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X2Y2

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 polynomial of degree 2 in $x$ and $y$. Along the way, we confirm that the singular point eliminant is the square of the modular discriminant and derive differential equations for the parameterising elliptic functions. We calculate the $g_2$ and $g_3$ invariants in terms of the coefficients of $F$ and in terms of the apolar invariants of the coefficient polynomials of $F$.

Let $F(x,y)$ be a polynomial of degree 2 in both $x$ and $y$. \begin{equation} \label{eq:F} F(x,y) = 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} \end{equation} In general the curve $F(x,y) = 0$ has genus 1. We want to compute the elliptic $g_2$ and $g_3$ invariants of this curve as a function of the coefficients $a_{ij}$.

Singular Point Eliminant

A simple condition for the curve \eqref{eq:F} to degenerate to genus 0 is that there exist a singular point on the curve where both partial derivatives of $F$ vanish. For such a point to exist the eliminant with respect to $x$ and $y$ of \begin{equation} F = 0,\quad\frac{\partial F}{\partial x} = 0,\quad\frac{\partial F}{\partial y} = 0 \end{equation} must be zero. The eliminant can be calculated using resultants as follows: \begin{equation} \Delta_{xy} = \resultant_y(\resultant_x(F, \frac{\partial F}{\partial x}), \space \resultant_x(F, \frac{\partial F}{\partial y})) \end{equation} where $\resultant_x(F,G)$ denotes the resultant of $F$ and $G$ with respect to $x$. This formula is well beyond manual calculation but can be performed in a few seconds using CAS.

$\Delta_{xy} = \left( \space 256 a_{00}^{5}a_{02}a_{20}a_{22}^{5} - 64a_{00}^{5}a_{02}a_{21}^{2}a_{22}^{4} - \ldots + a_{02}^{2}a_{10}^{2}a_{11}^{4}a_{12}^{2}a_{20}^{2} \space \right)^2$

When I first looked at this I was wondering why I had never seen a formula for this in a textbook. Click on the formula and you will see why.

$\Delta_{xy}$ is the square of the modular discriminant $\Delta$ for this curve, see Eq. (2.11) in [1]. Fortunately there is an easier path to the explicit calculation of the $g_2$ and $g_3$ invariants.

Parameterising the circle

Consider the special case of the curve $F = 0$, the circle \begin{equation} \label{eq:circle} x^2 + y^2 = 1 \end{equation} It can be parameterised by a pair of circular functions \begin{equation} x = {\cos\theta}, \quad y = {\sin\theta} \end{equation} These functions satisfy a pair of simultaneous differential equations \begin{equation} \frac {dx} {d\theta} = -y, \quad \frac {dy} {d\theta} = x \end{equation} With the help of \eqref{eq:circle} they can be converted to two ordinary differential equations \begin{equation} \left(\frac {dx} {d\theta} \right)^2 = 1 - x^2, \quad \left(\frac {dy} {d\theta} \right)^2 = 1 - y^2 \end{equation}

Parameterising $F(x,y) = 0$

A similar process can be applied to any curve of genus 1. Start by constructing the holomorphic differential on the curve. Then differential equations for parameterising elliptic functions can be obtained. The holomorphic differential on $F$ is given by \begin{equation} {dz} = \frac 1 {\large {\frac{\partial F}{\partial y}}} {dx} = - \frac 1 {\large {\frac{\partial F}{\partial x}}} {dy} \end{equation} This is equivalent to a pair of simultaneous differential equations for the parametrising elliptic functions \begin{equation} \label{eq:simdiff} \frac {dx} {dz} = \frac{\partial F}{\partial y}, \quad \frac {dy} {dz} = -\frac{\partial F}{\partial x} \end{equation} with known integral $F = 0$. They can be converted to an ordinary differential equation for $x$ by computing the eliminant with respect to $y$ of \begin{equation} {F = 0},\quad {\frac {dx} {dz} = \frac{\partial F}{\partial y}} \end{equation} and similarly for $y$ which gives a pair of ordinary differential equations for the parameterising elliptic functions \begin{equation}\label{eq:dxdy} \left(\frac {dx} {dz} \right)^2 = \discriminant_y(F), \quad \left(\frac {dy} {dz} \right)^2 = \discriminant_x(F) \end{equation} where $\discriminant_y(F)$ denotes the discriminant of $F$ with respect to $y$.

Boundary Conditions

To complete the parameterisation, boundary conditions $x(0) = x_0$ and $y(0) = y_0$ are required for the differential equations \eqref{eq:dxdy}. We are free to chose the value of either $x_0$ or $y_0$, but not both, because it is required that $F(x_0,y_0) = 0$.

We can choose $x_0$ so that $x(z)$ is even, that is $x(z) = x(-z)$. This is done by choosing $x_0$ to be a root of $\discriminant_y(F) = 0$ and hence $\displaystyle \frac{\partial F}{\partial y} (x_0,y_0) = 0$.

