Home Powered by MathJax

X3Y3

by

Gregg Kelly

In this section we explicitly calculate the $g_2$ and $g_3$ invariants of the curve $F(x,y) = 0$, where $F$ is a cubic polynomial in $x$ and $y$. We also obtain the differential equation for the general elliptic function of order 3.

Let $F(x,y)$ be a cubic polynomial in $x$ and $y$. \begin{equation} \label{eq:F} F(x,y) = a_{30} x^3 + a_{21} x^2 y + a_{12} x y^2 + a_{03} y^3 + a_{20} x^2 + a_{11} x y + a_{02} y^2 + a_{10} x + a_{01} y + a_{00} \end{equation} In general the curve $F(x,y) = 0$ has genus 1 and the holomorphic differential is given by \begin{equation} \label{eq:Fdiff} dz = \frac 1 {F_y(x,y)} dx = -\frac 1 {F_x(x,y)} dy \end{equation} We want to compute the $g_2$ and $g_3$ invariants of this differential as a function of the coefficients $a_{ij}$.

Reduction to Quadratic Curve

Let $r_1,r_2,r_3$ be roots of the cubic polynomial \begin{equation} a_{30} r^3 + a_{21} r^2 + a_{12} r + a_{03} = 0 \end{equation} Making the following linear transformation in equation \eqref{eq:F} \begin{equation} \label{eq:uvsub} \begin{split} x &= r_1 u + r_2 v \\ y &= u + v \end{split} \end{equation} gives a polynomial in $u$ and $v$ \begin{equation} \label{eq:G} G(u,v) = F(r_1 u + r_2 v, u + v) \end{equation} and it is easy to see that this cancels out the $u^3$ and $v^3$ terms and reduces the curve to the X2Y2 curve.

The reason for this can be understood in terms of the uniformizing elliptic functions. Curve \eqref{eq:F} is parameterised by two third order elliptic functions. The homogenous nature of the cubic terms means that they must share the same poles. So by taking a suitable linear combination of them we can cancel out one of the poles and end up with a second order elliptic function. We can do that in three ways, one for each pole. That gives three elliptic functions of order two, but because the sum of residues is zero only two of them will be independent.

Computation of the $g_2$ and $g_3$ Invariants

To facilitate the algebraic calculations it is convenient to replace the $a_{21},a_{12},a_{03}$ coefficients with their root expressions in \eqref{eq:G} \begin{equation} \label{eq:asub} \begin{split} a_{21} &= -(r_1 + r_2 + r_3) a_{30} \\ a_{12} &= (r_1 r_2 + r_1 r_3 + r_2 r_3) a_{30} \\ a_{03} &= - (r_1 r_2 r_3) a_{30} \end{split} \end{equation} which gives \begin{multline} G(u,v) = (-r_1^3 + 2r_1^2r_2 + r_1^2r_3 - r_1r_2^2 - 2r_1r_2r_3 + r_2^2r_3)a_{30}u^2v + (-r_1^2r_2 + r_1^2r_3 + 2r_1r_2^2 - 2r_1r_2r_3 - r_2^3 + r_2^2r_3)a_{30}uv^2 +\\ (a_{20}r_1^2 + a_{11}r_1 + a_{02})u^2 + (2a_{20}r_1r_2 + a_{11}r_1 + a_{11}r_2 + 2a_{02})uv + (a_{10}r_1 + a_{01})u + (a_{20}r_2^2 + a_{11}r_2 + a_{02})v^2 + (a_{10}r_2 + a_{01})v + a_{00} \end{multline} and the holomorphic differentials transformation can be obtained by using \eqref{eq:Fdiff} and differentiating \eqref{eq:uvsub} and \eqref{eq:G} to give \begin{equation} dz = \frac 1 {F_y(x,y)} dx = \frac {(r_1 - r_2)} {G_v(u,v)} du = \frac {(r_1 - r_2)} {\sqrt {D(u)}} du \end{equation} where $D(u) = \discrim_v(G(u,v))$. Computing the discriminant $D$ gives \begin{equation} \label{eq:D} \begin{aligned} D(u) =& \quad (r_1 - r_3)^2 a_{30}^2 (r_1-r_2)^4 u^4 + (-4a_{20}r_1r_3 - 2a_{11}r_1 - 2a_{11}r_3 - 4a_{02}) a_{30} (r_1-r_2)^3 u^3 \\ &+ (2a_{10}a_{30}r_1r_2 - 4a_{10}a_{30}r_1r_3 + 2a_{10}a_{30}r_2r_3 - 2a_{01}a_{30}r_1 + 4a_{01}a_{30}r_2 - 2a_{01}a_{30}r_3 - 4a_{02}a_{20} + a_{11}^2) (r_1-r_2)^2 u^2 \\ &+ (4a_{00}a_{30}r_1r_2 - 4a_{00}a_{30}r_1r_3 - 4a_{00}a_{30}r_2^2 + 4a_{00}a_{30}r_2r_3 + 4a_{01}a_{20}r_2 - 2a_{10}a_{11}r_2 + 2a_{01}a_{11} - 4a_{02}a_{10})(r_1-r_2) u \\ &- 4a_{00}a_{20}r_2^2 + a_{10}^2r_2^2 - 4a_{00}a_{11}r_2 + 2a_{01}a_{10}r_2 - 4a_{00}a_{02} + a_{01}^2 \end{aligned} \end{equation} Now to compute $g_2$ it is simply a matter of substituting the coefficients of \eqref{eq:D} into the standard formula $g_2 = \tfrac {1} {12} (12ae - 3bd + c^2)$, dividing by $(r_1 - r_2)^4$ and then restoring $a_{21},a_{12},a_{03}$ from the symmetric root expressions \eqref{eq:asub}.

