It is well-known that the compactification $\overline M_{1,1}$ of the moduli space of elliptic curves over $\mathbb C$ is a weighted projective line with weights $4$ and $6$. As far as I can tell, this is more or less directly linked to the fact that the ring of holomorphic modular forms for $SL_2(\mathbb Z)$ is a polynomial ring with generators $E_4$ and $E_6$. As a consequence, $\overline M_{1,1}$ is a smooth toric Deligne-Mumford stack.

It has been shown in Eichler-Zagier's book


that the ring of weak Jacobi forms of even weight is a polynomial ring with a double grading by weight and index (see Theorem 9.3 on page 108). The bigradings of the variables are $(4,0),(6,0),(-2,1),(0,1)$. It is thus tempting to consider a GIT quotient of $\mathbb C^4$ with respect to the corresponding action of $\mathbb (C^*)^2$ by looking at the Gale dual of the set of weights. This would be a smooth toric DM stack of dimension two and Picard number 2. Dimension two makes a choice of triangulation automatic.

Does one get the universal elliptic curve $\overline M_{1,2}$ this way? If so, is there a reference for this? My guess is that it is a little bit off, but I am not positive. (Note: I believe the fibers of the projection map to $\overline M_{1,1}$ are elliptic curves modulo Kummer involution, which is why they are rational).

  • 1
    $\begingroup$ Maybe, the second answer to question:moduli-of-pointed-curves could help you. Moreover, your question about $\bar{M}_{1,2}$ is also addressed in the paper "Massarenti. The Automorphisms group of $\bar{M}_{g,n}$" arXiv:1110.1464. $\endgroup$ – Francesco Dec 8 '13 at 11:09

The fibers aren't rational curves when viewed as stacks, because they have four points with extra automorphisms. I think this will be problematic.

The universal family of elliptic curves is the quotient of the scheme with projective variablesz $x,y,z$, affine variables $g_2,g_3$, and equation $y^2z=x^3-g_2xz^2-g_3z^3$ by the torus action that takes fixez $z$ and takes $x \to t^2x, y \to t^3 y, g_2 \to t^4g_2, g_3 \to t^6 g_3$. So if $\overline{M}_{1,2}$ were a toric variety then this should be as well, but the equation has too many terms to be toric.

Here the extra automorphism is the case $t=-1$. Taking the even weight forms should basically by equivalent to taking the quotient by this action, where you replace $y$ by $y^2$ and then eliminating that variable, which I think gives your ring $(4,0)=g_2$, $(6,0)=g_3$, $(-2,1)=z$, $(0,1)=x$. I'm not completely sure of this because I don't think you actually can eliminate $y$ and get the Jacobi ring because of the extra factor of $z$, but you certainly get a toric variety by doing this (you just need to add the variable $x^3/z$.) However this is not going to be the same stack as you get from the full ring.

| cite | improve this answer | |
  • 2
    $\begingroup$ The funny thing is that $\overline M_{1,1}$ is definitely toric, but $M_{1,1}$ is not! The point at infinity that corresponds to the nodal curve is not torus invariant (the torus action is not particularly natural). I understand your argument about the fibers having (in general) four stacky points. So the best one can hope for is that the coarse moduli space is a toric surface. That seems plausible. $\endgroup$ – Lev Borisov Dec 5 '13 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.