In this case $y(z)$ can never be even, because that would imply \begin{equation*} F(x_0,y_0) = \frac {\partial F}{\partial x} (x_0,y_0) = \frac {\partial F}{\partial y} (x_0,y_0) = 0 \end{equation*} and $F$ would have a singular point.

The other way to see that an even-even parameterisation is impossible is because if both uniformising functions are even they can be transformed into one another by a Möbius transform and therefore satisfy a degree 1 equation $Axy + Bx + Cy + D = 0$ and so $F$ would be reducible.

This can also be stated in terms on the centres of the uniformising functions: $\centre(x) \not\equiv \centre(y) \mod \tfrac 1 2 \Omega$ (because order 2 elliptic functions have even symmetry about points on the half-period grid).

Expanding

Now write $F(x,y)$ as polynomial in $x$ and as a polynomial in $y$. Put \begin{aligned} P_0(x) &= a_{20}x^2 + a_{10}x + a_{00} \\ P_1(x) &= a_{21}x^2 + a_{11}x + a_{01} \\ P_2(x) &= a_{22}x^2 + a_{12}x + a_{02} \\ \\ Q_0(y) &= a_{02}y^2 + a_{01}y + a_{00} \\ Q_1(y) &= a_{12}y^2 + a_{11}y + a_{10} \\ Q_2(y) &= a_{22}y^2 + a_{21}y + a_{20} \end{aligned} Then \begin{equation} F(x,y) = P_2(x) y^2 + P_1(x) y + P_0(x) = Q_2(y) x^2 + Q_1(y) x + Q_0(y) \end{equation} Substituting the $P_i(X)$ and $Q_i(Y)$ into \eqref{eq:dxdy} gives ordinary differential equations for the two second order parametrising elliptic functions.

\begin{equation} \label{eq:dxdy2} \left(\frac {dx} {dz} \right)^2 = P_1(x)^2 - 4 P_0(x) P_2(x), \quad \left(\frac {dy} {dz} \right)^2 = Q_1(y)^2 - 4 Q_0(y) Q_2(y) \end{equation}

Expanding \eqref{eq:dxdy2} gives \begin{multline} \label{eq:dx} \left(\frac {dx} {dz} \right)^2 = (-4a_{20}a_{22} + a_{21}^2)x^4 + (-4a_{10}a_{22} + 2a_{11}a_{21} - 4a_{12}a_{20})x^3 + (-4a_{00}a_{22} + 2a_{01}a_{21} - 4a_{02}a_{20} - 4a_{10}a_{12} + a_{11}^2)x^2 + \\ (-4a_{00}a_{12} + 2a_{01}a_{11} - 4a_{02}a_{10})x + (-4a_{00}a_{02} + a_{01}^2) \end{multline} \begin{multline} \label{eq:dy} \left(\frac {dy} {dz} \right)^2 = (-4a_{02}a_{22} + a_{12}^2)y^4 + (-4a_{01}a_{22} - 4a_{02}a_{21} + 2a_{11}a_{12})y^3 + (-4a_{00}a_{22} - 4a_{01}a_{21} - 4a_{02}a_{20} + 2a_{10}a_{12} + a_{11}^2)y^2 + \\ (-4a_{00}a_{21} - 4a_{01}a_{20} + 2a_{10}a_{11})y + (-4a_{00}a_{20} + a_{10}^2) \end{multline}

Computing $g_2$, $g_3$ and $j$

From Y2R4 we have \begin{equation} \bigint dz = \bigint \frac 1 {\sqrt{P_1(x)^2 - 4 P_0(x) P_2(x)}} dx = \bigint \frac 1 {\sqrt{a x^4 + b x^3 + c x^2 + d x + e}} dx = \bigint \frac 1 {\sqrt{4u^3 - g_2 u - g_3}} du \end{equation} where \begin{equation} \label{eq:g2g3} g_2 = \tfrac 1 {12} (12 a e - 3 b d + c^2),\quad g_3 = \tfrac 1 {432} (72 a c e - 27 a d^2 - 27 b^2 e + 9 b c d - 2 c^3) \end{equation}