Of course it does seem like a bit of miracle that resulting expression is actually divisible by $(r_1 - r_2)^4$, and that after division the resulting expression is symmetric in $r_1,r_2,r_3$.

\begin{equation} \label{eq:g2} \begin{aligned} g_2 = \tfrac 1 {12} &\Big[ 144 a_{00} a_{02} a_{12} a_{30} - 48 a_{00} a_{02} a_{21}^2 - 216 a_{00} a_{03} a_{11} a_{30} + 144 a_{00} a_{03} a_{20} a_{21} + 24 a_{00} a_{11} a_{12} a_{21 } \\ &- 48 a_{00} a_{12}^2 a_{20} - 48 a_{01}^2 a_{12} a_{30} + 16 a_{01}^2 a_{21}^2 + 24 a_{01} a_{02} a_{11} a_{30} - 16 a_{01} a_{02} a_{20} a_{21 } \\ &+ 144 a_{01} a_{03} a_{10} a_{30} - 48 a_{01} a_{03} a_{20}^2 - 16 a_{01} a_{10} a_{12} a_{21} - 8 a_{01} a_{11}^2 a_{21} + 24 a_{01} a_{11} a_{12} a_{20 } \\ &- 48 a_{02}^2 a_{10} a_{30} + 16 a_{02}^2 a_{20}^2 + 24 a_{02} a_{10} a_{11} a_{21} - 16 a_{02} a_{10} a_{12} a_{20} - 8 a_{02} a_{11}^2 a_{20 } \\ &- 48 a_{03} a_{10}^2 a_{21} + 24 a_{03} a_{10} a_{11} a_{20} + 16 a_{10}^2 a_{12}^2 - 8 a_{10} a_{11}^2 a_{12} + a_{11}^ 4 \Big] \end{aligned} \end{equation}

Similarly to compute $g_3$ substitute the coefficients of \eqref{eq:D} into the standard formula $g_3 = \tfrac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3)$, divide by $(r_1 - r_2)^6$ and then restore $a_{21},a_{12},a_{03}$ from the symmetric root expressions \eqref{eq:asub}.

