Modular curves of genus zero and normal forms for elliptic curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:16:14Z http://mathoverflow.net/feeds/question/17031 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves Modular curves of genus zero and normal forms for elliptic curves Qiaochu Yuan 2010-03-04T00:17:19Z 2011-03-07T14:35:00Z <p>This is maybe the first question I actually need to know the answer to!</p> <p>Let $N$ be a positive integer such that $\mathbb{H}/\Gamma(N)$ has genus zero. Then the function field of $\mathbb{H}/\Gamma(N)$ is generated by a single function. When $N = 2$, the cross-ratio $\lambda$ is such a function. A point of $\mathbb{H}/\Gamma(2 )$ at which $\lambda = \lambda_0$ is precisely an elliptic curve in Legendre normal form</p> <p>$$y^2 = x(x - 1)(x - \lambda_0)$$</p> <p>where the points $(0, 0), (1, 0)$ constitute a choice of basis for the $2$-torsion. When $N = 3$, there is a modular function $\gamma$ such that a point of $\mathbb{H}/\Gamma(3)$ at which $\gamma = \gamma_0$ is precisely an elliptic curve in Hesse normal form</p> <p>$$x^3 + y^3 + 1 + \gamma_0 xy = 0$$</p> <p>where (I think) the points $(\omega, 0), (\omega^3, 0), (\omega^5, 0)$ (where $\omega$ is a primitive sixth root of unity) constitute a choice of basis for the $3$-torsion. </p> <p><strong>Question:</strong> Does this picture generalize? That is, for every $N$ above does there exist a normal form for elliptic curves which can be written in terms of a generator of the function field of $\mathbb{H}/\Gamma(N)$ and which "automatically" equips the $N$-torsion points with a basis? (I don't even know if this is possible when $N = 1$, where the Hauptmodul is the $j$-invariant.) If not, what's special about the cases where it is possible?</p> http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves/17056#17056 Answer by BCnrd for Modular curves of genus zero and normal forms for elliptic curves BCnrd 2010-03-04T04:56:24Z 2010-03-04T04:56:24Z <p>As I mentioned in connection with an answer to another question, it is not generally true for elliptic curves $f:E \rightarrow S$ over a base $S$ that there is a global embedding of $E$ into $\mathbf{P}^2_ S$. For example, if $S = {\rm{Spec}} ( A )$ for a Dedekind domain $A$ whose class group is nontrivial, it could fail (and sometimes does fail). The necessary and sufficient condition is that the line bundle $\omega_{E/S} = f_{\ast}(\Omega^1_{E/S})$ on $S$ is trivial.</p> <p>Example: If $S$ is the complement of a non-empty finite set of rational points in the projective line over a field $k$ then it is the spectrum of a localization of $k[x]$ and hence has trivial Picard group. Thus, the obstructions vanish and a global embedding exists. This applies to the modular curve $Y(N)$ over $\mathbf{Q}$ (geometrically connected over $\mathbf{Q}(\zeta_ N)$ via the Weil $N$-torsion pairing) when $N = 3, 4, 5$. Of course, to then really find the normal form in explicit terms requires real work and not just this kind of "brain work". </p> <p>In case the elliptic curve is the universal one over some (fine) moduli scheme $S$ and this line bundle obstruction vanishes, such as if we know the stronger fact that ${\rm{Pic}}(S) = 1$, then such a global embedding must exist and its determination can then be regarded as a "normal form". </p> <p>On the other hand, consider universal elliptic curves $E \rightarrow Y$ over fine modular curves $Y$ whose "level structure" doesn't dominate one of the fine ones of genus 0, such as $Y(p)$ with a prime $p > 5$. To figure out if there is a Weierstrass form for $E$ over the entire affine base $Y$ (i.e., the projective plane doesn't need to be replaced with a projective space bundle, as is needed for general families of elliptic curves) one