Substituting the coefficients of \eqref{eq:dx}, or the coefficients of \eqref{eq:dy}, into \eqref{eq:g2g3} gives \begin{aligned} g_2 = \tfrac 1 {12} &\big( {\phantom +} 16a_{00}^2a_{22}^2 - 16a_{00}a_{01}a_{21}a_{22} + 224a_{00}a_{02}a_{20}a_{22} - 48a_{00}a_{02}a_{21}^2 - 16a_{00}a_{10}a_{12}a_{22} - 8a_{00}a_{11}^2a_{22} + 24a_{00}a_{11}a_{12}a_{21} - 48a_{00}a_{12}^2a_{20} \\ & - 48a_{01}^2a_{20}a_{22} + 16a_{01}^2a_{21}^2 - 16a_{01}a_{02}a_{20}a_{21} + 24a_{01}a_{10}a_{11}a_{22} - 16a_{01}a_{10}a_{12}a_{21} - 8a_{01}a_{11}^2a_{21} + 24a_{01}a_{11}a_{12}a_{20} + 16a_{02}^2a_{20}^2 \\ & - 48a_{02}a_{10}^2a_{22} + 24a_{02}a_{10}a_{11}a_{21} - 16a_{02}a_{10}a_{12}a_{20} - 8a_{02}a_{11}^2a_{20} + 16a_{10}^2a_{12}^2 - 8a_{10}a_{11}^2a_{12} + a_{11}^4 \big) \\ \\ g_3 = \tfrac 1 {216} &\big( {\phantom +}64a_{00}^3a_{22}^3 - 96a_{00}^2a_{01}a_{21}a_{22}^2 - 2112a_{00}^2a_{02}a_{20}a_{22}^2 + 576a_{00}^2a_{02}a_{21}^2a_{22} - 96a_{00}^2a_{10}a_{12}a_{22}^2 - 48a_{00}^2a_{11}^2a_{22}^2 + 144a_{00}^2a_{11}a_{12}a_{21}a_{22} + 576a_{00}^2a_{12}^2a_{20}a_{22} \\ & - 216a_{00}^2a_{12}^2a_{21}^2 + 576a_{00}a_{01}^2a_{20}a_{22}^2 - 96a_{00}a_{01}^2a_{21}^2a_{22} + 960a_{00}a_{01}a_{02}a_{20}a_{21}a_{22} - 288a_{00}a_{01}a_{02}a_{21}^3 + 144a_{00}a_{01}a_{10}a_{11}a_{22}^2 - 48a_{00}a_{01}a_{10}a_{12}a_{21}a_{22} \\ & - 24a_{00}a_{01}a_{11}^2a_{21}a_{22} - 720a_{00}a_{01}a_{11}a_{12}a_{20}a_{22} + 144a_{00}a_{01}a_{11}a_{12}a_{21}^2 + 144a_{00}a_{01}a_{12}^2a_{20}a_{21} - 2112a_{00}a_{02}^2a_{20}^2a_{22} + 576a_{00}a_{02}^2a_{20}a_{21}^2 + 576a_{00}a_{02}a_{10}^2a_{22}^2 \\ & - 720a_{00}a_{02}a_{10}a_{11}a_{21}a_{22} + 960a_{00}a_{02}a_{10}a_{12}a_{20}a_{22} + 144a_{00}a_{02}a_{10}a_{12}a_{21}^2 + 480a_{00}a_{02}a_{11}^2a_{20}a_{22} + 72a_{00}a_{02}a_{11}^2a_{21}^2 - 720a_{00}a_{02}a_{11}a_{12}a_{20}a_{21} + 576a_{00}a_{02}a_{12}^2a_{20}^2 \\ & - 96a_{00}a_{10}^2a_{12}^2a_{22} - 24a_{00}a_{10}a_{11}^2a_{12}a_{22} + 144a_{00}a_{10}a_{11}a_{12}^2a_{21} - 288a_{00}a_{10}a_{12}^3a_{20} + 12a_{00}a_{11}^4a_{22} - 36a_{00}a_{11}^3a_{12}a_{21} + 72a_{00}a_{11}^2a_{12}^2a_{20} - 288a_{01}^3a_{20}a_{21}a_{22} \\ & + 64a_{01}^3a_{21}^3 + 576a_{01}^2a_{02}a_{20}^2a_{22} - 96a_{01}^2a_{02}a_{20}a_{21}^2 - 216a_{01}^2a_{10}^2a_{22}^2 + 144a_{01}^2a_{10}a_{11}a_{21}a_{22} + 144a_{01}^2a_{10}a_{12}a_{20}a_{22} - 96a_{01}^2a_{10}a_{12}a_{21}^2 + 72a_{01}^2a_{11}^2a_{20}a_{22} \\ & - 48a_{01}^2a_{11}^2a_{21}^2 + 144a_{01}^2a_{11}a_{12}a_{20}a_{21} - 216a_{01}^2a_{12}^2a_{20}^2 - 96a_{01}a_{02}^2a_{20}^2a_{21} + 144a_{01}a_{02}a_{10}^2a_{21}a_{22} - 720a_{01}a_{02}a_{10}a_{11}a_{20}a_{22} + 144a_{01}a_{02}a_{10}a_{11}a_{21}^2 \\ & - 48a_{01}a_{02}a_{10}a_{12}a_{20}a_{21} - 24a_{01}a_{02}a_{11}^2a_{20}a_{21} + 144a_{01}a_{02}a_{11}a_{12}a_{20}^2 + 144a_{01}a_{10}^2a_{11}a_{12}a_{22} - 96a_{01}a_{10}^2a_{12}^2a_{21} - 36a_{01}a_{10}a_{11}^3a_{22} - 24a_{01}a_{10}a_{11}^2a_{12}a_{21} \\ & + 144a_{01}a_{10}a_{11}a_{12}^2a_{20} + 12a_{01}a_{11}^4a_{21} - 36a_{01}a_{11}^3a_{12}a_{20} + 64a_{02}^3a_{20}^3 + 576a_{02}^2a_{10}^2a_{20}a_{22} - 216a_{02}^2a_{10}^2a_{21}^2 + 144a_{02}^2a_{10}a_{11}a_{20}a_{21} - 96a_{02}^2a_{10}a_{12}a_{20}^2 - 48a_{02}^2a_{11}^2a_{20}^2 \\ & - 288a_{02}a_{10}^3a_{12}a_{22} + 72a_{02}a_{10}^2a_{11}^2a_{22} + 144a_{02}a_{10}^2a_{11}a_{12}a_{21} - 96a_{02}a_{10}^2a_{12}^2a_{20} - 36a_{02}a_{10}a_{11}^3a_{21} - 24a_{02}a_{10}a_{11}^2a_{12}a_{20} + 12a_{02}a_{11}^4a_{20} + 64a_{10}^3a_{12}^3 - 48a_{10}^2a_{11}^2a_{12}^2 \\ & + 12a_{10}a_{11}^4a_{12} - a_{11}^6 \big) \end{aligned}