\begin{equation} \label{eq:g3} \begin{aligned} g_3 = \tfrac 1 {216} &\Big[ 5832 a_{00}^2 a_{03}^2 a_{30}^2 - 3888 a_{00}^2 a_{03} a_{12} a_{21} a_{30} + 864 a_{00}^2 a_{03} a_{21}^3 + 864 a_{00}^2 a_{12}^3 a_{30} - 216 a_{00}^2 a_{12}^2 a_{21}^ 2 \\ &- 3888 a_{00} a_{01} a_{02} a_{03} a_{30}^2 + 1296 a_{00} a_{01} a_{02} a_{12} a_{21} a_{30} - 288 a_{00} a_{01} a_{02} a_{21}^3 + 1296 a_{00} a_{01} a_{03} a_{11} a_{21} a_{30} + 1296 a_{00} a_{01} a_{03} a_{12} a_{20} a_{30 } \\ &- 864 a_{00} a_{01} a_{03} a_{20} a_{21}^2 - 864 a_{00} a_{01} a_{11} a_{12}^2 a_{30} + 144 a_{00} a_{01} a_{11} a_{12} a_{21}^2 + 144 a_{00} a_{01} a_{12}^2 a_{20} a_{21} + 864 a_{00} a_{02}^3 a_{30}^ 2 \\ &- 864 a_{00} a_{02}^2 a_{11} a_{21} a_{30} - 864 a_{00} a_{02}^2 a_{12} a_{20} a_{30} + 576 a_{00} a_{02}^2 a_{20} a_{21}^2 + 1296 a_{00} a_{02} a_{03} a_{10} a_{21} a_{30} + 1296 a_{00} a_{02} a_{03} a_{11} a_{20} a_{30 } \\ &- 864 a_{00} a_{02} a_{03} a_{20}^2 a_{21} - 864 a_{00} a_{02} a_{10} a_{12}^2 a_{30} + 144 a_{00} a_{02} a_{10} a_{12} a_{21}^2 + 648 a_{00} a_{02} a_{11}^2 a_{12} a_{30} + 72 a_{00} a_{02} a_{11}^2 a_{21}^ 2 \\ &- 720 a_{00} a_{02} a_{11} a_{12} a_{20} a_{21} + 576 a_{00} a_{02} a_{12}^2 a_{20}^2 - 3888 a_{00} a_{03}^2 a_{10} a_{20} a_{30} + 864 a_{00} a_{03}^2 a_{20}^3 + 1296 a_{00} a_{03} a_{10} a_{11} a_{12} a_{30 } \\ &- 864 a_{00} a_{03} a_{10} a_{11} a_{21}^2 + 1296 a_{00} a_{03} a_{10} a_{12} a_{20} a_{21} - 540 a_{00} a_{03} a_{11}^3 a_{30} + 648 a_{00} a_{03} a_{11}^2 a_{20} a_{21} - 864 a_{00} a_{03} a_{11} a_{12} a_{20}^ 2 \\ &+ 144 a_{00} a_{10} a_{11} a_{12}^2 a_{21} - 288 a_{00} a_{10} a_{12}^3 a_{20} - 36 a_{00} a_{11}^3 a_{12} a_{21} + 72 a_{00} a_{11}^2 a_{12}^2 a_{20} + 864 a_{01}^3 a_{03} a_{30}^ 2 \\ &- 288 a_{01}^3 a_{12} a_{21} a_{30} + 64 a_{01}^3 a_{21}^3 - 216 a_{01}^2 a_{02}^2 a_{30}^2 + 144 a_{01}^2 a_{02} a_{11} a_{21} a_{30} + 144 a_{01}^2 a_{02} a_{12} a_{20} a_{30 } \\ &- 96 a_{01}^2 a_{02} a_{20} a_{21}^2 - 864 a_{01}^2 a_{03} a_{10} a_{21} a_{30} - 864 a_{01}^2 a_{03} a_{11} a_{20} a_{30} + 576 a_{01}^2 a_{03} a_{20}^2 a_{21} + 576 a_{01}^2 a_{10} a_{12}^2 a_{30 } \\ &- 96 a_{01}^2 a_{10} a_{12} a_{21}^2 + 72 a_{01}^2 a_{11}^2 a_{12} a_{30} - 48 a_{01}^2 a_{11}^2 a_{21}^2 + 144 a_{01}^2 a_{11} a_{12} a_{20} a_{21} - 216 a_{01}^2 a_{12}^2 a_{20}^ 2 \\ &+ 144 a_{01} a_{02}^2 a_{10} a_{21} a_{30} + 144 a_{01} a_{02}^2 a_{11} a_{20} a_{30} - 96 a_{01} a_{02}^2 a_{20}^2 a_{21} + 1296 a_{01} a_{02} a_{03} a_{10} a_{20} a_{30} - 288 a_{01} a_{02} a_{03} a_{20}^ 3 \\ &- 720 a_{01} a_{02} a_{10} a_{11} a_{12} a_{30} + 144 a_{01} a_{02} a_{10} a_{11} a_{21}^2 - 48 a_{01} a_{02} a_{10} a_{12} a_{20} a_{21} - 36 a_{01} a_{02} a_{11}^3 a_{30} - 24 a_{01} a_{02} a_{11}^2 a_{20} a_{21 } \\ &+ 144 a_{01} a_{02} a_{11} a_{12} a_{20}^2 - 864 a_{01} a_{03} a_{10}^2 a_{12} a_{30} + 576 a_{01} a_{03} a_{10}^2 a_{21}^2 + 648 a_{01} a_{03} a_{10} a_{11}^2 a_{30} - 720 a_{01} a_{03} a_{10} a_{11} a_{20} a_{21 } \\ &+ 144 a_{01} a_{03} a_{10} a_{12} a_{20}^2 + 72 a_{01} a_{03} a_{11}^2 a_{20}^2 - 96 a_{01} a_{10}^2 a_{12}^2 a_{21} - 24 a_{01} a_{10} a_{11}^2 a_{12} a_{21} + 144 a_{01} a_{10} a_{11} a_{12}^2 a_{20 } \\ &+ 12 a_{01} a_{11}^4 a_{21} - 36 a_{01} a_{11}^3 a_{12} a_{20} - 288 a_{02}^3 a_{10} a_{20} a_{30} + 64 a_{02}^3 a_{20}^3 + 576 a_{02}^2 a_{10}^2 a_{12} a_{30 } \\ &- 216 a_{02}^2 a_{10}^2 a_{21}^2 + 72 a_{02}^2 a_{10} a_{11}^2 a_{30} + 144 a_{02}^2 a_{10} a_{11} a_{20} a_{21} - 96 a_{02}^2 a_{10} a_{12} a_{20}^2 - 48 a_{02}^2 a_{11}^2 a_{20}^ 2 \\ &- 864 a_{02} a_{03} a_{10}^2 a_{11} a_{30} + 144 a_{02} a_{03} a_{10}^2 a_{20} a_{21} + 144 a_{02} a_{03} a_{10} a_{11} a_{20}^2 + 144 a_{02} a_{10}^2 a_{11} a_{12} a_{21} - 96 a_{02} a_{10}^2 a_{12}^2 a_{20 } \\ &- 36 a_{02} a_{10} a_{11}^3 a_{21} - 24 a_{02} a_{10} a_{11}^2 a_{12} a_{20} + 12 a_{02} a_{11}^4 a_{20} + 864 a_{03}^2 a_{10}^3 a_{30} - 216 a_{03}^2 a_{10}^2 a_{20}^ 2 \\ &- 288 a_{03} a_{10}^3 a_{12} a_{21} + 72 a_{03} a_{10}^2 a_{11}^2 a_{21} + 144 a_{03} a_{10}^2 a_{11} a_{12} a_{20} - 36 a_{03} a_{10} a_{11}^3 a_{20} + 64 a_{10}^3 a_{12}^ 3 \\ &- 48 a_{10}^2 a_{11}^2 a_{12}^2 + 12 a_{10} a_{11}^4 a_{12} - a_{11}^ 6 \Big] \end{aligned} \end{equation} We have succeeded in computing the $g_2$ and $g_3$ invariants in terms of the coefficients $a_{ij}$. We would also like them in terms of the invariants of the coefficient polynomials.

