In this section we summarise and extend the symmetric addition formulae for general genus 1 curves of degree 3 and 4 calculated on the previous pages.
We will start by looking at the cubic curve \begin{equation} \label{eq:y2x3} y^2 \space = \space a x^3 \space + b x^2 \space + \space c x \space + \space d \end{equation} If $f$ and $g$ are the parameterising elliptic functions then $f$ has a double pole at say, $z = z_0$ and $g$ has a triple pole also at $z = z_0$. It follows that $1,f(z),g(z)$ is a basis for the three dimensional vector space of elliptic functions with a pole of order 3 or less at $z_0$. Therefore by the Extended Frobenius-Stickelberger formula, when $z_1 + z_2 + z_3 = 3 z_0$ we have \begin{equation} \begin{vmatrix} 1 & f(z_1) & g(z_1) \\ 1 & f(z_2) & g(z_2) \\ 1 & f(z_3) & g(z_3) \\ \end{vmatrix} \space = \space 0 \end{equation} Writing everything in terms of the coordinates $x_i=f(z_i)$ and $y_i=g(z_i)$ we have \begin{equation} \label{eq:y2x3add1} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix} \space = \space 0 \end{equation} This is the level 1 addition formulae for curve \eqref{eq:y2x3}. It implies the three points lie on a straight line. The equation for the straight line is simply \begin{equation} \label{eq:line} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x & y \\ \end{vmatrix} \space = \space 0 \end{equation}
The level 2 addition formula for $x_3$ and $y_3$ can be obtained by solving \eqref{eq:y2x3} and \eqref{eq:line} for $x$ and $y$.
To eliminate $y$ we view these equations as polynomials in $y$ and compute their resultant. This is done by multiplying the two $y$-conjugates of the level 1 formula \eqref{eq:line} together, reducing modulo \eqref{eq:y2x3} to eliminate $y$, and then solving for $x$. The last step can be performed simply by evaluating at $x=0$ to obtain the numerator and $x = \infty$ to obtain the denominator. Similarly the formula for $y_3$ is obtained by multiplying the three $x$-conjugates of \eqref{eq:line} together, eliminating $x$ and then solving for $y$ using the same technique. This process yields the following formulae
\begin{equation} \label{eq:y2x3add2} x_3 \space = \space \frac 1 {a x_1 x_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & \sqrt{d} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & -\sqrt{d} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 0 & 1 \\ \end{vmatrix}^2}, \qquad\qquad y_3 \space = \space - \frac 1 {y_1 y_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & e_1 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & e_2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & e_3 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 0 & 1 \\ \end{vmatrix}^3} \end{equation}
where $e_1,e_2,e_3$ are the three roots of the right hand side of \eqref{eq:y2x3}. Notice that although formulae \eqref{eq:y2x3add2} contains square roots and cubic roots, they appear symmetrically, so it is a rational function of $x_1,y_1,x_2,y_2$ and a rational function of the coefficients $a,b,c,d$ of curve \eqref{eq:y2x3}.
Equations \eqref{eq:y2x3add2} can be interpreted as addition formulae for $\wp(z_1+z_2)$ and $\wp'(z_1+z_2)$ see here.
If in \eqref{eq:y2x3add2} you expand the determinants by the cofactors of their last row and simplify using the relation \eqref{eq:y2x3}, you will eventually end up with the formula \begin{equation} x_3 \space = \space - x_1 - x_2 - \frac b a \space + \space \frac 1 a \left(\frac {y_1 - y_2} {x_1 - x_2}\right)^2 \end{equation} which is essentially the classic addition formula for the Weierstrass $\wp$ function.
If we extend the basis like $\{\space 1,x,y,x^2,xy,x^3,x^2y,x^4,x^3y\space\ldots\space\}$ then we have $\{\space 0,2,3,4,5,6,7,8,9\space\ldots\space\}$ poles. So we can extend \eqref{eq:y2x3add1} etc. to $n$ variables where $n \ge 2$. For example for $n = 4$ the level 1 addition formula is \begin{equation} \label{eq:y2x3add1v4} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & x_4 & y_4 & x_4^2 \\ \end{vmatrix} \space = \space 0 \end{equation} and the level 2 formula is
\begin{equation} \label{eq:y2x3add2v4} x_4 \space = \space \frac 1 {x_1x_2x_3} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2\\ 1 & x_2 & y_2 & x_2^2\\ 1 & x_3 & y_3 & x_3^2\\ 1 & 0 & \sqrt{d} & 0 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & 0 & -\sqrt{d} & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 0 & 0 & 0 & 1 \\ \end{vmatrix}^2}, \qquad\qquad y_4 \space = \space \frac {a^2} {y_1y_2y_3} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & e_1 & 0 & e_1^2 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & e_2 & 0 & e_2^2 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & e_3 & 0 & e_3^2 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 0 & 0 & 0 & 1 \\ \end{vmatrix}^3} \end{equation}
This formula can be interpreted as addition formulae for $\wp(z_1+z_2+z_3)$ and $\wp'(z_1+z_2+z_3)$.
