Given a monic degree-four polynomial $f(z)$ without double roots, the equation $w^3 = f(z)$ defines a plane curve whose closure is a smooth curve of genus 3 with an action of $\Bbb Z/3$ (on $w$). The only isomorphisms between them which are $\Bbb Z/3$-equivariant are induced by the action itself and the maps $z \mapsto z+t$.

The Jacobian of said curve is 3-dimensional abelian variety with an action of $\Bbb Z/3$. It's also principally polarized -- this roughly corresponds to the cohomology pairing on $H^1$ of the curve.

There is a basis $$ \frac{dz}{w^2}, \frac{z dz}{w^2}, \text{ and }\frac{dz}{w} $$ of the 1-forms. This splits the formal completion of the Jacobian into two $\Bbb Z/3$-invariant summands, one 2-dimensional, one 1-dimensional. If $p \equiv 1 \mod 3$, then this splitting extends to the $p$-divisible group in the desired manner. (The formal group is actually additive for other primes.)

This defines for you a map of moduli from the moduli of these curves to a corresponding Shimura variety of $3$-dimensional polarized abelian surfaces with an action of the ring $\Bbb Z[\zeta_3]$. The action $[-1]$ on the abelian varieties doesn't lift to any of these curves (none of them are hyperelliptic), and so the map of moduli factors through a $\Bbb Z/2$-quotient.

However, I think (?) that the Torelli theorem implies that after you take this quotient, the result is an open substack. It's not the whole Shimura surface. Roughly, there are situations where the polynomial $f(z)$ degenerates, but the abelian variety does not. In terms of the Deligne-Mumford compactification, if you have only double roots then the smooth curve can be made to degenerate down to a stable curve of arithmetic genus $3$, whose "Jacobian" $Pic^0$ is still an abelian variety (but which now decomposes as the product of some 1- and 2-dimensional abelian varieties with $\Bbb Z/3$-action).

If you close up under this procedure, you get the whole Shimura surface. The surface is not compact in this case; there is one "point at infinity." There are two compactifications: the Satake-Baily-Borel compactification adds points to the moduli parametrizing so-called semi-abelian schemes (this compactification is singular), and the smooth compactification expands it so that these points are actually (quotients of) elliptic curves.

In this case, the moduli is pretty simple because it's described entirely in terms of the $a_i$. Away from the primes $2$ and $3$, you can eliminate the coefficient $a_1$ explicitly by a translation, and the Satake-Baily-Borel compactification is the weighted projective stack $$ Proj(\Bbb Z[1/6, a_2, a_4, a_3^2]) $$ with $|a_i| = 3i$. (The element $a_3$ gets a square because of this $\pm 1$ issue.) The moduli of curves hits the portion where the quartic discriminant of $f(z)$ doesn't vanish, and the uncompactified part is the portion where $f(z)$ has no triple zeros. This is the complement of the part defined by the ideal $(a_2^2 + 12 a_4,27 a_3^2 + 8 a_2^3)$.

The largest automorphism group occuring here is for the curves $w^3 = x^4 + a_3 x$. The resulting abelian scheme has $18$ (equivariant) automorphisms. At $p=7$ this is a height-$3$ point (but there's another one with only 6 automorphisms).

Unless I'm mistaken, the height-$3$ locus has mass $(p-1)(p^2-1)/1296$.

Given a monic degree-four polynomial $f(z)$ without double roots, the equation $w^3 = f(z)$ defines a plane curve whose closure is a smooth curve of genus 3 with an action of $\Bbb Z/3$ (on $w$). The only isomorphisms between them which are $\Bbb Z/3$-equivariant are induced by the action itself and the maps $z \mapsto z+t$.

The Jacobian of said curve is 3-dimensional abelian variety with an action of $\Bbb Z/3$. It's also principally polarized -- this roughly corresponds to the cohomology pairing on $H^1$ of the curve.

There is a basis $$ \frac{dz}{w^2}, \frac{z dz}{w^2}, \text{ and }\frac{dz}{w} $$ of the 1-forms. This splits the formal completion of the Jacobian into two $\Bbb Z/3$-invariant summands, one 2-dimensional, one 1-dimensional. If $p \equiv 1 \mod 3$, then this splitting extends to the $p$-divisible group in the desired manner. (The formal group is actually additive concentrated in even heights for primes congruent to 2.)