Invariance Under Linear Fractional Transformations

Unlike the X2Y2 curve \eqref{eq:F} does not retain the same form under individual Möbius transformations in $x$ and $y$. But it does retain the same form under joint linear fractional transformations \begin{equation} \label{eq:linfrac} u = \frac {a_1x + b_1y + c_1} {a_3x + b_3y + c_3}, \qquad v = \frac {a_2x + b_2y + c_2} {a_3x + b_3y + c_3} \end{equation} If we view $F$ as a ternary form in $x$, $y$ and $z$, by homogenising, then the linear fractional transformations \eqref{eq:linfrac} correspond to linear transformations on \begin{equation} \label{eq:F3} F(x,y,z) = a_{300} x^3 + a_{210} x^2 y + a_{120} x y^2 + a_{030} y^3 + a_{201} x^2 z + a_{111} x y z + a_{021} y^2 z + a_{102} x z^2 + a_{012} y z^2 + a_{003} z^3 \end{equation} Therefore the $g_2$ and $g_3$ invariants must be invariants of this cubic ternary form. It has been known since the mid nineteenth century that they generate all invariants of ternary cubics. In that context they are known as the Aronhold invariants of the ternary cubic form (see for example Pg. 128 of [1]).

In addition \eqref{eq:F3} can be written as a polynomial in $z$ with coefficients a cubic, quadratic, linear and constant binary form in $x$ and $y$. Because this polynomial retains the same form under transformations \eqref{eq:linfrac} with $c_1=c_2=c_3=0$ the ternary invariants must also be composed from the simultaneous invariants of those three binary forms and the constant. And similarly \eqref{eq:F3} can be written as a polynomial in $x$ or $y$ so there are three ways this can be done. \begin{equation} \label{eq:A} A = \begin{pmatrix} a_{003} & a_{012} & a_{021} & a_{030} \\ a_{102} & a_{111} & a_{120} & 0 \\ a_{201} & a_{210} & 0 & 0 \\ a_{300} & 0 & 0 & 0 \end{pmatrix} \end{equation} That is the ternary invariants can be composed from the simultaneous binary invariants made up from the rows of $A$, or from the columns of $A$, or from the diagonals of $A$.

