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This is maybe the first question I actually need to know the answer to!

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

$$y^2 = x(x - 1)(x - \lambda_0)$$

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

$$x^3 + y^3 + 1 + \gamma_0 xy = 0$$

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.

Question: 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?

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This is a great question. I feel like I should know the answer off the top of my head, but apparently I don't. What I can contribute at the moment is the full list of N such that X(N) has genus 0: 1,2,3,4,5. So you are missing at most normal forms for N = 4 and N = 5. To try to work this out for myself, I would start by taking the curve in Kubert-Tate normal form and see what additional relations come from having full N-torsion. (Note that you will necessarily have to extend the ground field to Q(\zeta_N) to see full N-torsion.) –  Pete L. Clark Mar 4 '10 at 0:35
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By the way, in more sophisticated terms, since X(N) is a fine moduli space over Q(\zeta_N) for N > 2, what you are looking for is an equation for the universal elliptic curve over this rational (genus zero) curve. So such normal forms definitely do exist. Looking back at an old paper of mine, I found the one for N = 3 together with the remark that the one for N = 4 is "well known". Too bad I forgot to write it down! –  Pete L. Clark Mar 4 '10 at 0:38
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When N = 1, as you say the j-invariant is the Hauptmodul. So you just want to write down an elliptic curve with j-invariant some given quantity j. There is a standard recipe for this; see e.g. Silverman's book. However, when N = 1 and also when N = 2 the moduli space is not fine, so the family is not "universal" in the strict sense of moduli spaces, and also there will be multiple nonisomorphic elliptic curves over a non-algebraically closed ground field. For instance, as I believe came up here recently, full 2-torsion is not quite enough to put an elliptic curve in Legendre normal form. –  Pete L. Clark Mar 4 '10 at 1:22
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There is a paper of Daniel Kubert's from early 70s in which he writes out many normal forms (using 2 parameters) for elliptic curves with torsion points of various small orders. These could help you. Also, you can't get around the problem of not having a single universal family parameterized by j ; the curves with j=0 and 1728 have extra automorphisms, and there is no getting around it. –  Emerton Mar 4 '10 at 3:03
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Although I'm quite late, I want to remark that the N=4-case is treated in a paper of Shioda: projecteuclid.org/… –  Lennart Meier Aug 22 '12 at 8:42

4 Answers 4

up vote 6 down vote accepted

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 $p>3$.

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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.

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".

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".

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.

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Thanks for the response! I don't exactly have much background here, but from what I can tell you're confirming and elaborating on what Pete mentioned in the comments. I guess I should mention that I would be perfectly happy with an embedding into a higher-dimensional projective space as long as its defining equations could be written explicitly in terms of a Hauptmodul. Does one exist for N = 1, 2? –  Qiaochu Yuan Mar 4 '10 at 5:18
    
I'm only saying there is a way to define "normal form", a conceptual reason for its existence in some cases, and obstruction in general. (There are ways to work with elliptic curves other than Weierstrass equations. Useful!) The "Legendre elliptic curve" over Q(lambda) has no Legendre form over Q(lambda) relative to other ordered 2-torsion bases. Why? If you understand that, you'll better understand why N = 2 is subtle (N=1 is more so). Hint: what extra structure on an elliptic curve with 2-torsion basis corresponds to a compatible "Lengendre structure"? x = a/t^2 + ...., look at a. –  BCnrd Mar 4 '10 at 6:17
    
"For example, if S=Spec(A) for a Dedekind domain A whose class group is nontrivial, it could fail (and sometimes does fail)." When A is the ring of integers in a number field K, a result of Silverman says that every ideal class c occurs as the obstruction for some quadratic twist of any fixed elliptic curve E|K. Cf. MR0804199 (86k:11030) Silverman, Joseph H., Weierstrass equations and the minimal discriminant of an elliptic curve. Mathematika 31 (1984), no. 2, 245--251 (1985). –  Chandan Singh Dalawat Mar 4 '10 at 10:35

Hello,

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 $Y_N$.

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:

1) [Rubin and Silverberg] for $N=3$ and $5$ in Families of elliptic curves with constant mod p representations

and

2) [Silverberg] for $N=4$ in ``Explicit families of elliptic curves with prescribed mod $N$ representations'', in Modular forms and Fermat's last theorem, Cornell, Silverman, Stevens (Editors), Springer, p. 447 - 461.

I hope that helps,

Alvaro

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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 On-Line Encyclopedia of Integer Sequences

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):

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 $0, a_i(z), b_i(z), a_i(z)+b_i(z)$.

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.

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Qiaochu specified that he was looking at the genus zero cases. (Unless that was edited in after you gave this answer?) –  David Speyer Mar 4 '10 at 16:40
    
@David: sure, and the first step in solving the genus 0 cases is knowing which ones they are –  David Lehavi Mar 5 '10 at 6:02
    
I think the confusion is that Qiaochu asked about $X(N)$, and David Lehavi seems to be thinking about $X_0(N)$. –  Jamie Weigandt Oct 8 '10 at 1:10

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