This defines for you a map of moduli from the moduli of these curves to a corresponding Shimura variety of $3$-dimensional polarized abelian surfaces with an action of the ring $\Bbb Z[\zeta_3]$. The action $[-1]$ on the abelian varieties doesn't lift to any of these curves (none of them are hyperelliptic), and so the map of moduli factors through a $\Bbb Z/2$-quotient.

However, I think (?) that the Torelli theorem implies that after you take this quotient, the result is an open substack. It's not the whole Shimura surface. Roughly, there are situations where the polynomial $f(z)$ degenerates, but the abelian variety does not. In terms of the Deligne-Mumford compactification, if you have only double roots then the smooth curve can be made to degenerate down to a stable curve of arithmetic genus $3$, whose "Jacobian" $Pic^0$ is still an abelian variety (but which now decomposes as the product of some 1- and 2-dimensional abelian varieties with $\Bbb Z/3$-action).

If you close up under this procedure, you get the whole Shimura surface. The surface is not compact in this case; there is one "point at infinity." There are two compactifications: the Satake-Baily-Borel compactification adds points to the moduli parametrizing so-called semi-abelian schemes (this compactification is singular), and the smooth compactification expands it so that these points are actually (quotients of) elliptic curves.

In this case, the moduli is pretty simple because it's described entirely in terms of the $a_i$. Away from the primes $2$ and $3$, you can eliminate the coefficient $a_1$ explicitly by a translation, and the Satake-Baily-Borel compactification is the weighted projective stack $$ Proj(\Bbb Z[1/6, a_2, a_4, a_3^2]) $$ with $|a_i| = 3i$. (The element $a_3$ gets a square because of this $\pm 1$ issue.) The moduli of curves hits the portion where the quartic discriminant of $f(z)$ doesn't vanish, and the uncompactified part is the portion where $f(z)$ has no triple zeros. This is the complement of the part defined by the ideal $(a_2^2 + 12 a_4,27 a_3^2 + 8 a_2^3)$.

The largest automorphism group occuring here is for the curves $w^3 = x^4 + a_3 x$. The resulting abelian scheme has $18$ (equivariant) automorphisms. At $p=7$ this is a height-$3$ point (but there's another one with only 6 automorphisms).

Unless I'm mistaken, the height-$3$ locus has mass $(p-1)(p^2-1)/1296$.

Given a monic degree-four polynomial $f(z)$ without double roots, the equation $w^3 = f(z)$ defines a plane curve whose closure is a smooth curve of genus 3 with an action of $\Bbb Z/3$ (on $w$). The only isomorphisms between them which are $\Bbb Z/3$-equivariant are induced by the action itself and the maps $z \mapsto z+t$.

The Jacobian of said curve is 3-dimensional abelian variety with an action of $\Bbb Z/3$. It's also principally polarized -- this roughly corresponds to the cohomology pairing on $H^1$ of the curve.

There is a basis $$ \frac{dz}{w^2}, \frac{z dz}{w^2}, \text{ and }\frac{dz}{w} $$ of the 1-forms. This splits the formal completion of the Jacobian into two $\Bbb Z/3$-invariant summands, one 2-dimensional, one 1-dimensional. If $p \equiv 1 \mod 3$, then this splitting extends to the $p$-divisible group in the desired manner. (The formal group is actually additive for other primes.)

This defines for you a map of moduli from the moduli of these curves to a corresponding Shimura variety of $3$-dimensional polarized abelian surfaces with an action of the ring $\Bbb Z[\zeta_3]$. The action $[-1]$ on the abelian varieties doesn't lift to any of these curves (none of them are hyperelliptic), and so the map of moduli factors through a $\Bbb Z/2$-quotient.

However, I think (?) that the Torelli theorem implies that after you take this quotient, the result is an open substack. It's not the whole Shimura surface. Roughly, there are situations where the polynomial $f(z)$ degenerates, but the abelian variety does not. In terms of the Deligne-Mumford compactification, if you have only double roots then the smooth curve can be made to degenerate down to a stable curve of arithmetic genus $3$, whose "Jacobian" $Pic^0$ is still an abelian variety (but which now decomposes as the product of some 1- and 2-dimensional abelian varieties with $\Bbb Z/3$-action).

If you close up under this procedure, you get the whole Shimura surface. The surface is not compact in this case; there is one "point at infinity." There are two compactifications: the Satake-Baily-Borel compactification adds points to the moduli parametrizing so-called semi-abelian schemes (this compactification is singular), and the smooth compactification expands it so that these points are actually (quotients of) elliptic curves.