Thus we have explicitly calculated the $g_2$ and $g_3$ invariants for the curve \eqref{eq:F} in terms of it's coefficients. Technically it is more correct to say we have calculated them for the differential $dz$ on the curve. We can verify that the discriminant of \eqref{eq:dx} is equal to the discriminant of \eqref{eq:dy} and \begin{equation} \Delta = g_2^3 - 27 g_3^2 = \tfrac 1 {256} \discriminant_x(\discriminant_y(F)) = \tfrac 1 {256} \discriminant_y(\discriminant_x(F)) \end{equation} and finally the $j$-invariant is \begin{equation} j = 1728 {g_2^3 \over \Delta} \end{equation} We can also verify that the eliminant calculated above is the square of the modular discriminant $\Delta_{xy} = \Delta^2$.

Apolar Invariants

If a Möbius transformation is applied to $x$ in \eqref{eq:F} then the parameterising elliptic function for $x$ will be similarly transformed, but the parameterising elliptic function for $y$ will be unchanged. Therefore the coefficients of equation \eqref{eq:dy} must be simultaneous invariants of the $P_i(x)$ polynomials. In fact they can be very simply expressed in terms of the 6 apolar invariants a.k.a. the second transvectants. The $k$-th transvectant for two homogeneous polynomials (with the leading constant factor omitted), is given by (see for example Pg. 16 of [2]) \begin{equation} \label{eq:transvectant} \transvectant{A,B}_k = \sum_{i=0}^k (-1)^i {k\choose i} \frac {\partial^k A}{ {\partial x}^{k-i} {\partial y}^i} \frac {\partial^k B}{ {\partial x}^i {\partial y}^{k-i} } \end{equation} Put $p_{ij} = \tfrac 1 2 \transvectant{P_i,P_j}_2$ then \begin{aligned} p_{00} &= 4a_{02}a_{22} - a_{12}^2 \\ p_{11} &= 4a_{01}a_{21} - a_{11}^2 \\ p_{22} &= 4a_{00}a_{20} - a_{10}^2 \\ p_{01} &= 2a_{22}a_{01} - a_{12}a_{11} + 2a_{02}a_{21} \\ p_{02} &= 2a_{22}a_{00} - a_{12}a_{10} + 2a_{02}a_{20} \\ p_{12} &= 2a_{21}a_{00} - a_{11}a_{10} + 2a_{01}a_{20} \end{aligned} and \eqref{eq:dy} becomes \begin{equation} \left(\frac {dy} {dz} \right)^2 = - \left( p_{00}y^4 + 2p_{01}y^3 + (2p_{02} + p_{11})y^2 + 2p_{12}y + p_{22} \right) \end{equation} And a simple substitution into \eqref{eq:g2g3} gives the $g_2$ and $g_3$ invariants in terms of the apolar invariants of $P_i(x)$

\begin{aligned} g_2 &= \tfrac 1 {12} ( 12p_{00}p_{22} - 12p_{01}p_{12} + 4p_{02}^2 + 4p_{02}p_{11} + p_{11}^2 ) \\\\ g_3 &= \tfrac 1 {216} (-72p_{00}p_{02}p_{22} - 36p_{00}p_{11}p_{22} + 54p_{00}p_{12}^2 + 54p_{01}^2p_{22} - 36p_{01}p_{02}p_{12} - 18p_{01}p_{11}p_{12} + 8p_{02}^3 + 12p_{02}^2p_{11} + 6p_{02}p_{11}^2 + p_{11}^3 ) \end{aligned}

Similarly the $g_2$ and $g_3$ invariants can be expressed in terms of the apolar invariants of $Q_i(y)$.