For $n = 5$ the level 1 addition formula is \begin{equation} \label{eq:y2x3add1v5} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & x_5 & y_5 & x_5^2 & x_5y_5 \\ \end{vmatrix} \space = \space 0 \end{equation} and the level 2 formula is
\begin{equation} \label{eq:y2x3add2v5} x_5 = \frac 1 {ax_1x_2x_3x_4} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & 0 & \sqrt{d} & 0 & 0 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & 0 & -\sqrt{d} & 0 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{vmatrix}^2}, \qquad y_5 = \frac {-a} {y_1y_2y_3y_4} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & e_1 & 0 & e_1^2 & 0 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & e_2 & 0 & e_2^2 & 0 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 1 & e_3 & 0 & e_3^2 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & x_1y_1 \\ 1 & x_2 & y_2 & x_2^2 & x_2y_2 \\ 1 & x_3 & y_3 & x_3^2 & x_3y_3 \\ 1 & x_4 & y_4 & x_4^2 & x_4y_4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{vmatrix}^3} \end{equation}
This can be interpreted as addition formulae for $\wp(z_1+z_2+z_3+z_4)$ and $\wp'(z_1+z_2+z_3+z_4)$.
By the way, although the $a$ factors in equations \eqref{eq:y2x3add2}, \eqref{eq:y2x3add2v4} and \eqref{eq:y2x3add2v5} may seem to be jumping around, there is a simple pattern to them which has been obscured by simplifying the denominators.
Now we extend the addition formula to the general cubic curve given by \begin{equation} \label{eq:x3y3} 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} \space = \space 0 \end{equation} In general the elliptic functions $f$ and $g$ parameterising this curve share three simple poles. Therefore, similar to the previous case, $1,f,g$ are a basis for the vector space of elliptic functions with a pole of order at most one at these three points. From which it follows this curve has the same level 1 addition formula \eqref{eq:y2x3add1} as the previous cubic curve.
Let $\alpha_1,\alpha_2,\alpha_3$, $\beta_1,\beta_2,\beta_3$ and $\gamma_1,\gamma_2,\gamma_3$ be the roots of the three polynomials \begin{equation} \label{eq:abc} \begin{aligned} a_{30} t^3 + a_{20} t^2 + a_{10} t + a_{00} \space = \space 0 \\ a_{03} t^3 + a_{02} t^2 + a_{01} t + a_{00} \space = \space 0 \\ a_{30} t^3 + a_{21} t^2 + a_{12} t + a_{03} \space = \space 0 \\ \end{aligned} \end{equation} then the level 2 addition formulae are given by
\begin{equation} \label{eq:x3y3add2} x_3 \space = \space \frac {a_{03}} {a_{30} x_1 x_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & \beta_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & \beta_2 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & \beta_3 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & \gamma_1 & 1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & \gamma_2 & 1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & \gamma_3 & 1 \\ \end{vmatrix}},\qquad\qquad y_3 \space = \space \frac {a_{30}} {a_{03} y_1 y_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & \alpha_1 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & \alpha_2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & \alpha_3 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 1 & 1/\gamma_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 1 & 1/\gamma_2 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 1 & 1/\gamma_3 \\ \end{vmatrix}} \end{equation}
Once again, because the roots appear symmetrically, these formulae are rational functions of the coefficients of the cubic curve \eqref{eq:x3y3}.
If we extend the basis in groups of three like $\{\space 1,x,y \space:\space x^2,xy,y^2 \space:\space x^3,x^2y,xy^2 \space:\space x^4,x^3y,x^2y^2\space\ldots\space\}$ then we have $\{\space 0,3,3 \space:\space 6,6,6 \space:\space 9,9,9 \space:\space 12,12,12\space\ldots\space\}$ poles. So we can extend \eqref{eq:x3y3add2} etc. to $3n$ variables. Note that $y^3$ etc. don't appear in the basis because they are linearly dependent on preceding elements due to relation \eqref{eq:x3y3}.