Symmetric Expressions for $g_2$ and $g_3$

The symmetry in the $g_2$ and $g_3$ formulae \eqref{eq:g2} and \eqref{eq:g3} is more obvious when written using the symmetric subscripts in \eqref{eq:F3}. \begin{equation} \label{eq:g2s} \begin{aligned} g_2 = \tfrac 1 {12} \Big[ +144& \space ( a_{003} a_{021} a_{120} a_{300} + a_{003} a_{030} a_{201} a_{210} + a_{012} a_{030} a_{102} a_{300} ) \\ -48& \space ( a_{003} a_{021} a_{210}^2 + a_{003} a_{120}^2 a_{201} + a_{012}^2 a_{120} a_{300} + a_{012} a_{030} a_{201}^2 + a_{021}^2 a_{102} a_{300} + a_{030} a_{102}^2 a_{210} ) \\ +16& \space ( a_{012}^2 a_{210}^2 + a_{021}^2 a_{201}^2 + a_{102}^2 a_{120}^2 ) \\ -16& \space ( a_{012} a_{021} a_{201} a_{210} + a_{012} a_{102} a_{120} a_{210} + a_{021} a_{102} a_{120} a_{201} ) \\ -216& \space a_{111} a_{003} a_{030} a_{300} \\ +24& \space a_{111} ( a_{003} a_{120} a_{210} + a_{012} a_{021} a_{300} + a_{012} a_{120} a_{201} + a_{021} a_{102} a_{210} + a_{030} a_{102} a_{201} ) \\ -8& \space a_{111}^2 ( a_{012} a_{210} + a_{021} a_{201} + a_{102} a_{120} ) \space + \space a_{111}^ 4 \space \Big] \end{aligned} \end{equation} \begin{equation} \label{eq:g3s} \begin{aligned} g_3 = \tfrac 1 {216} \Big[ \hphantom{+} 5832& \space \hphantom{(} a_{003}^2 a_{030}^2 a_{300}^2 \\ -3888& \space ( a_{003}^2 a_{030} a_{120} a_{210} a_{300} + a_{003} a_{012} a_{021} a_{030} a_{300}^2 + a_{003} a_{030}^2 a_{102} a_{201} a_{300} ) \\ +1296& \space \hphantom{(} a_{003} a_{021} a_{030} a_{102} a_{210} a_{300} \\ +864& \space ( a_{003}^2 a_{030} a_{210}^3 + a_{003}^2 a_{120}^3 a_{300} + a_{003} a_{021}^3 a_{300}^2 + a_{003} a_{030}^2 a_{201}^3 + a_{012}^3 a_{030} a_{300}^2 + a_{030}^2 a_{102}^3 a_{300} ) \\ +1296& \space ( a_{003} a_{012} a_{021} a_{120} a_{210} a_{300} + a_{003} a_{012} a_{030} a_{120} a_{201} a_{300} + a_{003} a_{030} a_{102} a_{120} a_{201} a_{210} + a_{012} a_{021} a_{030} a_{102} a_{201} a_{300} ) \\ -864& \space ( a_{003} a_{012} a_{030} a_{201} a_{210}^2 + a_{003} a_{021}^2 a_{120} a_{201} a_{300} + a_{003} a_{021} a_{030} a_{201}^2 a_{210} + a_{003} a_{021} a_{102} a_{120}^2 a_{300} + a_{012}^2 a_{030} a_{102} a_{210} a_{300} + a_{012} a_{030} a_{102}^2 a_{120} a_{300} ) \\ -216& \space ( a_{003}^2 a_{120}^2 a_{210}^2 + a_{012}^2 a_{021}^2 a_{300}^2 + a_{030}^2 a_{102}^2 a_{201}^2 ) \\ +576& \space ( a_{003} a_{021}^2 a_{201} a_{210}^2 + a_{003} a_{021} a_{120}^2 a_{201}^2 + a_{012}^2 a_{030} a_{201}^2 a_{210} + a_{012}^2 a_{102} a_{120}^2 a_{300} + a_{012} a_{030} a_{102}^2 a_{210}^2 + a_{021}^2 a_{102}^2 a_{120} a_{300} ) \\ -288& \space ( a_{003} a_{012} a_{021} a_{210}^3 + a_{003} a_{102} a_{120}^3 a_{201} + a_{012}^3 a_{120} a_{210} a_{300} + a_{012} a_{021} a_{030} a_{201}^3 + a_{021}^3 a_{102} a_{201} a_{300} + a_{030} a_{102}^3 a_{120} a_{210} ) \\ +144& \space ( a_{003} a_{012} a_{120}^2 a_{201} a_{210} + a_{003} a_{021} a_{102} a_{120} a_{210}^2 + a_{012}^2 a_{021} a_{120} a_{201} a_{300} + a_{012} a_{021}^2 a_{102} a_{210} a_{300} + a_{012} a_{030} a_{102} a_{120} a_{201}^2 + a_{021} a_{030} a_{102}^2 a_{201} a_{210} ) \\ -216& \space ( a_{012}^2 a_{120}^2 a_{201}^2 + a_{021}^2 a_{102}^2 a_{210}^2 ) \\ -96& \space ( a_{012}^2 a_{021} a_{201} a_{210}^2 + a_{012}^2 a_{102} a_{120} a_{210}^2 + a_{012} a_{021}^2 a_{201}^2 a_{210} + a_{012} a_{102}^2 a_{120}^2 a_{210} + a_{021}^2 a_{102} a_{120} a_{201}^2 + a_{021} a_{102}^2 a_{120}^2 a_{201} ) \\ +64& \space ( a_{012}^3 a_{210}^3 + a_{021}^3 a_{201}^3 + a_{102}^3 a_{120}^3 ) \\ -48& \space \hphantom{(} a_{012} a_{021} a_{102} a_{120} a_{201} a_{210} \\ +1296& \space a_{111} \space ( a_{003} a_{012} a_{030} a_{210} a_{300} + a_{003} a_{021} a_{030} a_{201} a_{300} + a_{003} a_{030} a_{102} a_{120} a_{300} ) \\ -864& \space a_{111} \space ( a_{003} a_{012} a_{120}^2 a_{300} + a_{003} a_{021}^2 a_{210} a_{300} + a_{003} a_{030} a_{102} a_{210}^2 + a_{003} a_{030} a_{120} a_{201}^2 + a_{012}^2 a_{030} a_{201} a_{300} + a_{021} a_{030} a_{102}^2 a_{300} ) \\ -720& \space a_{111} \space ( a_{003} a_{021} a_{120} a_{201} a_{210} + a_{012} a_{021} a_{102} a_{120} a_{300} + a_{012} a_{030} a_{102} a_{201} a_{210} ) \\ +144& \space a_{111} \space ( a_{003} a_{012} a_{120} a_{210}^2 + a_{003} a_{102} a_{120}^2 a_{210} + a_{012}^2 a_{021} a_{210} a_{300} + a_{012} a_{021}^2 a_{201} a_{300} + a_{021} a_{030} a_{102} a_{201}^2 + a_{030} a_{102}^2 a_{120} a_{201} ) \\ +144& \space a_{111} \space (a_{012}^2 a_{120} a_{201} a_{210} + a_{012} a_{021} a_{102} a_{210}^2 + a_{012} a_{021} a_{120} a_{201}^2 + a_{012} a_{102} a_{120}^2 a_{201} + a_{021}^2 a_{102} a_{201} a_{210} + a_{021} a_{102}^2 a_{120} a_{210} ) \\ +648& \space a_{111}^2 \space ( a_{003} a_{021} a_{120} a_{300} + a_{003} a_{030} a_{201} a_{210} + a_{012} a_{030} a_{102} a_{300} ) \\ +72& \space a_{111}^2 \space ( a_{003} a_{021} a_{210}^2 + a_{003} a_{120}^2 a_{201} + a_{012}^2 a_{120} a_{300} + a_{012} a_{030} a_{201}^2 + a_{021}^2 a_{102} a_{300} + a_{030} a_{102}^2 a_{210} ) \\ -48& \space a_{111}^2 \space ( a_{012}^2 a_{210}^2 + a_{021}^2 a_{201}^2 + a_{102}^2 a_{120}^2 ) \\ -24& \space a_{111}^2 \space ( a_{012} a_{021} a_{201} a_{210} + a_{012} a_{102} a_{120} a_{210} + a_{021} a_{102} a_{120} a_{201} ) \\ -540& \space a_{111}^3 \space \hphantom{(} a_{003} a_{030} a_{300} \\ -36& \space a_{111}^3 \space ( a_{003} a_{120} a_{210} + a_{012} a_{021} a_{300} + a_{030} a_{102} a_{201} ) \\ -36& \space a_{111}^3 \space ( a_{012} a_{120} a_{201} + a_{021} a_{102} a_{210} ) \\ +12& \space a_{111}^4 \space ( a_{012} a_{210} + a_{021} a_{201} + a_{102} a_{120} ) - \space a_{111}^ 6 \space \Big] \end{aligned} \end{equation}