Determinant

The determinant of the $a_{ij}$ is also a simultaneous invariant of the $P_i(x)$. It can also be expressed by a transvectant formula $p_{012} = \tfrac 1 8 \transvectant{P_0,\transvectant{P_1,P_2}_1}_2$ \begin{equation} p_{012} = a_{00}a_{11}a_{22} - a_{00}a_{12}a_{21} - a_{01}a_{10}a_{22} + a_{01}a_{12}a_{20} + a_{02}a_{10}a_{21} - a_{02}a_{11}a_{20} \end{equation} Unlike the apolar invariants it is simultaneously a simultaneous invariant of the$P_i(x)$ and the $Q_i(y)$. That is $p_{012} = \tfrac 1 8 \transvectant{Q_0,\transvectant{Q_1,Q_2}_1}_2$. Because of this it might be expected to appear in the formulae for the $g_2$ and $g_3$ invariants. However since it is of degree 3 and skew-symmetric it could only appear as a square in the $g_3$ formula. But there's a syzygy expressing it's square in terms of the 6 apolar invariants and it is therefore redundant. \begin{equation} 4p_{012}^2 = p_{00}p_{11}p_{22} - p_{00}p_{12}^2 - p_{01}^2p_{22} + 2p_{01}p_{02}p_{12} - p_{02}^2p_{11} \end{equation}

Birational Transformation

The differential equation for the parametrising elliptic function $x$ is equivalent to birationally mapping the curve $F(x,y) = 0$ to the curve \begin{equation} \label{vu} v^2 = P_1(u)^2 - 4 P_0(u) P_2(u) \end{equation} The birational mapping is given by $(u,v) \rightarrow (x,\frac {dx} {dz})$ which using \eqref{eq:simdiff} can be written \begin{equation} \label{uv} \begin{aligned} u &= x \\ v &= 2 y P_2(x) + P_1(x) \end{aligned} \end{equation} Substituting \eqref{uv} into \eqref{vu} recovers the original curve \begin{equation} y^2 P_2(x) + y P_1(x) + P_0(x) = 0 \end{equation}

Level 1 Addition Formula

A four variable addition formula for curve \eqref{eq:F} is easily obtained from the uniformising elliptic functions.

In general the two uniformising functions $x(z)$ and $y(z)$ each have two distinct poles and so the functions $1, x(z), y(z), x(z)y(z)$ form a basis for the four dimensional vector space of all elliptic functions with poles of order at most one, at each of those four points. By applying the Extended Frobenius-Stickelberger Formula we therefore have the following level one addition formula. \begin{equation} \label{eq:add4} \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} In the special case that $a_{22}=0$ the uniformising elliptic functions have one pole in common and the vector space reduces to three dimensions, and the addition formula can be reduced to \begin{equation} \label{eq:add3} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix} \space = \space 0 \end{equation} We can also get a similar reduction when one of the other corner coefficients $a_{20}, a_{02}, a_{00}$ vanishes by using the reciprocials of the uniformising functions.

Intersecting Hyperbola

Writing equation \eqref{eq:add4} as \begin{equation} G(x,y) \enspace = \enspace \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} \end{equation} we see that this formula can be viewed as the equation of an hyperbola intersecting the curve $F(x,y) = 0$ at three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$.

Eliminating $y$ between the two equations $F(x,y)=0$ and $G(x,y)=0$, using a resultant, we get a fourth degree polynomial in $x$. The four roots of this polynomial are the $x$-coordinates of the four points of intersection of these two curves.

Now since three points of intersection $\{(x_i,y_i): i=1,2,3\}$ are already known, we can infer the existance of a polynomial identity of the form \begin{equation} \label{eq:residx} \resultant_y(F(x,y),G(x,y)) \enspace = \enspace (x - x_1)(x - x_2)(x - x_3)(D x - N) \enspace + \enspace \sum_{i=1}^3 Q_i(x) F(x_i,y_i) \end{equation} where $D,N$ are polynomial functions in $x_1,x_2,x_3,y_1,y_2,y_3$ and $Q_1,Q_2,Q_3$ are fourth degree polynomials in $x$ whose coefficients are polynomial functions in $x_1,x_2,x_3,y_1,y_2,y_3$ and the $x$-coordinate of the fourth point of intersection is given by $x_4=N/D$.