When we remove the cubic terms in \eqref{eq:x3y3} by putting $a_{30} = a_{03} = a_{00} = 0$ it reduces to \begin{equation} \label{eq:x2y2reduced} a_{21} x^2 y + a_{12} x y^2 + a_{20} x^2 + a_{11} x y + a_{02} y^2 + a_{10} x + a_{01} y \space = \space 0 \end{equation} The parameterising elliptic functions in this case are order 2, each has two simple poles, one of which they share, and two simple zeroes, one of which they share. That is a total of three simple poles as in the general case. The level 2 addition formula reduces to the simpler form
\begin{equation} x_3 \space = \space \frac 1 {x_1 x_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ a_{02} & 0 & -a_{01} \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & a_{12} & -a_{21} \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 1 & 0 \\ \end{vmatrix}},\qquad\qquad y_3 \space = \space \frac 1 {y_1 y_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ a_{20} & -a_{10} & 0 \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & 0 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & a_{12} & -a_{21} \\ \end{vmatrix} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 0 & 0 & 1 \\ \end{vmatrix}} \end{equation}
Now we extend these formulae to quartic curves. We start with the curve \begin{equation} \label{eq:y2x4} y^2 \space + \space \left(px^2 \space + \space qx \space + r\right)y \space = \space ax^4 \space + \space bx^3 \space + \space cx^2 \space + \space dx \space + \space e \end{equation} This curve is genus one, provided its discriminant with respect to $y$ is a quartic polynomial in $x$ with 4 distinct roots, or a cubic polynomial in $x$ with 3 distinct roots. In this case the parameterising elliptic function $f$ is order 2 with two simple poles. And $g$ is order 4 with two double poles at the same location. Therefore $1,f,f^2,g$ form a basis for the four dimensional space of elliptic functions with poles of order at most two at these two locations. And the extended Frobenius-Stickelberger formula gives the following level 1 addition formula \begin{equation} \label{eq:y2x4add1} \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & x_4 & x_4^2 & y_4 \\ \end{vmatrix} \space = \space 0 \end{equation} Unlike the previous cubic curves this formula involves four points. And it implies these four points lie, not on a straight line but, on a parabola. The equation of that parabola is simply \begin{equation} \label{eq:parabola} \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & x & x^2 & y \\ \end{vmatrix} \space = \space 0 \end{equation} Let $\alpha_1,\alpha_2,\beta_1,\beta_2,e_1,e_2,e_3,e_4$ be the roots of the following polynomials \begin{equation} \begin{aligned} t^2 + p t - a &= 0 \\ t^2 + r t - e &= 0 \\ a t^4 + b t^3 + c t^2 + d t + e &= 0 \end{aligned} \end{equation} Computing the level 2 addition formula gives
\begin{equation} \label{eq:y2x4add2} x_4 \space = \space \frac 1 {x_1 x_2 x_3} \frac {\begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & 0 & 0 & \beta_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & 0 & 0 & \beta_2 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 0 & 0 & 1 & \alpha_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 0 & 0 & 1 & \alpha_2 \\ \end{vmatrix}}, \qquad\qquad y_4 \space = \space \frac {a^2} {y_1 y_2 y_3} \frac {\begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & e_1 & e_1^2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & e_2 & e_2^2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & e_3 & e_3^2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & e_4 & e_4^2 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 0 & 0 & 1 & \alpha_1 \\ \end{vmatrix}^2 \thinspace \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 0 & 0 & 1 & \alpha_2 \\ \end{vmatrix}^2} \end{equation}
And again the roots appear symmetrically implying these formulae are rational functions of the coefficients of \eqref{eq:y2x4}.
If we extend the basis in groups of two like $\{\space 1,x \space:\space x^2,y \space:\space x^3,xy \space:\space x^4,x^2y \space:\space \ldots\space\}$ then we have $\{\space 0,2 \space:\space 4,4 \space:\space 6,6 \space:\space 8,8 \space:\space\ldots\space\}$ poles. So we can extend \eqref{eq:y2x4add2} etc. to $2n$ variables. Note that $y^2$ and higher monomials don't appear in the basis because they are linearly dependent on preceding elements due to relation \eqref{eq:y2x4}.
Putting $p = q = r = 0$ in \eqref{eq:y2x4} gives a four point addition formula for the quartic curve \begin{equation} y^2 \space = \space ax^4 \space + \space bx^3 \space + \space cx^2 \space + \space dx \space + \space e \end{equation} given by \eqref{eq:y2x4add2} with $\alpha_1,\alpha_2 = \pm \sqrt{a}$ and $\beta_1,\beta_2 = \pm \sqrt{e}$.