Hessian Curve

As can be seen from the formula above the invariants take a particularly simple form when all the coefficients with a 0, 1 and 2 in their indices vanish. This is the Hessian curve. \begin{equation} \label{eq:Hess} F(x,y) = x^3 + y^3 - 3\alpha xy + 1 \end{equation} A simple substitution into \eqref{eq:g2s} and \eqref{eq:g3s} gives \begin{equation} g_2 = \tfrac {27} {4} (\alpha^4 + 8\alpha),\qquad g_3 = \tfrac {27} {8} (-\alpha^6 + 20\alpha^3 + 8),\qquad \Delta= 19683(\alpha^3 - 1)^3,\qquad J = \frac {(\alpha^4 + 8 \alpha)^3} {64(\alpha^3 - 1)^3} \end{equation}

Reduced Cubic Curve

There are 15 generators (see [2]) of the simultaneous invariants of a linear, quadratic and cubic binary form. So it seems of little advantage expressing the ternary invariants directly in terms of them. However if we drop out the linear form and the generator with degree 7 we get a more manageable set of 4 non-trivial generators. They are respectively the discriminant of the quadratic, the simultaneous invariant of the quadratic and cubic of degree 3, the discriminant of the cubic, and the resultant of the quadratic and cubic. Let

\begin{equation} \label{eq:basis} \begin{aligned} \Delta_1 &= h \\ \Delta_2 &= b^2 - 4 a c \\ \Delta_3 &= 6 a q s - 2 a r^2 - 9 b p s + b q r + 6 c p r - 2 c q^2 \\ \Delta_4 &= -27 p^2 s^2 + 18 p q r s - 4 p r^3 - 4 q^3 s + q^2 r^2 \\ \Delta_5 &= a^3 s^2 - a^2 b r s - 2 a^2 c q s + a^2 c r^2 + a b^2 q s + 3 a b c p s - a b c q r - 2 a c^2 p r + a c^2 q^2 - b^3 p s + b^2 c p r - b c^2 p q + c^3 p^2 \\ \end{aligned} \end{equation}