This inference can be confirmed by computing the Normal Form of the LHS with respect to the Gröbner basis $\{F(x_i,y_i) : i=1,2,3\}$ using CAS. The computation yields explicit formulae for $N,D,Q_1,Q_2,Q_3$ which have hundreds of terms. From \eqref{eq:residx} under the condition that $F(x_i,y_i)=0$ for $i=1,2,3$ we therefore have \begin{equation} \label{eq:residx4} \mathfrak{R}(x) \enspace = \enspace D(x - x_1)(x - x_2)(x - x_3)(x - x_4) \end{equation} where $\mathfrak{R}(x) = \resultant_y(F(x,y),G(x,y))$. Then by equating coefficients we can obtain several different simple explicit formulae for $x_4$ in terms of resultants, for example \begin{equation} -\left[\frac {\coeff(\mathfrak{R}, x^3)} {\coeff(\mathfrak{R}, x^4)} \space + \space (x_1 + x_2 + x_3) \right], \qquad \frac 1 {x_1 + x_2 + x_3} \left[\frac {\coeff(\mathfrak{R}, x^2)} {\coeff(\mathfrak{R}, x^4)} \space - \space (x_1 x_2 + x_1 x_3 + x_2 x_3)\right], \qquad - \frac 1 {x_1 x_2 + x_1 x_3 + x_2 x_3} \left[\frac {\coeff(\mathfrak{R}, x)} {\coeff(\mathfrak{R}, x^4)} \space + \space x_1 x_2 x_3\right] \end{equation} and finally this formula where we simply evaluate \eqref{eq:residx4} at $x=0$ and $x=\infty$ to isolate the first and last coefficients

\begin{equation} \label{eq:resadd} x_4 \enspace = \enspace \frac 1 {x_1 x_2 x_3} \cdot \frac {\mathfrak{R}(0)} {\lim\limits_{t \rightarrow \infty} t^{-4} \thinspace \mathfrak{R}(t)} \qquad\qquad y_4 \enspace = \enspace \frac 1 {y_1 y_2 y_3} \cdot \frac {\mathfrak{S}(0)} {\lim\limits_{t \rightarrow \infty} t^{-4} \thinspace \mathfrak{S}(t)} \end{equation}

where $\mathfrak{S}(y) = \resultant_x(F(x,y),G(x,y))$

Level 2 Addition Formula

If the two algebraic $y$ roots of $F(x,y) = 0$ are $\psi_1(x)$ and $\psi_2(x)$ then the resultant $\mathfrak{R}$ may be written using the product formula \begin{equation} \label{eq:resx} \mathfrak{R}(x) \enspace = \enspace \operatorname{lead}_y(F(x,y)) \cdot G(x,\psi_1(x)) \cdot G(x,\psi_2(x)) \enspace = \enspace P_2(x) \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 & \psi_1(x) & x \psi_1(x) \\ \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 & x & \psi_2(x) & x \psi_2(x) \\ \end{vmatrix} \end{equation} where $\operatorname{lead}_y$ means the leading coefficient as a polynomial in $y$. And similarly if the two algebraic $x$ roots of $F(x,y) = 0$ are $\phi_1(y)$ and $\phi_2(y)$ then \begin{equation} \label{eq:resy} \mathfrak{S}(y) \enspace = \enspace \operatorname{lead}_x(F(x,y)) \cdot G(\phi_1(y),y) \cdot G(\phi_2(y),y) \enspace = \enspace Q_2(y) \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 & \phi_1(y) & y & \phi_1(y) y \\ \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 & \phi_2(y) & y & \phi_2(y) y \\ \end{vmatrix} \end{equation} By evaluating \eqref{eq:resadd} using \eqref{eq:resx} and \eqref{eq:resy}, we obtain the four variable level 2 addition formula \begin{equation} \label{eq:x4y4} x_4 \space = \space \frac {a_{02}} {a_{22} \thinspace 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} \thinspace 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} where $\alpha,\beta,\gamma,\delta$ are the four pairs of roots of $Q_0(\alpha) = 0, \space Q_2(\beta) = 0, \space P_0(\gamma) = 0, \space P_2(\delta) = 0$.

By partially expanding the determinants by their last row formula \eqref{eq:x4y4} may be written as rational functions in the coefficients of $F$

\begin{equation} \label{eq:x4y4rat} x_4 \space = \space \frac {a_{02} E_1^2 - a_{01} E_1 E_3 + a_{00} E_3^2} {x_1 x_2 x_3 \left(a_{22} E_2^2 - a_{21} E_2 E_4 + a_{20} E_4^2\right)},\qquad y_4 \space = \space \frac {a_{20} E_1^2 + a_{10} E_1 E_2 + a_{00} E_2^2} {y_1 y_2 y_3 \left(a_{22} E_3^2 + a_{12} E_3 E_4 + a_{02} E_4^2\right)} \end{equation}

where $E_1,E_2,E_3,E_4$ are the (unsigned) minors of the last row of \eqref{eq:add4}.

Geometric Interpretation

Suppose you have three distinct rational points $(x_1,y_1) \ldots (x_3,y_3)$ on curve \eqref{eq:F}, and the curve has rational coefficients. Pass an axis-aligned hyperbola, like $A + Bx + Cy + Dxy = 0$, through those three points. It will intersect the curve at a fourth point $(x_4,y_4)$ given by \eqref{eq:x4y4rat}, and that fourth point will also be a rational point.