Now let us examine the genus one quartic curve in which $x$ and $y$ appear individually with degree at most two. \begin{equation} \label{eq:x2y2} a_{22} x^2 y^2 + a_{21} x^2 y + a_{20} x^2 + a_{12} x y^2 + a_{11} x y + a_{10} x + a_{02} y^2 + a_{01} y + a_{00} \space = \space 0 \end{equation} The parameterising elliptic functions $f$ and $g$ are both of order 2 and have two poles none of which are common. Therefore $1,f,g,fg$ is a basis for the four dimensional vector space of elliptic functions which have a pole of order at most one at each of the four locations. By the extended Frobenius Stickelberger formula the level 1 addition formula is given by \begin{equation} \label{eq:x2y2add1} \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & x_4 & y_4 & x_4 y_4 \\ \end{vmatrix} \space = \space 0 \end{equation} Therefore if four points satisfy the addition formula they will lie on a hyperbola given by \begin{equation} \label{eq:hyperbola} \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & x & y & x y \\ \end{vmatrix} \space = \space 0 \end{equation} Let $\alpha_1,\alpha_2$, $\beta_1,\beta_2$, $\gamma_1,\gamma_2$ and $\delta_1,\delta_2$ be the respective roots of the four quadratic equations \begin{equation} \begin{aligned} a_{02} t^2 \space + \space a_{01} t \space + \space a_{00} \space = \space 0 \\ a_{22} t^2 \space + \space a_{21} t \space + \space a_{20} \space = \space 0 \\ a_{20} t^2 \space + \space a_{10} t \space + \space a_{00} \space = \space 0 \\ a_{22} t^2 \space + \space a_{12} t \space + \space a_{02} \space = \space 0 \\ \end{aligned} \end{equation} then the level 2 addition formula are given by
\begin{equation} \label{eq:x2y2add2} x_4 \space = \space \frac {a_{02}} {a_{22} x_1 x_2 x_3} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & 0 & \alpha_1 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & 0 & \alpha_2 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 0 & 1 & 0 & \beta_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 0 & 1 & 0 & \beta_2 \\ \end{vmatrix}},\qquad y_4 \space = \space \frac {a_{20}} {a_{22} y_1 y_2 y_3} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & \gamma_1 & 0 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 1 & \gamma_2 & 0 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 0 & 0 & 1 & \delta_1 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & x_3 & y_3 & x_3 y_3 \\ 0 & 0 & 1 & \delta_2 \\ \end{vmatrix}} \end{equation}
As before these equations are symmetric in the roots and therefore are rational functions of the coefficients of \eqref{eq:x2y2}.
If we extend the basis in groups of four like $\{\space 1,x,y,xy\space:\space x^2,x^2y,xy^2,y^2\space:\space x^3,x^3y,xy^3,y^3\space\ldots\space\}$ then we have $\{\space 0,2,4,4\space:\space 6,8,8,8\space:\space 10,12,12,12\space\ldots\space\}$ poles. So we can extend \eqref{eq:x2y2add1} etc. to $4n$ variables. Note that $x^2y^2$ and higher monomials don't appear in the basis because they are linearly dependent on preceding elements due to relation \eqref{eq:x2y2}.
Now let's look at a genus one sextic curve with three-fold complex multiplication \begin{equation} \label{eq:y3x6} y^3 \space = \space \left(a x^3 \space + b x^2 \space + \space c x \space + \space d\right)^2 \end{equation} From this equation we may infer that the parameterising elliptic function for $y$ is always a perfect square ie. it has an exact square root. \begin{equation} \label{eq:exactroot} \sqrt{y} \space = \space \frac {a x^3 \space + b x^2 \space + \space c x \space + \space d} {y} \end{equation} Therefore $x$ and $\sqrt{y}$ satisfy \eqref{eq:x3y3} and the level 1 addition formula is \begin{equation} \label{eq:y3x6add1} \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & x_3 & \sqrt{y_3} \\ \end{vmatrix} \space = \space 0 \end{equation} So if three points satisfy the addition formula then they lie on a parabola tangential to the $x$-axis. And from \eqref{eq:x3y3add2} the level 2 formula is
\begin{equation} \label{eq:y3x6add2} x_3 \space = \space \frac 1 {x_1 x_2} \frac {\begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & 0 & \sqrt[3]{d} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & 0 & \zeta \thinspace\sqrt[3]{d} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & 0 & \zeta^{\small 2} \thinspace\sqrt[3]{d} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \sqrt[3]{a} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \zeta \thinspace\sqrt[3]{a} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \zeta^{\small 2} \thinspace\sqrt[3]{a} \\ \end{vmatrix}}, \qquad\qquad y_3 \space = \space \left[ \frac {a} {\sqrt{y_1 y_2}} \frac {\begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & e_1 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & e_2 & 0 \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 1 & e_3 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \sqrt[3]{a} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \zeta \thinspace\sqrt[3]{a} \\ \end{vmatrix} \space \begin{vmatrix} 1 & x_1 & \sqrt{y_1} \\ 1 & x_2 & \sqrt{y_2} \\ 0 & 1 & \zeta^{\small 2} \thinspace\sqrt[3]{a} \\ \end{vmatrix}} \right]^2 \end{equation}
where $\zeta$ is a primitive cube root of unity and $e_1,e_2,e_3$ are the three roots of the right hand side of \eqref{eq:y3x6}.