Together they can be used to compute the $g_2$ and $g_3$ invariants of the reduced cubic curve \begin{equation} \label{eq:ReducedCubic} F(x,y) = hy^3 + (ax^2 + bx + c)y + (px^3 + qx^2 + rx + s) \end{equation} Applying a Gröbner basis calculation to \eqref{eq:g2} and \eqref{eq:g3}, with $a_{ij}$ replaced by $a,b,c\dots$ and using the basis \eqref{eq:basis} yields

\begin{equation} \label{eq:g2g3} \begin{aligned} g_2 &= \tfrac 1 {12} \left(\Delta_2^2 + 24 \Delta_1 \Delta_3\right) \\\\ g_3 &= \tfrac 1 {216} \left(-\Delta_2^3 - 36 \Delta_1 \Delta_2 \Delta_3 + 864 \Delta_1 \Delta_5 - 216 \Delta_1^2 \Delta_4 \right) \end{aligned} \end{equation}

Differential Equation For Order 3 Elliptic Functions

The reduced cubic equation can be used to derive the differential equation for the general elliptic function of order 3. In \eqref{eq:ReducedCubic} put $h = 1$ and write it as \begin{equation} \label{eq:reducedcubic} F(x,y) = y^3 + Q(x) y + R(x) = 0 \end{equation} Then from the holomorphic differential \eqref{eq:Fdiff} we have \begin{equation} \label{eq:diff} \frac {dx} {dz} = 3 y^2 + Q(x) \end{equation} Eliminating $y$ from \eqref{eq:reducedcubic} and \eqref{eq:diff} gives an ordinary differential equation for the general elliptic function of order 3

\begin{equation} \left(\frac {dx} {dz}\right)^3 + 3 Q(x) \left(\frac {dx} {dz}\right)^2 = 27 R(x)^2 + 4 Q(x)^3 \end{equation}

where $Q(x) = ax^2 + bx + c$ and $R(x) = px^3 + qx^2 + rx + s$ and the $g_2$ and $g_3$ invariants are given by equation \eqref{eq:g2g3} with $\Delta_1 = 1$.

Associated Genus 1 Curve

Observe that the genus 1 curve associated with this differential equation \begin{equation} y^3 + 3 Q(x) y^2 = 27 R(x)^2 + 4 Q(x)^3 \end{equation} reduces to the $I_3$ curve derived in Y3R3 when $Q(x) = 0$.

Addition Formula

Curve \eqref{eq:F} has a relatively simple addition formula \begin{equation} \label{eq:add} \begin{aligned} x_3 \space &= \space \frac {a_{03}(x_1y_2-x_2 y_1)^3 \space + \space a_{02}(x_1y_2-x_2y_1)^2(x_1-x_2) \space + \space a_{01}(x_1y_2-x_2y_1)(x_1-x_2)^2 \space + \space a_{00}(x_1-x_2)^3} {x_1 x_2 \thinspace \Big[a_{30}(x_1-x_2)^3 \space + \space a_{21}(x_1-x_2)^2(y_1-y_2) \space + \space a_{12}(x_1-x_2)(y_1-y_2)^2 \space + \space a_{03}(y_1-y_2)^3 \Big]} \\\\ y_3 \space &= \space \frac {a_{30}(x_1y_2-x_2 y_1)^3 \space + \space a_{20}(x_1y_2-x_2 y_1)^2(y_2-y_1) \space + \space a_{10}(x_1y_2-x_2 y_1)(y_2-y_1)^2 \space + \space a_{00}(y_2-y_1)^3} {y_1 y_2 \thinspace \Big[a_{30}(x_1-x_2)^3 \space + \space a_{21}(x_1-x_2)^2(y_1-y_2) \space + \space a_{12}(x_1-x_2)(y_1-y_2)^2 \space + \space a_{03}(y_1-y_2)^3 \Big]} \end{aligned} \end{equation} See Cubic Addition Formulae for the details.

References

[1] I. Dolgachev Classical Algebraic Geometry: A Modern View Cambridge University Press, September 2012

[2] M. Popoviciu Draisma 2014 Invariants of binary forms. Doctoral Dissertation: Philosophisch-Naturwissenschaftlichen Fakultät der Universität Basel