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Will Sawin
Will Sawin
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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}_{1,1}$$\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.

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 $\overline{M}_{1,1}$ 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.

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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Will Sawin
Will Sawin
  • 148.4k
  • 9
  • 324
  • 563

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 $\overline{M}_{1,1}$ 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.