As usual, because of \eqref{eq:exactroot} and the fact that the other roots appear symmetrically, $x_3$ is a rational function of $x_1,y_1,x_2,y_2,a,b,c,d$. And $y_3$ is not just a rational function, but the square of a rational function.
Finally let's look at the Jacobian elliptic functions $\sn$, $\cn$ and $\dn$.
Instead of viewing these functions as parameterising a curve in two variables, we can view them as parameterising a space curve in three dimensions given by the intersection of the two surfaces \begin{equation} \label{eq:x2y2z2} x^2 \space + \space y^2 \space = \space 1\quad \text{and} \quad k^2 x^2 \space + \space z^2 \space = \space 1 \end{equation} The paramerisation is given by \begin{equation} x = \sn(u), \qquad y = \cn(u), \qquad z = \dn(u) \end{equation} The level 1 addition formula in this case can also be derived from the extended Frobenius Stickelberger formula. But it requires a more detailed knowledge of the position of the poles and periods of the Jacobian elliptic functions and so won't be elaborated here. It is given by \begin{equation} \label{eq:x2y2z2add1} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \\ \end{vmatrix} \space = \space 0 \end{equation} Instead of having three points in two dimensions lying on a line, like the cubic curve, we have four points in three dimensions lying on a plane given by \begin{equation} \label{eq:plane} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x & y & z & 1 \\ \end{vmatrix} \space = \space 0 \end{equation}
The corresponding level 2 addition formula for $x_4$ is then given by multiplying the four $y_4,z_4$-conjugates of \eqref{eq:x2y2z2add1} together and solving for $x_4$. The formulae for $y_4$ and $z_4$ are derived similarly.
\begin{equation} \label{eq:x2y2z2add2x}
x_4 \space = \space \frac 1 {x_1 x_2 x_3} \frac
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
0 & 1 & 1 & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
0 & -1 & 1 & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
0 & 1 & -1 & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
0 & -1 & -1 & 1 \\
\end{vmatrix}}
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & -ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & -ik & 0 \\
\end{vmatrix}}
\end{equation}
\begin{equation} \label{eq:x2y2z2add2y}
y_4 \space = \space \frac 1 {y_1 y_2 y_3} \frac
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & 0 & k' & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
-1 & 0 & k' & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & 0 & -k' & 1 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
-1 & 0 & -k' & 1 \\
\end{vmatrix}}
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & -ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & -ik & 0 \\
\end{vmatrix}}
\end{equation}
\begin{equation} \label{eq:x2y2z2add2z}
z_4 \space = \space \frac 1 {z_1 z_2 z_3} \frac
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & k' & 0 & k \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
-1 & k' & 0 & k \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -k' & 0 & k \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
-1 & -k' & 0 & k \\
\end{vmatrix}}
{\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & i & -ik & 0 \\
\end{vmatrix} \space
\begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
1 & -i & -ik & 0 \\
\end{vmatrix}}
\end{equation}
where as usual $k^2 + {k'}^2 = 1$.
As with the other level 2 formulae these functions are symmetric in the roots and are therefore rational functions of the coefficents of \eqref{eq:x2y2z2} ie. of $k^2$.
They can be interpreted as addition formulae for $\sn(u_1+u_2+u_3)$, $\cn(u_1+u_2+u_3)$ and $\dn(u_1+u_2+u_3)$.
By definition $\sn(0) = 0$ and $\cn(0) = \dn(0) = 1$. Therefore if we put $(x_3,y_3,z_3) = (0,1,1)$ in \eqref{eq:x2y2z2add2x}, \eqref{eq:x2y2z2add2y} and \eqref{eq:x2y2z2add2z} we should be able to obtain the usual addition formula for $\sn$, $\cn$ and $\dn$. I haven't verified this.
References
[1] A Normal Form For Elliptic Curves Bull. Amer. Math. Soc. 44 (2007), 393-422
[2] Zur Theorie der elliptischen Functionen J. reine angew. Math 83 (1877), 175–179