and most curves for $N\leq 250$. It also contains a few additional miscellaneous examples, e.g., with $r_E=0$ and some of larger conductor with $r_F=2$. Most rows took only a few seconds to compute, though ones with $m_F$ large in some cases took much longer; the total CPU time to compute the entire table was less than 10 hours. Unless otherwise noted, we used $y=10^{-4}$, $d=500$, $b'=20$, and $r=0.1$ with 53 bits of precision, as in Section~\ref{sec:numerical}. We also repeated all computations with at least one additional value of $r\neq 0.1$, and in every case got the same result (usually we used $r=0.2$). %TODO (didn't really do this yet in all cases): In each case, we also %did the computation using $r=0.2$ as a double check. \subsection{Discussion} One numerical observation in the table is that in every case $2\mid [E(\Q)_{/\tor}: \Z P_{E,F}].$ This can likely be proved in some cases by using that $r_{\an}(E)=1$ implies that the sign in the functional equation for $L(E,s)$ is $-1$, so at least one nontrivial Atkin-Lehner involution $W_q$ acts as $+1$ on $E$, which means that the map $X_0(N) \to E$ factors through $X_0(N)\to X_0(N)/W_q$. %[[TODO: just prove this.]] %Another observation in the data is that $P_{E,F}$ is usually nonzero %for $N<100$, but is frequently $0$ for larger $N$. There are four cases in which the index $[E(\Q)/_{\tor}: \Z P_{E,F}]$ is divisible by a prime $\ell\geq 5$. They are (\CR{106b}, \CR{106c}, $\ell=11$), (\CR{118a}, \CR{118d}, $\ell=7$), (\CR{121b}, \CR{121d}, $\ell=7$), and (\CR{158b}, \CR{158c}, $\ell=7$). In each case, the prime divisor $\ell$ of the index does not appear to have anything to do with the invariants of $E$ and $F$, individually. \begin{center}\small \begin{tabular}{|c|c|c|c|c||c|c|c||c|c|}\hline $E$& $\eps_p$'s & $r_E$ & $E(\Q)$ & $m_E$ & $F$& $r_F$ & $m_F$ & $P_{E,F}$ & Notes\\\hline\hline \CR{37a}&$ + $& $1$ & $(0, -1)$ & $2$ & \CR{37b} & $0$ & $2$ & $-6P_{1}$ & \\\hline \CR{37b}&$ - $& $0$ & $(8, 18)_{3}$ & $2$ & \CR{37a} & $1$ & $2$ & $P_{1}$ & \\\hline \CR{57a}&$ ++ $& $1$ & $(2, 1)$ & $4$ & \CR{57c} & $0$ & $12$ & $8P_{1}$ & \\\hline \CR{57a}&$ ++ $& $1$ & $(2, 1)$ & $4$ & \CR{57b} & $0$ & $3$ & $-8P_{1}$ & \\\hline \CR{57b}&$ -+ $& $0$ & $(7/4, -11/8)_{2},(1, -1)_{2}$ & $3$ & \CR{57a} & $1$ & $4$ & $0$ & \\\hline \CR{57b}&$ -+ $& $0$ & $(7/4, -11/8)_{2},(1, -1)_{2}$ & $3$ & \CR{57c} & $0$ & $12$ & $0$ & \\\hline \CR{57c}&$ -+ $& $0$ & $(2, 4)_{5}$ & $12$ & \CR{57a} & $1$ & $4$ & $3P_{1}$ & \\\hline \CR{57c}&$ -+ $& $0$ & $(2, 4)_{5}$ & $12$ & \CR{57b} & $0$ & $3$ & $P_{1}$ & \\\hline \CR{58a}&$ ++ $& $1$ & $(0, -1)$ & $4$ & \CR{58b} & $0$ & $4$ & $8P_{1}$ & \\\hline \CR{58b}&$ -+ $& $0$ & $(-1, 2)_{5}$ & $4$ & \CR{58a} & $1$ & $4$ & $3P_{1}$ & \\\hline \CR{77a}&$ ++ $& $1$ & $(2, 3)$ & $4$ & \CR{77b} & $0$ & $20$ & $24P_{1}$ & (1) \\\hline % deg1=1500, min_imag=1e-5, equiv_prec=15 \CR{77a}&$ ++ $& $1$ & $(2, 3)$ & $4$ & \CR{77c} & $0$ & $6$ & $-4P_{1}$ & \\\hline \CR{89a}&$ + $& $1$ & $(0, -1)$ & $2$ & \CR{89b} & $0$ & $5$ & $4P_{1}$ & \\\hline \CR{91a}&$ ++ $& $1$ & $(0, 0)$ & $4$ & \CR{91b} & $1$ & $4$ & $4P_{1}$ & \\\hline \CR{91b}&$ - - $& $1$ & $(-1, 3),(1, 0)_{3}$ & $4$ & \CR{91a} & $1$ & $4$ & $P_{2}$ & \\\hline \CR{92b}&$ - - $& $1$ & $(1, 1)$ & $6$ & \CR{92a} & $0$ & $2$ & $0$ & \\\hline \CR{99a}&$ ++ $& $1$ & $(2, 0),(-1, 0)_{2}$ & $4$ & \CR{99b}& $0$ & $12$ & $-4P_{1}$ & \\\hline \CR{99a}&$ ++ $& $1$ & $(2, 0),(-1, 0)_{2}$ & $4$ & \CR{99c} & $0$ & $12$ & $0$ & \\\hline \CR{99a}&$ ++ $& $1$ & $(2, 0),(-1, 0)_{2}$ & $4$ & \CR{99d}& $0$ & $6$ & $2P_{1}$ & \\\hline \CR{102a}&$ +++ $& $1$ & $(2, -4),(0, 0)_{2}$ & $8$ & \CR{102b} & $0$ & $16$ & $-8P_{1}$ & (1) \\\hline %deg1=1500, min_imag=1e-5 \CR{102a}&$ +++ $& $1$ & $(2, -4),(0, 0)_{2}$ & $8$ & \CR{102c} & $0$ & $24$ & $32P_{1}$ & \\\hline \CR{106b}&$ ++ $& $1$ & $(2, 1)$ & $8$ & \CR{106a} & $0$ & $6$ & $-4P_{1}$ & \\\hline \CR{106b}&$ ++ $& $1$ & $(2, 1)$ & $8$ & \CR{106c} & $0$ & $48$ & $-88P_{1}$ & \\\hline \CR{106b}&$ ++ $& $1$ & $(2, 1)$ & $8$ & \CR{106d} & $0$ & $10$ & $12P_{1}$ & \\\hline \CR{112a}&$ ++ $& $1$ & $(0, -2),(-2, 0)_{2}$ & $8$ & \CR{112b} & $0$ & $4$ & $0$ & \\\hline \CR{112a}&$ ++ $& $1$ & $(0, -2),(-2, 0)_{2}$ & $8$ & \CR{112c} & $0$ & $8$ & $0$ & \\\hline \CR{118a}&$ ++ $& $1$ & $(0, -1)$ & $4$ & \CR{118b} & $0$ & $12$ & $-8P_{1}$ & (1) \\\hline %deg1=1500, min_imag=1e-5 \CR{118a}&$ ++ $& $1$ & $(0, -1)$ & $4$ & \CR{118c} & $0$ & $6$ & $4P_{1}$ & \\\hline \CR{118a}&$ ++ $& $1$ & $(0, -1)$ & $4$ & \CR{118d} & $0$ & $38$ & $-28P_{1}$ & \\\hline \CR{121b}&$ + $& $1$ & $(4, 5)$ & $4$ & \CR{121a} & $0$ & $6$ & $4P_{1}$ & \\\hline \CR{121b}&$ + $& $1$ & $(4, 5)$ & $4$ & \CR{121c} & $0$& $6$ & $4P_{1}$ & \\\hline \CR{121b}&$ + $& $1$ & $(4, 5)$ & $4$ & \CR{121d} & $0$ & $24$ & $-28P_{1}$ & (2) \\\hline % actually: min_imag=1e-5, deg1=3000, and yes such a large deg1 needed - - finds all 24 points right off. - - takes 10 minutes. \CR{123a}&$ - - $& $1$ & $(-4, 1),(-1, 4)_{5}$ & $20$ & \CR{123b} & $1$ & $4$ & $0$ & \\\hline \CR{123b}&$ ++ $& $1$ & $(1, 0)$ & $4$ & \CR{123a} & $1$ & $20$ & $4P_{1}$ & \\\hline \CR{124a}&$ - - $& $1$ & $(-2, 1),(0, 1)_{3}$ & $6$ & \CR{124b} & $0$ & $6$ & $0$ & \\\hline \CR{128a}&$ + $& $1$ & $(0, 1),(-1, 0)_{2}$ & $4$ & \CR{128b} & $0$ & $8$ & $0$ & \\\hline \CR{128a}&$ + $& $1$ & $(0, 1),(-1, 0)_{2}$ & $4$ & \CR{128c} & $0$ & $4$ & $0$ & \\\hline \CR{128a}&$ + $& $1$ & $(0, 1),(-1, 0)_{2}$ & $4$ & \CR{128d} & $0$ & $8$ & $0$ & \\\hline \CR{129a}&$ ++ $& $1$ & $(1, -5)$ & $8$ & \CR{129b} & $0$ & $15$ & $-8P_{1}$ & \\\hline \CR{130a}&$ +- - $& $1$ & $(-6, 10),(-1, 10)_{6}$ & $24$ & \CR{130b} & $0$ & $8$ & $0$ & \\\hline \CR{130a}&$ +- - $& $1$ & $(-6, 10),(-1, 10)_{6}$ & $24$ & \CR{130c} & $0$ & $80$ & $0$ & \\\hline % a lot of rounds \CR{135a}&$ ++ $& $1$ & $(4, -8)$ & $12$ & \CR{135b} & $0$ & $36$ & $0$ & (1) \\\hline%deg1=1500, min_imag=1e-5 \CR{136a}&$ - - $& $1$ & $(-2, 2),(0, 0)_{2}$ & $8$ & \CR{136b} & $0$ & $8$ & $0$ & \\\hline \CR{138a}&$ +++ $& $1$ & $(1, -2),(-2, 1)_{2}$ & $8$ & \CR{138b} & $0$ & $16$ & $-16P_{1}$ & (1) \\\hline %actualy -- deg1=1000 \CR{138a}&$ +++ $& $1$ & $(1, -2),(-2, 1)_{2}$ & $8$ & \CR{138c} & $0$ & $8$ & $-8P_{1}$ & \\\hline \CR{141a}&$ - - $& $1$ & $(-3, -5)$ & $28$ & \CR{141b} & $0$ & $12$ & $0$ & \\\hline \CR{141a}&$ - - $& $1$ & $(-3, -5)$ & $28$ & \CR{141c} & $0$ & $6$ & $0$ & \\\hline \CR{141a}&$ - - $& $1$ & $(-3, -5)$ & $28$ & \CR{141d} & $1$ & $4$ & $0$ & \\\hline \end{tabular} \begin{tabular}{|c|c|c|c|c||c|c|c||c|c|}\hline $E$& $\eps_p$'s& $r_E$ & $E(\Q)$ & $m_E$ & $F$& $r_F$ & $m_F$ & $P_{E,F}$ & Notes\\\hline\hline \CR{141a}&$$ - - $$& $1$ & $(-3, -5)$ & $28$ & \CR{141e} & $0$ & $12$ & $0$ & \\\hline \CR{141d}&$$ ++ $$& $1$ & $(0, -1)$ & $4$ & \CR{141a} & $1$ & $28$ & $-12P_{1}$ & \\\hline \CR{141d}&$$ ++ $$& $1$ & $(0, -1)$ & $4$ & \CR{141b} & $0$ & $12$ & $4P_{1}$ & \\\hline \CR{141d}&$$ ++ $$& $1$ & $(0, -1)$ & $4$ & \CR{141c} & $0$ & $6$ & $4P_{1}$ & \\\hline \CR{141d}&$$ ++ $$& $1$ & $(0, -1)$ & $4$ & \CR{141e} & $0$ & $12$ & $4P_{1}$ & \\\hline \CR{142a}&$$ - - $$& $1$ & $(1, 1)$ & $36$ & \CR{142b} & $1$ & $4$ & $0$ & \\\hline \CR{142a}&$$ - - $$& $1$ & $(1, 1)$ & $36$ & \CR{142c} & $0$ & $9$ & $0$ & \\\hline \CR{142a}&$ - - $& $1$ & $(1, 1)$ & $36$ & \CR{142d} & $0$ & $4$ & $0$ & \\\hline \CR{142a}&$ - - $& $1$ & $(1, 1)$ & $36$ & \CR{142e} & $0$ & $324$ & $0$ & (2) \\\hline % deg1=3000, min_imag=1e-5/2; takes 5196 seconds \CR{142b}&$ ++ $& $1$ & $(-1, 0)$ & $4$ & \CR{142a} & $1$ & $36$ & $4P_{1}$ & (1) \\\hline % actually, deg1=1000 \CR{142b}&$ ++ $& $1$ & $(-1, 0)$ & $4$ & \CR{142c} & $0$ & $9$ & $-4P_{1}$ & \\\hline \CR{142b}&$ ++ $& $1$ & $(-1, 0)$ & $4$ & \CR{142d} & $0$ & $4$ & $4P_{1}$ & \\\hline \CR{142b}&$ ++ $& $1$ & $(-1, 0)$ & $4$ & \CR{142e} & $0$ & $324$ & $8P_{1}$ & (2) \\\hline % deg1=3000, min_imag=1e-5/2; takes 5380 seconds \CR{152a}&$ ++ $& $1$ & $(-1, -2)$ & $8$ & \CR{152b} & $0$ & $8$ & $0$ & \\\hline \CR{153a}&$ ++ $& $1$ & $(0, 1)$ & $8$ & \CR{153b} & $1$ & $16$ & $8P_{1}$ & \\\hline \CR{153a}&$ ++ $& $1$ & $(0, 1)$ & $8$ & \CR{153c} & $0$ & $8$ & $8P_{1}$ & \\\hline \CR{153a}&$ ++ $& $1$ & $(0, 1)$ & $8$ & \CR{153d} & $0$ & $24$ & $0$ & \\\hline \CR{153b}&$ - - $& $1$ & $(5, -14)$ & $16$ & \CR{153a} & $1$ & $8$ & $0$ & \\\hline \CR{153b}&$ - - $& $1$ & $(5, -14)$ & $16$ & \CR{153d} & $0$ & $24$ & $0$ & \\\hline \CR{154a}&$ +++ $& $1$ & $(5, 3),(-6, 3)_{2}$ & $24$ & \CR{154b} & $0$ & $24$ & $-24P_{1}$ & \\\hline \CR{154a}&$ +++ $& $1$ & $(5, 3),(-6, 3)_{2}$ & $24$ & \CR{154c} & $0$ & $16$ & $16P_{1}$ & \\\hline \CR{155a}&$ - - $& $1$ & $(5/4, 31/8),(0, 2)_{5}$ & $20$ & \CR{155b} & $0$ & $8$ & $0$ & \\\hline \CR{155a}&$ - - $& $1$ & $(5/4, 31/8),(0, 2)_{5}$ & $20$ & \CR{155c} & $1$ & $4$ & $0$ & \\\hline \CR{155c}&$ ++ $& $1$ & $(1, -1)$ & $4$ & \CR{155a} & $1$ & $20$ & $-12P_{1}$ & \\\hline \CR{155c}&$ ++ $& $1$ & $(1, -1)$ & $4$ & \CR{155b} & $0$ & $8$ & $4P_{1}$ & \\\hline \CR{156a}&$ -+- $& $1$ & $(1, 1),(2, 0)_{2}$ & $12$ & \CR{156b} & $0$ & $12$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{158a}&$ - - $& $1$ & $(-1, -4)$ & $32$ & \CR{158b} & $1$ & $8$ & $0$ & \\\hline \CR{158a}&$ - - $& $1$ & $(-1, -4)$ & $32$ & \CR{158c} & $0$ & $48$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{158a}&$ - - $& $1$ & $(-1, -4)$ & $32$ & \CR{158d} & $0$ & $40$ & $0$ & \\\hline \CR{158a}&$ - - $& $1$ & $(-1, -4)$ & $32$ & \CR{158e} & $0$ & $6$ & $0$ & \\\hline \CR{158b}&$ ++ $& $1$ & $(0, -1)$ & $8$ & \CR{158a} & $1$ & $32$ & $-8P_{1}$ & \\\hline \CR{158b}&$ ++ $& $1$ & $(0, -1)$ & $8$ & \CR{158c} & $0$ & $48$ & $-56P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{158b}&$ ++ $& $1$ & $(0, -1)$ & $8$ & \CR{158d} & $0$ & $40$ & $0$ & \\\hline \CR{158b}&$ ++ $& $1$ & $(0, -1)$ & $8$ & \CR{158e} & $0$ & $6$ & $-8P_{1}$ & \\\hline \CR{160a}&$ ++ $& $1$ & $(2, -2),(1, 0)_{2}$ & $8$ & \CR{160b} & $0$ & $8$ & $0$ & \\\hline \CR{162a}&$ ++ $& $1$ & $(-2, 4),(1, 1)_{3}$ & $12$ & \CR{162b} & $0$ & $6$ & $0$ & \\\hline \CR{162a}&$ ++ $& $1$ & $(-2, 4),(1, 1)_{3}$ & $12$ & \CR{162c} & $0$ & $6$ & $0$ & \\\hline \CR{162a}&$ ++ $& $1$ & $(-2, 4),(1, 1)_{3}$ & $12$ & \CR{162d} & $0$ & $12$ & $0$ & \\\hline \CR{170a}&$ +- - $& $1$ & $(0, 2),(1, -1)_{2}$ & $16$ & \CR{170d} & $0$ & $12$ & $0$ & \\\hline \CR{170a}&$ +- - $& $1$ & $(0, 2),(1, -1)_{2}$ & $16$ & \CR{170e} & $0$ & $20$ & $0$ & \\\hline \CR{171b}&$ - - $& $1$ & $(2, -5)$ & $8$ & \CR{171a} & $0$ & $12$ & $0$ & \\\hline \CR{171b}&$ - - $& $1$ & $(2, -5)$ & $8$ & \CR{171c} & $0$ & $96$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{171b}&$ - - $& $1$ & $(2, -5)$ & $8$ & \CR{171d} & $0$ & $32$ & $0$ & \\\hline \CR{175a}&$ - - $& $1$ & $(2, -3)$ & $8$ & \CR{175b} & $1$ & $16$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{175a}&$ - - $& $1$ & $(2, -3)$ & $8$ & \CR{175c} & $0$ & $40$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{175b}&$ ++ $& $1$ & $(-3, 12)$ & $16$ & \CR{175a} & $1$ & $8$ & $16P_{1}$ & \\\hline \CR{175b}&$ ++ $& $1$ & $(-3, 12)$ & $16$ & \CR{175c} & $0$ & $40$ & $16P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{176c}&$ - - $& $1$ & $(1, -2)$ & $8$ & \CR{176b} & $0$ & $8$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \end{tabular} \begin{tabular}{|c|c|c|c|c||c|c|c||c|c|}\hline $E$& $\eps_p$'s & $r_E$ & $E(\Q)$ & $m_E$ & $F$& $r_F$ & $m_F$ & $P_{E,F}$ & Notes\\\hline\hline \CR{176c}&$ - - $& $1$ & $(1, -2)$ & $8$ & \CR{176a} & $0$ & $16$ & $0$ & \\\hline \CR{176c}&$ - - $& $1$ & $(1, -2)$ & $8$ & \CR{176b} & $0$ & $8$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{184a}&$ - - $& $1$ & $(0, 1)$ & $8$ & \CR{184c} & $0$ & $12$ & $0$ & \\\hline \CR{184a}&$ - - $& $1$ & $(0, 1)$ & $8$ & \CR{184d} & $0$ & $24$ & $0$ & \\\hline \CR{184b}&$ ++ $& $1$ & $(2, -1)$ & $8$ & \CR{184a} & $1$ & $8$ & $0$ & \\\hline \CR{184b}&$ ++ $& $1$ & $(2, -1)$ & $8$ & \CR{184c} & $0$ & $12$ & $0$ & \\\hline \CR{184b}&$ ++ $& $1$ & $(2, -1)$ & $8$ & \CR{184d} & $0$ & $24$ & $0$ & \\\hline \CR{185a}&$ ++ $& $1$ & $(4, -13)$ & $48$ & \CR{185b} & $1$ & $8$ & $8P_{1}$ & \\\hline \CR{185a}&$ ++ $& $1$ & $(4, -13)$ & $48$ & \CR{185c} & $1$ & $6$ & $24P_{1}$ & \\\hline $\cdots$ &$ $& & & & & & & $\cdots$ \\\hline \CR{185b}&$ - - $& $1$ & $(0, 2)$ & $8$ & \CR{185c} & $1$ & $6$ & $0$ & \\\hline \CR{185c}&$ ++ $& $1$ & $(-5/4, 3/8),(-1, 0)_{2}$ & $6$ & \CR{185b} & $1$ & $8$ & $2P_{1}$ & \\\hline \CR{189a}&$ ++ $& $1$ & $(-1, -2)$ & $12$ & \CR{189b} & $1$ & $12$ & $-12P_{1}$ & \\\hline \CR{189a}&$ ++ $& $1$ & $(-1, -2)$ & $12$ & \CR{189c} & $0$ & $12$ & $12P_{1}$ & \\\hline \CR{189b}&$ - - $& $1$ & $(-3, 9),(3, 0)_{3}$ & $12$ & \CR{189a} & $1$ & $12$ & $0$ & \\\hline \CR{189b}&$ - - $& $1$ & $(-3, 9),(3, 0)_{3}$ & $12$ & \CR{189c} & $0$ & $12$ & $0$ & \\\hline \CR{190a}&$ -+- $& $1$ & $(13, -47)$ & $88$ & \CR{190b} & $1$ & $8$ & $0$ & \\\hline \CR{190a}&$ -+- $& $1$ & $(13, -47)$ & $88$ & \CR{190c} & $0$ & $24$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{190b}&$ +++ $& $1$ & $(1, 2)$ & $8$ & \CR{190c} & $0$ & $24$ & $16P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{192a}&$ ++ $& $1$ & $(3, 2),(-1, 0)_{2}$ & $8$ & \CR{192b} & $0$ & $8$ & $0$ & \\\hline \CR{192a}&$ ++ $& $1$ & $(3, 2),(-1, 0)_{2}$ & $8$ & \CR{192c} & $0$ & $8$ & $0$ & \\\hline \CR{192a}&$ ++ $& $1$ & $(3, 2),(-1, 0)_{2}$ & $8$ & \CR{192d} & $0$ & $8$ & $0$ & \\\hline \CR{196a}&$ - - $& $1$ & $(0, -1)$ & $6$ & \CR{196b} & $0$ & $42$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{198a}&$ +- - $& $1$ & $(-1, -4),(-4, 2)_{2}$ & $32$ & \CR{198b} & $0$ & $32$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{198a}&$ +- - $& $1$ & $(-1, -4),(-4, 2)_{2}$ & $32$ & \CR{198c} & $0$ & $32$ & $0$ & \\\hline \CR{198a}&$ +- - $& $1$ & $(-1, -4),(-4, 2)_{2}$ & $32$ & \CR{198d} & $0$ & $32$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{198a}&$ +- - $& $1$ & $(-1, -4),(-4, 2)_{2}$ & $32$ & \CR{198e} & $0$ & $160$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{200b}&$ - - $& $1$ & $(-1, 1),(-2, 0)_{2}$ & $8$ & \CR{200c} & $0$ & $24$ & $0$ & \\\hline \CR{200b}&$ - - $& $1$ & $(-1, 1),(-2, 0)_{2}$ & $8$ & \CR{200d} & $0$ & $40$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{200b}&$ - - $& $1$ & $(-1, 1),(-2, 0)_{2}$ & $8$ & \CR{200e} & $0$ & $24$ & $0$ & \\\hline \CR{201a}&$ ++ $& $1$ & $(1, -2)$ & $12$ & \CR{201b} & $1$ & $12$ & $4P_{1}$ & \\\hline \CR{201b}&$ - - $& $1$ & $(-1, 2)$ & $12$ & \CR{201a} & $1$ & $12$ & $0$ & \\\hline \CR{201c}&$ ++ $& $1$ & $(16, -7)$ & $60$ & \CR{201a} & $1$ & $12$ & $-24P_{1}$ & \\\hline \CR{201c}&$ ++ $& $1$ & $(16, -7)$ & $60$ & \CR{201b} & $1$ & $12$ & $8P_{1}$ & \\\hline \CR{203b}&$ - - $& $1$ & $(2, -5)$ & $8$ & \CR{203a} & $0$ & $48$ & $0$ & \\\hline \CR{203b}&$ - - $& $1$ & $(2, -5)$ & $8$ & \CR{203c} & $0$ & $12$ & $0$ & \\\hline \CR{205a}&$ - - $& $1$ & $(-1, 8),(2, 1)_{4}$ & $12$ & \CR{205b} & $0$ & $16$ & $0$ & \\\hline \CR{205a}&$ - - $& $1$ & $(-1, 8),(2, 1)_{4}$ & $12$ & \CR{205c} & $0$ & $8$ & $0$ & \\\hline \CR{208a}&$ - - $& $1$ & $(4, -8)$ & $16$ & \CR{208c} & $0$ & $12$ & $0$ & \\\hline \CR{208a}&$ - - $& $1$ & $(4, -8)$ & $16$ & \CR{208d} & $0$ & $48$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{208b}&$ ++ $& $1$ & $(4, 4)$ & $16$ & \CR{208a} & $1$ & $16$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{208b}&$ ++ $& $1$ & $(4, 4)$ & $16$ & \CR{208c} & $0$ & $12$ & $0$ & \\\hline \CR{208b}&$ ++ $& $1$ & $(4, 4)$ & $16$ & \CR{208d} & $0$ & $48$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{212a}&$ - - $& $1$ & $(2, 2)$ & $12$ & \CR{212b} & $0$ & $21$ & $0$ & \\\hline \CR{214a}&$ - - $& $1$ & $(0, -4)$ & $28$ & \CR{214b} & $1$ & $12$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{214a}&$ - - $& $1$ & $(0, -4)$ & $28$ & \CR{214d} & $0$ & $12$ & $0$ & \\\hline \CR{214b}&$ ++ $& $1$ & $(0, 0)$ & $12$ & \CR{214a} & $1$ & $28$ & $-8P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{214b}&$ ++ $& $1$ & $(0, 0)$ & $12$ & \CR{214d} & $0$ & $12$ & $-4P_{1}$ & \\\hline \end{tabular} \begin{tabular}{|c|c|c|c|c||c|c|c||c|c|}\hline $E$& $\eps_p$'s & $r_E$ & $E(\Q)$ & $m_E$ & $F$& $r_F$ & $m_F$ & $P_{E,F}$ & Notes\\\hline\hline \CR{214c}&$ ++ $& $1$ & $(11, 10)$ & $60$ & \CR{214a} & $1$ & $28$ & $-4P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{214c}&$ ++ $& $1$ & $(11, 10)$ & $60$ & \CR{214d} & $0$ & $12$ & $16P_{1}$ & \\\hline \CR{214c}&$ ++ $& $1$ & $(11, 10)$ & $60$ & \CR{214b} & $1$ & $12$ & $12P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{216a}&$ ++ $& $1$ & $(-2, -6)$ & $24$ & \CR{216b} & $0$ & $24$ & $0$ & \\\hline \CR{219a}&$ ++ $& $1$ & $(2, -1)$ & $12$ & \CR{219c} & $1$ & $60$ & $-12P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{219a}&$ ++ $& $1$ & $(2, -1)$ & $12$ & \CR{219b} & $1$ & $12$ & $-4P_{1}$ & \\\hline \CR{216a}&$ ++ $& $1$ & $(-2, -6)$ & $24$ & \CR{216d} & $0$ & $72$ & $0$ & \\\hline \CR{219b}&$ - - $& $1$ & $(-3/4, -1/8),(0, 1)_{3}$ & $12$ & \CR{219a} & $1$ & $12$ & $0$ & \\\hline \CR{219b}&$ - - $& $1$ & $(-3/4, -1/8),(0, 1)_{3}$ & $12$ & \CR{219c} & $1$ & $60$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{219c}&$ ++ $& $1$ & $(-6, 7),(10, -5)_{2}$ & $60$ & \CR{219a} & $1$ & $12$ & $-12P_{1}$ & \\\hline \CR{219c}&$ ++ $& $1$ & $(-6, 7),(10, -5)_{2}$ & $60$ & \CR{219b} & $1$ & $12$ & $4P_{1}$ & \\\hline \CR{220a}&$ - -+ $& $1$ & $(-7, 11),(15, 55)_{6}$ & $36$ & \CR{220b} & $0$ & $12$ & $0$ & \\\hline \CR{224a}&$ ++ $& $1$ & $(1, 2),(0, 0)_{2}$ & $8$ & \CR{224b} & $0$ & $8$ & $0$ & \\\hline \CR{225a}&$ ++ $& $1$ & $(1, 1)$ & $8$ & \CR{225b} & $0$ & $40$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{225e}&$ - - $& $1$ & $(-5, 22)$ & $48$ & \CR{225a} & $1$ & $8$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{225e}&$ - - $& $1$ & $(-5, 22)$ & $48$ & \CR{225b} & $0$ & $40$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{228b}&$ -+- $& $1$ & $(3, 6)$ & $24$ & \CR{228a} & $0$ & $18$ & $0$ & \\\hline \CR{232a}&$ ++ $& $1$ & $(2, -4)$ & $16$ & \CR{232b} & $0$ & $16$ & $0$ & \\\hline \CR{234c}&$ +++ $& $1$ & $(1, -2),(-2, 1)_{2}$ & $16$ & \CR{234b} & $0$ & $48$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{234c}&$ +++ $& $1$ & $(1, -2),(-2, 1)_{2}$ & $16$ & \CR{234e} & $0$ & $20$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{235a}&$ - - $& $1$ & $(-2, 3)$ & $12$ & \CR{235c} & $0$ & $18$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{236a}&$ - - $& $1$ & $(1, -1)$ & $6$ & \CR{236b} & $0$ & $14$ & $0$ & \\\hline \CR{238a}&$ - -+ $& $1$ & $(24, 100),(-8, 4)_{2}$ & $112$ & \CR{238b} & $1$ & $8$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{238a}&$ - -+ $& $1$ & $(24, 100),(-8, 4)_{2}$ & $112$ & \CR{238c} & $0$ & $16$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{238a}&$ - -+ $& $1$ & $(24, 100),(-8, 4)_{2}$ & $112$ & \CR{238d} & $0$ & $16$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{238b}&$ +++ $& $1$ & $(1, 1),(0, 0)_{2}$ & $8$ & \CR{238a} & $1$ & $112$ & $12P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{238b}&$ +++ $& $1$ & $(1, 1),(0, 0)_{2}$ & $8$ & \CR{238c} & $0$ & $16$ & $-4P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{238b}&$ +++ $& $1$ & $(1, 1),(0, 0)_{2}$ & $8$ & \CR{238d} & $0$ & $16$ & $4P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{240c}&$ +++ $& $1$ & $(1, 2),(0, 0)_{2}$ & $16$ & \CR{240a} & $0$ & $16$ & $0$ & \\\hline \CR{240c}&$ +++ $& $1$ & $(1, 2),(0, 0)_{2}$ & $16$ & \CR{240d} & $0$ & $16$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{243a}&$ + $& $1$ & $(1, 0)$ & $6$ & \CR{243b} & $0$ & $9$ & $0$ & (1) \\\hline% deg1=1500, min_imag=1e-5 \CR{245a}&$ - - $& $1$ & $(7, 17)$ & $48$ & \CR{245c} & $1$ & $32$ & $0$ & \\\hline \CR{246d}&$ +++ $& $1$ & $(3, -6),(4, -2)_{2}$ & $48$ & \CR{246a} & $0$ & $84$ & $24P_{1}$ & (1) \\\hline% deg1=1500, min_imag=1e-5 %$\cdots$ &$ $& & & & & & & $\cdots$ \\\hline \CR{446a}&$ ++ $& 1 & $(4,-6)$ & $24$ & \CR{446d}& 2 & $88$ & 0 & (2) \\\hline \CR{446b}&$ - - $& 1& $(5,-10)$ & $56$ & \CR{446d} & 2 & $88$ & 0 & (2) \\\hline \CR{681a} & $++$ & $1$ & $(4, 4)$ & $32$ & \CR{681c} & $2$ & $96$ & $-24P_{1}$ & (2) \\\hline \CR{446d}&$+-$ & 2 & - & $88$ & \CR{446a} & 1 & 12 & 0 & (1) \\\hline \CR{446d}&$+-$ & 2 & - & $88$ & \CR{446b} & 1 & 56 & 0 & (1) \\\hline \end{tabular} \end{center} \vspace{2em} \noindent{}{\bf Notes:} \begin{itemize}\setlength{\itemsep}{-.5ex} \item[{\bf (1)}] We used $y=10^{-5}$, $d=1500$, which typically takes about 4 minutes. \item[{\bf (2)}] We used $y=10^{-5}/2$, $d=3000$, which takes up to 2 hours. \end{itemize} \bibliography{biblio} \end{document}