Back To The Circle

This geometric interpretation can be applied to the circle to obtain a two parabola's theorem. Namely if you pass an $x$ parabola and a $y$ parabola through three points on a circle, they will both intersect the circle at the same fourth point. And the coordinates of the fourth point are rational functions of the coordinates of the first three points. For the unit circle the two parabola's through $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are given by the level 1 addition formulae \begin{equation} \label{eq:twoparabolas} \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,\quad \begin{vmatrix} 1 & y_1 & y_1^2 & x_1 \\ 1 & y_2 & y_2^2 & x_2 \\ 1 & y_3 & y_3^2 & x_3 \\ 1 & y & y^2 & x \\ \end{vmatrix} \space = \space 0 \end{equation} The coordinates of the fourth point $(x_4,y_4)$ are given by solving two linear equations (the level 2 addition formulae) for $x$ and $y$. \begin{equation*} (x_4,y_4) = \left(\frac {\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3} {\alpha_1 + \alpha_2 + \alpha_3}, \frac {\beta_1 y_1 + \beta_2 y_2 + \beta_3 y_3} {\beta_1 + \beta_2 + \beta_3}\right) \end{equation*} where \begin{gather*} \alpha_1 = (x_1-x_2)(x_1-x_3)(y_2-y_3)^2, \quad \alpha_2 = (x_2-x_1)(x_2-x_3)(y_1-y_3)^2, \quad \alpha_3 = (x_3-x_1)(x_3-x_2)(y_1-y_2)^2 \\\\ \beta_1 = (y_1-y_2)(y_1-y_3)(x_2-x_3)^2, \quad \beta_2 = (y_2-y_1)(y_2-y_3)(x_1-x_3)^2, \quad \beta_3 = (y_3-y_1)(y_3-y_2)(x_1-x_2)^2 \end{gather*} In fact it's not just true for an $x$ and $y$ parabola, it's true for any axis-aligned conic, as can be seen by taking an arbitrary linear combination of the two formulae \eqref{eq:twoparabolas}. \begin{equation} \label{eq:axisalignedconic} \begin{vmatrix} 1 & x_1 & y_1 & S x_1^2 \space + \space T y_1^2 \\ 1 & x_2 & y_2 & S x_2^2 \space + \space T y_2^2 \\ 1 & x_3 & y_3 & S x_3^2 \space + \space T y_3^2 \\ 1 & x & y & S x^2 \space + \space T y^2 \\ \end{vmatrix} \space = \space 0 \end{equation}

Edwards Curve

It's interesting to look a one special case of the general curve \eqref{eq:F} and then relate it to the above formulae. The Edwards[3] curve is given by \begin{equation} x^2 \space + \space y^2 \space = \space a^2 \space + \space a^2 x^2 y^2 \end{equation} and has the remarkably simple addition formulae \begin{equation} x_3 \enspace = \enspace \frac 1 a \cdot \frac {x_1 y_2 \space + \space x_2 y_1} {1 \space + \space x_1 x_2 y_1 y_2} \qquad\qquad y_3 \enspace = \enspace \frac 1 a \cdot \frac {y_1 y_2 \space - \space x_1 x_2} {1 \space - \space x_1 x_2 y_1 y_2} \end{equation} For reasons which will become apparent transform it slightly by putting $(x,y) \mapsto (ax,ay)$ and $k = a^2$, so that the curve becomes \begin{equation} \label{eq:edwards} k^2x^2y^2 \space - \space x^2 \space - \space y^2 \space + \space 1 \space = \space 0 \end{equation} with the simple addition formulae \begin{equation} \label{eq:eadd} x_3 \enspace = \enspace \frac {x_1 y_2 \space + \space x_2 y_1} {1 \space + \space k^2 x_1 x_2 y_1 y_2} \qquad\qquad y_3 \enspace = \enspace \frac {y_1 y_2 \space - \space x_1 x_2} {1 \space - \space k^2 x_1 x_2 y_1 y_2} \end{equation}

Parameterisation of Edwards Curve using Jacobian elliptic functions

Using equation \eqref{eq:dxdy2} and the identities $\sn^2 u \space + \space \cn^2 u = 1$ and $k^2\sn^2 u \space + \space \dn^2 u = 1$ gives the parameterisation of \eqref{eq:edwards} in terms of the Jacobian elliptic functions \begin{equation} \label{eq:jac} x \space = \space \sn u \qquad\qquad y \space = \space \frac {\cn u} {\dn u} \end{equation} The essential reason for this simple algebraic relation between these two functions is because they are both order two elliptic functions with respect to the same period lattice $[4K,2iK']$.

