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).

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    $\begingroup$ 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. $\endgroup$
    – user17778
    Dec 8, 2013 at 11:09

1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$ Dec 5, 2013 at 1:09

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