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I believe there are only finitely many congruence subgroups $\Gamma\le SL_2(\mathbb{Z})$ such that the compactification of $\mathcal{H}/\Gamma$ is genus 1.

Is there somewhere I can find a list of these genus 1 modular curves and look at their $j$-invariants (and ideally also reduction types over their field of definition). Here, I'm referring to the moduli-theoretic model of the elliptic curve, which is necessarily a quotient of some $X(n)$ (which is defined over $\mathbb{Q}(\zeta_n)$) and hence the $j$-invariant will lie in $\mathbb{Q}(\zeta_n)$.

In particular I'm interested in whether or not you can say anything about the integrality of the $j$-invariants of these genus 1 modular curves, especially in the case when the congruence subgroup is torsion-free.

For example, are the $j$-invariants all integral? Do they all have additive/multiplicative reduction? Do they have CM?

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  • $\begingroup$ For $\Gamma=\Gamma_0(N)$ you can read everything off from (the first few lines of) Cremona's tables. There will be other possibilities for $\Gamma$ though, I guess. $\endgroup$
    – znt
    Commented Oct 12, 2016 at 20:54
  • $\begingroup$ Note that specifying $\Gamma \leq SL_{2}(\mathbb{Z})$ gives a curve $\mathcal{H}/\Gamma$ up to isomorphism over $\mathbb{C}$. This does not specify a $\mathbb{Q}$-structure on the curve. To do so, you need to choose an subgroup $\Gamma \leq GL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that $\det : \Gamma \to (\mathbb{Z}/n\mathbb{Z})^{\times}$ is surjective. For this reason, it is not immediately obvious to me that all of the $j$-invariants are rational, as (I'm pretty sure) that not all of the $\Gamma \leq SL_{2}(\mathbb{Z})$ have models over $\mathbb{Q}$. $\endgroup$
    – Jeremy Rouse
    Commented Oct 12, 2016 at 23:26
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    $\begingroup$ Not all integral ($X_0(27)$ is the Fermat cubic, with $j=0$), nor all non-integral (if $p\|N$ and $X_0(N)$ has genus $1$ then it has multiplicative reduction mod $p$ so there is a factor of $p$ in the denominator; example: $N=p=11$). $\endgroup$
    – Noam D. Elkies
    Commented Oct 13, 2016 at 1:44
  • $\begingroup$ @JeremyRouse That's a good point. Here I mean the moduli-theoretic model. I've edited the question accordingly. $\endgroup$
    – stupid_question_bot
    Commented Oct 13, 2016 at 1:46
  • $\begingroup$ @Jeremy Rouse: you're quite right of course, but let me flag the following spanner which sometimes people throw in to add to the confusion. If $\Gamma$ is $\Gamma(N)$ ($N>2$) then using a full level $N$ structure in the $GL(2)$ theory of canonical models of course spits out a connected but geometrically disconnected curve, defined over $Q$ but whose geom irred components are defined over $Q(\zeta_N)$. However some people use $Y(N)$ to mean the moduli space of ell curve + symplectic full level $N$ structure, which is geom connected and does give a model for the quotient over $Q$. Urk. $\endgroup$
    – znt
    Commented Oct 13, 2016 at 6:51

1 Answer 1

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First, there is no comprehensive list of all models of modular curves of genus $1$. (There is a list of congruence subgroups of $SL_{2}(\mathbb{Z})$ here.) Many cases have been computed, including the $X_{0}(n)$, $X_{1}(n)$, $X(n)$, as well as many cases that have prime power level. (With David Zureick-Brown, I did all of the cases whose level is a power of $2$ that might have rational points. See our paper here). Another really handy tool is the code that David Zywina has written (see the paper here) that will count points on any modular curve $X_{G}$ (provided $G \subseteq GL_{2}(\mathbb{Z}/n\mathbb{Z})$ has surjective determinant). This will allow one to determine the isogeny class of ${\rm Jac}~X_{G}$, but this is not quite enough to determine the $j$-invariant.

Based on the examples that are known, it is not clear to me there is a relationship between the integrality of the $j$-invariant and the torsion structure. Here are some examples:

$X_{0}(11): y^{2} + y = x^{3} - x^{2} - 10x - 20$, $j$-invariant is $-122023936/161051$ is not integral, but the subgroup $\Gamma_{0}(11)$ is torsion-free (well, its image in ${\rm PSL}_{2}(\mathbb{Z})$ is).

$X_{0}(17): y^{2} + xy + y = x^{3} - x^{2} - x - 14$, $j$-invariant is $-35937/83521$ is not integral, and the subgroup $\Gamma_{0}(17)$ is not torsion-free (because it has elliptic elements of order $2$).

$X_{0}(27): y^{2} + y = x^{3} - 7$, $j$-invariant is zero (and the curve is isomorphic to the Fermat cubic). The subgroup is torsion-free.

$X_{0}(49): y^{2} + xy = x^{3} - x^{2} - 2x - 1$, $j$-invariant is $-3375$ (and has CM by the ring of integers in $\mathbb{Q}(\sqrt{-7})$). The subgroup is not torsion-free, as it contains elliptic elements of order $3$.

Many of these curves have CM. ($X_{0}(32)$, $X_{0}(36)$, $X_{0}(49)$, $X_{{\rm ns}}^{+}(11)$, $X(6)$, most, but not all of the $2$-power level genus $1$ curves.)

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  • $\begingroup$ In your paper with DZB that you linked, in the first table, would you happen to know if the curves with non-integral $j$-invariant corresponded to torsion-free subgroups of $SL_2(\mathbb{Z})$? $\endgroup$
    – stupid_question_bot
    Commented Oct 13, 2016 at 2:02
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    $\begingroup$ Didn't know of Pauli's list; thanks. Might make a good project for a student or a team of students to get explicit models and parametrizations of all these curves. $\endgroup$
    – Noam D. Elkies
    Commented Oct 13, 2016 at 3:37
  • $\begingroup$ In reference to the paper with DZB, all the modular curves of genus $1$ that we compute models for have good reduction away from $2$, and all elliptic curves with good reduction away from $2$ have integral $j$-invariants. $\endgroup$
    – Jeremy Rouse
    Commented Oct 13, 2016 at 16:56

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