Addition Formulae for the Edwards Curve

Using this parametrisation, numerous addition formula can be deduced from various tables of addition formulae for Jacobian elliptic functions. For example the following simple addition formula can be deduced from these formulae for $\sn(u+v)$ and $\cn(u+v)$ / $\dn(u+v)$

\begin{equation} \label{eq:jadd} x_3 \enspace = \enspace \frac {x_1 y_1 \space + \space x_2 y_2} {x_1 x_2 \space + \space y_1 y_2} \qquad\qquad y_3 \enspace = \enspace \frac {x_1 y_1 \space - \space x_2 y_2} {x_1 y_2 \space - \space x_2 y_1} \end{equation}

The equivalence of \eqref{eq:eadd} and \eqref{eq:jadd} is easily confirmed by equating the two $x_3$ terms to give \begin{equation} x_2 y_2 \left(k^2 x_1^2 y_1^2 \space - \space x_1^2 \space - \space y_1^2 \space + \space 1\right) \enspace + \enspace x_1 y_1 \left(k^2 x_2^2 y_2^2 \space - \space x_2^2 \space - \space y_2^2 \space + \space 1\right) \enspace = \enspace 0 \end{equation} and the two $y_3$ terms to give \begin{equation} -x_2 y_2 \left(k^2 x_1^2 y_1^2 \space - \space x_1^2 \space - \space y_1^2 \space + \space 1\right) \enspace + \enspace x_1 y_1 \left(k^2 x_2^2 y_2^2 \space - \space x_2^2 \space - \space y_2^2 \space + \space 1\right) \enspace = \enspace 0 \end{equation} We can also reverse the flow and obtain this Jacobian addition formula which is not listed \begin{equation} \sn(u + v) \enspace = \enspace \frac {\sn u \cn v \dn u \space + \space \sn v \cn u \dn v} {\dn u \dn v \space + \space k^2 \sn u \sn v \cn u \cn v} \end{equation}

Parameterisation using Jacobian elliptic functions in Glaisher notation

Utilising the Glaisher notation we can write the parameterising functions more simply as \begin{equation} \label{eq:glaisher} x \space = \space \sn u \qquad\qquad y \space = \space \operatorname{cd} u \end{equation} and then observe that any pair of Glaisher functions $x = \operatorname{pq} u, \space y = \operatorname{rs} u$ whose four letters cover $n,s,c,d$ will parameterise a similar curve.

As observed here the Glaisher functions are essentially the squares root of an arbitrary second order elliptic function $f$ so we can also parametrise the curve \begin{equation} A x^2 y^2 \space + \space B x^2 \space + \space C y^2 \space + \space D \space = \space 0 \end{equation} like this \begin{equation} x \space = \space \sqrt{\frac {f(z) - e_1} {f(z) - e_2} } \qquad\qquad y \space = \space \sqrt{\frac {f(z) - e_3} {f(z) - e_4} } \end{equation} giving (with the coefficients are to be determined) \begin{equation} A(f - e_1)(f - e_3) \space + \space B (f - e_1)(f - e_4) \space + \space C (f - e_2)(f - e_3)\space + \space D (f - e_2)(f - e_4) \enspace = \enspace 0 \end{equation}

Three point addition formulae for the Edwards curve

The three point addition formulae can be obtained by substituting the coefficients of \eqref{eq:edwards} into \eqref{eq:x4y4} \begin{equation} \label{eq:sadd3} x_4 \space = \space \frac {1} {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 & 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 & -1 & 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 & k & 0 & 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 & -k & 0 & 1 \\ \end{vmatrix}}\qquad\qquad y_4 \space = \space \frac {1} {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 & 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 & -1 & 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 & k & 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 & -k & 1 \\ \end{vmatrix}} \end{equation} Putting $z=0$ in \eqref{eq:jac} gives $(x,y)=(0,1)$. Substituting these values for $(x_3,y_3)$ in \eqref{eq:sadd3} and noting $y=1+\bigO(x^2)$ near $x=0$ gives \begin{equation} \label{eq:sadd2} x_4 \space = \space \frac {1} {x_1 x_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 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 & 0 & 1 & 0 \\ 1 & 0 & -1 & 0 \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & 0 & 1 & 0 \\ 0 & k & 0 & 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 & 0 & 1 & 0 \\ 0 & -k & 0 & 1 \\ \end{vmatrix}}\qquad\qquad y_4 \space = \space \frac {1} {y_1 y_2} \frac {\begin{vmatrix} 1 & x_1 & y_1 & x_1 y_1 \\ 1 & x_2 & y_2 & x_2 y_2 \\ 1 & 0 & 1 & 0 \\ 1 & 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 & 0 & 1 & 0 \\ 1 & -1 & 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 & 0 & 1 & 0 \\ 0 & 0 & k & 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 & 0 & 1 & 0 \\ 0 & 0 & -k & 1 \\ \end{vmatrix}} \end{equation}

References

[1] J. C. Eilbeck, S. Matsutani, Y. Ônishi 2011 Addition formulae for Abelian functions associated with specialized curves. Phil. Trans. R. Soc. A (2011) 369, 1245–1263

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

[3] Harold M. Edwards 2007 A normal form for elliptic curves Bull. Amer. Math. Soc. 44 (2007), 393-422