Home Prev Next

In terms of the uniformising elliptic functions $f(z)$ and $g(z)$ of the curve $F(x,y)=0$, the symmetric level 2 addition formulae we have calculated in earlier sections have the form, when $n=3$ \begin{equation} \label{eq:symmetric0} f(z_3) \space = \space X(f(z_1),g(z_1),f(z_2),g(z_2))\qquad\qquad g(z_3) \space = \space Y(f(z_1),g(z_1),f(z_2),g(z_2)) \end{equation} where $z_1 + z_2 + z_3 = s$ is the "centre" point and $X$ and $Y$ are rational functions of their arguments and the coefficients of $F$.

It is convenient to write this addition formula as \begin{equation} \label{eq:symmetric} f(s - z_1 - z_2) \space = \space X(f(z_1),g(z_1),f(z_2),g(z_2))\qquad\qquad g(s - z_1 - z_2) \space = \space Y(f(z_1),g(z_1),f(z_2),g(z_2)) \end{equation}

From \eqref{eq:symmetric} we would like to derive a "classic" addition formulae of the form \begin{equation} \label{eq:classic} f(z_1 + z_2) \space = \space U(f(z_1),g(z_1),f(z_2),g(z_2))\qquad\qquad g(z_1 + z_2) \space = \space V(f(z_1),g(z_1),f(z_2),g(z_2)) \end{equation} where and $U$ and $V$ are rational functions.

If $s=0$ and $f$ and $g$ have even and odd symmetry respectively, as is the case for the Weiertrass curve with $f(z)=\wp(z)$ and $g(z)=\wp'(z)$, then it is trivially simple, using \eqref{eq:symmetric} we have \begin{equation} f(z_1 + z_2) \space = \space X(f(z_1),g(z_1),f(z_2),g(z_2))\qquad\qquad g(z_1 + z_2) \space = \space -Y(f(z_1),g(z_1),f(z_2),g(z_2)) \end{equation} and the classic and symmetric addition formulae are almost identical.

What is not immediately obvious is that we can carry out a similar process even when $s \ne 0$ and $f$ and $g$ have no special symmetries.

Substitute $z_1 \leftarrow s-z_1-z_2$ and $z_2 \leftarrow 0$ in \eqref{eq:symmetric} to get \begin{equation} \label{eq:cfs} \begin{aligned} f(z_1 + z_2) \space = \space X(f(s - z_1 - z_2),g(s - z_1 - z_2),f(0),g(0)) \\\\ g(z_1 + z_2) \space = \space Y(f(s - z_1 - z_2),g(s - z_1 - z_2),f(0),g(0)) \end{aligned} \end{equation} then apply \eqref{eq:symmetric} once more to the RHS of \eqref{eq:cfs} to get \begin{equation} \label{eq:classicfromsymmetric} \begin{aligned} f(z_1+z_2)\space &= \space X(X(f(z_1), g(z_1), f(z_2), g(z_2)), \thinspace Y(f(z_1), g(z_1), f(z_2), g(z_2)), \thinspace f(0), g(0)) \\\\ g(z_1+z_2)\space &= \space Y(X(f(z_1), g(z_1), f(z_2), g(z_2)), \thinspace Y(f(z_1), g(z_1), f(z_2), g(z_2)), \thinspace f(0), g(0)) \end{aligned} \end{equation} That is to say using the symmetric level 2 addition formula we are able to express $f(z_1+z_2)$ and $g(z_1+z_2)$ as rational functions of $f(z_1),g(z_1),f(z_2),g(z_2),f(0),g(0)$ and the coefficients of $F$.

Also observe that putting $z_1=z_2=0$ in \eqref{eq:symmetric} gives us the coordinates of the centre point $s$ in terms of the coordinates of the zero point $f(0),g(0)$ \begin{equation} \label{eq:centre} f(s) \space = \space X(f(0),g(0),f(0),g(0))\qquad\qquad g(s) \space = \space Y(f(0),g(0),f(0),g(0)) \end{equation} then by putting $z_1=z$ and $z_2=s$ in \eqref{eq:symmetric} gives us a "negation" formula \begin{equation} \label{eq:negation} f(-z) \space = \space X(f(z),g(z),f(s),g(s))\qquad\qquad g(-z) \space = \space Y(f(z),g(z),f(s),g(s)) \end{equation} That is to say using the symmetric level 2 addition formula we are able to express $f(-z)$ and $g(-z)$ as rational functions of $f(z),g(z),f(0),g(0)$ and the coefficients of $F$.

If the symmetric addition formula \eqref{eq:symmetric0} has $n$ variables instead of 3, the same results hold, all we need to do is set the surplus $z$ variables to zero.

The above formulae have a simple geometric interpretation. It is most easily described when the interpolating curves associated with the level 1 symmetric addition formula are straight lines as is the case for the Weierstrass curve.

To make things clearer we use the parameterisation $f(z) = \wp(z + z_0)$ and $g(z) = \wp'(z + z_0)$ so that the zero point $O$ corresponding to $z=0$ is an arbitrarily chosen point $(\wp(z_0),\wp'(z_0))$.

Draw a straight line through two points $P_1$ and $P_2$ on the curve and it will intersect the curve at a third point $P_3$. If $P_1$ and $P_2$ correspond to $z_1$ and $z_2$ repectively then $P_3$ will correspond to $z_3 = s - z_1 - z_2$ where $s = 3z_0$.

If we draw a line through $P_3$ and $O$ it will intersect the curve at a fourth point say $P_4$ which corresponds to $z_1 + z_2$. This is the classic addition formula $P_4 = P_1 \oplus P_2$.

The centre point $S$ corresponding to $z = s$ is the point where a tangent drawn at the zero point $O$ intersects the curve. To find the negation of a point $P_1$ draw a line through $P_1$ and the centre point $S$, it will intersect the curve at point $P_2$ which corresponds to $z = -z_1$. This is the negation formula $P_2 = \ominus P_1$.

To illustrate the four variable case, consider the quartic curve parameterised by the elliptic functions $f$ and $f'$ satisfying the differential equation \begin{equation} f'(z)^2 \enspace = \enspace R\left(f(z)\right) \end{equation} with boundary condition $f(0) = x_0$ so that $z=0$ at some arbitrary point $(x_0,y_0)$ on the quartic. Then the classic addition formula arises from two applications of the symmetric addition formula.

In this diagram the $x,y$-coordinates of the points are given by $\left(x_i,y_i\right) = \left(f(z_i),f'(z_i)\right)$. If $R$ has four real roots and negative leading coefficient then it will look something like this:

1. Run a parabola through the points $z_1,z_2,z_3$ to give a point $z_4$ such that $z_1 + z_2 + z_3 + z_4 \equiv s$ where $s$ is twice the sum of the position of the poles of $f$.
2. Run a second parabola tangential to the quartic at $0$ and passing through $z_4$ to give a point $z_5$ such that $0 + 0 + z_4 + z_5 \equiv s$.
3. From this it follows that $z_5 \equiv z_1 + z_2 + z_3$.

References

[1] Harold M. Edwards A Normal Form For Elliptic Curves Bull. Amer. Math. Soc. 44 (2007), 393-422

[2] F.G. Frobenius and L. Stickelberger Zur Theorie der elliptischen Functionen J. reine angew. Math 83 (1877), 175–179