In this case, the moduli is pretty simple because it's described entirely in terms of the $a_i$. Away from the primes $2$ and $3$, you can eliminate the coefficient $a_1$ explicitly by a translation, and the Satake-Baily-Borel compactification is the weighted projective stack $$ Proj(\Bbb Z[1/12, a_2, a_4, a_3^2]) $$ with $|a_i| = 3i$. (The element $a_3$ gets a square because of this $\pm 1$ issue.) The moduli of curves hits the portion where the quartic discriminant of $f(z)$ doesn't vanish, and the uncompactified part is the portion where $f(z)$ has no triple zeros. This is the complement of the part defined by the ideal $(a_2^2 + 12 a_4,27 a_3^2 + 8 a_2^3)$.

The largest automorphism group occuring here is for the curves $w^3 = x^4 + a_2 x$. The resulting abelian scheme has $18$ (equivariant) automorphisms. At $p=7$ this is a height-$3$ point (but there's another one with only 6 automorphisms).

Unless I'm mistaken, the height-$3$ locus has mass $(p-1)(p^2-1)/1296$.

Given a monic degree-four polynomial $f(z)$ without double roots, the equation $w^3 = f(z)$ defines a plane curve whose closure is a smooth curve of genus 3 with an action of $\Bbb Z/3$ (on $w$). The only isomorphisms between them which are $\Bbb Z/3$-equivariant are induced by the action itself and the maps $z \mapsto z+t$.

The Jacobian of said curve is 3-dimensional abelian variety with an action of $\Bbb Z/3$. It's also principally polarized -- this roughly corresponds to the cohomology pairing on $H^1$ of the curve.

There is a basis $$ \frac{dz}{w^2}, \frac{z dz}{w^2}, \text{ and }\frac{dz}{w} $$ of the 1-forms. This splits the formal completion of the Jacobian into two $\Bbb Z/3$-invariant summands, one 2-dimensional, one 1-dimensional. If $p \equiv 1 \mod 3$, then this splitting extends to the $p$-divisible group in the desired manner. (The formal group is actually additive for other primes.)

This defines for you a map of moduli from the moduli of these curves to a corresponding Shimura variety of $3$-dimensional polarized abelian surfaces with an action of the ring $\Bbb Z[\zeta_3]$. The action $[-1]$ on the abelian varieties doesn't lift to any of these curves (none of them are hyperelliptic), and so the map of moduli factors through a $\Bbb Z/2$-quotient.

However, I think (?) that the Torelli theorem implies that after you take this quotient, the result is an open substack. It's not the whole Shimura surface. Roughly, there are situations where the polynomial $f(z)$ degenerates, but the abelian variety does not. In terms of the Deligne-Mumford compactification, if you have only double roots then the smooth curve can be made to degenerate down to a stable curve of arithmetic genus $3$, whose "Jacobian" $Pic^0$ is still an abelian variety (but which now decomposes as the product of some 1- and 2-dimensional abelian varieties with $\Bbb Z/3$-action).

If you close up under this procedure, you get the whole Shimura surface. The surface is not compact in this case; there is one "point at infinity." There are two compactifications: the Satake-Baily-Borel compactification adds points to the moduli parametrizing so-called semi-abelian schemes (this compactification is singular), and the smooth compactification expands it so that these points are actually (quotients of) elliptic curves.

In this case, the moduli is pretty simple because it's described entirely in terms of the $a_i$. Away from the primes $2$ and $3$, you can eliminate the coefficient $a_1$ explicitly by a translation, and the Satake-Baily-Borel compactification is the weighted projective stack $$ Proj(\Bbb Z[1/6, a_2, a_4, a_3^2]) $$ with $|a_i| = 3i$. (The element $a_3$ gets a square because of this $\pm 1$ issue.) The moduli of curves hits the portion where the quartic discriminant of $f(z)$ doesn't vanish, and the uncompactified part is the portion where $f(z)$ has no triple zeros. This is the complement of the part defined by the ideal $(a_2^2 + 12 a_4,27 a_3^2 + 8 a_2^3)$.

The largest automorphism group occuring here is for the curves $w^3 = x^4 + a_3 x$. The resulting abelian scheme has $18$ (equivariant) automorphisms. At $p=7$ this is a height-$3$ point (but there's another one with only 6 automorphisms).

Unless I'm mistaken, the height-$3$ locus has mass $(p-1)(p^2-1)/1296$.