has to determine precisely if $\omega_{E/Y}$ is trivial. This amounts to the existence of modular functions which "transform" under the corresponding "congruence subgroup" (such as $\Gamma(p)$) like a weight-1 form and have no zeros or poles on the upper half-plane (if working over $\mathbf{C}$), and so can be analyzed concretely by thinking about Klein form in the case of full-level problems. </p> http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves/17066#17066 Answer by David Lehavi for Modular curves of genus zero and normal forms for elliptic curves David Lehavi 2010-03-04T09:03:08Z 2010-03-07T16:43:45Z <p>The first thing you'd need in order to define a normal form is unirationality of the moduli space (otherwise you don't even have the correct number of parameters). In dimension 1, this means that you (at least) need the modular curve to be of genus 0, at which point we may look at the The <a href="http://www.research.att.com/~njas/sequences/A001617" rel="nofollow">On-Line Encyclopedia of Integer Sequences</a></p> <p>Here is how you can do n-torsion assuming you know the m-torsion solution and m divides n (and of course, the moduli space is genus 0):</p> <p>Let z be the moduli space parameter, and E(z) be the universal plane curve. Let $a_i(z),b_i(z)$ be n-torsion points on E(z) which span the set of n-torsion points; let $l_i(z)$ in the dual projective plane be the line connecting $a_i(z),b_i(z)$. Then the locus of $l_i(z)$ is a plane curve, which is -- by our assumption -- rational. Now use your favorite "Italian" method to find an explicit rationalization of a rational plane curve. The coordinates of the universal projective plane are determined by the four points $0, a_i(z), b_i(z), a_i(z)+b_i(z)$.</p> <p>Note that from the original list of 2..10,12,13,16,18,25 you are now left with the task of finding solutions to 5,7,13.</p> http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves/17104#17104 Answer by Álvaro Lozano-Robledo for Modular curves of genus zero and normal forms for elliptic curves Álvaro Lozano-Robledo 2010-03-04T16:34:22Z 2010-03-04T16:34:22Z <p>Hello,</p> <p>I believe the following results that appear in papers of Rubin and Silverberg can be very useful here. Let $N=3,4,$ or $5$ and let $Y_N$ be the (non-compact) modular curve over $\mathbb{Q}$ which parametrizes $(E,P,C)$ where $E$ is an elliptic curve, $P$ is a point of order $N$ on $E$ and $C$ is a cyclic subgroup of order $N$ on $E$, and $C$ and $P$ generate $E[N]$. The curve $Y_N$ is isomorphic to one connected component of $Y(N)$, and $Y_N(\mathbb{C})$ is isomorphic to $\mathbb{H}/\Gamma(N)$. Let $X_N$ be the compactification of $Y_N$.</p> <p>Rubin and Silverberg describe explicit isomorphisms $f_N:X_N \cong \mathbb{P}^1$, with $f(u) = (A_u,P_u,C_u)$ and give equations for $A_u$, here:</p> <p>1) [Rubin and Silverberg] for $N=3$ and $5$ in <a href="http://www.math.ohio-state.edu/mri/preprints/1994/94_11.html" rel="nofollow">Families of elliptic curves with constant mod p representations</a></p> <p>and</p> <p>2) [Silverberg] for $N=4$ in <em>Explicit families of elliptic curves with prescribed mod $N$ representations</em>'', in Modular forms and Fermat's last theorem, Cornell, Silverman, Stevens (Editors), Springer, p. 447 - 461.</p> <p>I hope that helps,</p> <p>Alvaro</p> http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves/57673#57673 Answer by Soroosh for Modular curves of genus zero and normal forms for elliptic curves Soroosh 2011-03-07T14:35:00Z 2011-03-07T14:35:00Z <p>I think the answer to your question is the content of Velu's thesis: Courbes elliptiques munies d'un sous-groupe $Z/NZ\times \mu_N$. In there, he explicitly writes down the universal elliptic curve over $X(p)$ for $p>3$. </p>