./ABEL.html,./FXY2.html

In this section we compute more general rational mappings to the Weierstrass curve by solving differential equations using power series. In the process we uncover a general algorithm for doing this for any curve of genus 1.

Consider the curve

\begin{equation*} y^2 = R(x) = ax^4+bx^3+cx^2+dx+e \end{equation*}

and the associated differential equation

\begin{equation*} f'^2 = R(f) \end{equation*}

Previously we solved fde by computing the Möbius transform mapping this curve to Weierstrass form. Here is a more direct method which can be generalised to other curves.

Even in this relatively simple case these calculations require computing eight terms of the power series and solving a system of equations in eight unknowns. As a result they are best carried out using CAS.

Power Series Solution

By repeatedly differentiating fde it is easy to compute the first few terms of the power series solution with boundary condition $f(0)=x_0$ and $f'(0)=y_0$ as follows

\begin{aligned} f(z) = x_0 + y_0 z + \tfrac 1 4 R_1 z^2 + \tfrac 1 6 y_0 R_2 z^3 + \tfrac 1 {48} \left(6 R_0 R_3 + R_1 R_2\right) z^4 + \tfrac 1 {240} y_0 \left(24 R_0 R_4 + 9 R_1 R_3 + 2 R_2^2\right) z^5 + \\ \tfrac 1 {2880} \left(144 R_0 R_1 R_4 + 60 R_0 R_2 R_3 + 9 R_1^2 R_3 + 2 R_1 R_2^2\right) z^6 + \tfrac 1 {5040} y_0 \left(132 R_0 R_2 R_4 + 45 R_0 R_3^2 + 45 R_1^2 R_4 + 27 R_1 R_2 R_3 + R_2^3\right) z^7 + \bigO(z^8) \end{aligned}

where $R_i$ are the coefficients of the Taylor expansion $R$ evaluated at $x_0$ ie. $\displaystyle R_i= \frac 1 {i!} R^{[i]}(x_0)$. In a similar way when $x_0=\infty$ we get the Laurent series

\begin{aligned} f(z) = {a^{-1/2}} z^{-1} - \tfrac 1 4 b a^{-1} - \tfrac 1 {48} \left(8ac - 3b^2\right) a^{-3/2} z - \tfrac 1 {64} \left(8a^2d - 4abc + b^3\right) a^{-2} z^2 -\tfrac 1 {11520} \left(1152a^3e - 288a^2bd - 224a^2c^2 + 240ab^2c - 45b^4\right)a^{-5/2}z^3 + \\ \tfrac 1 {3072} \left(8a^2d - 4abc + b^3\right)\left(8ac - 3b^2\right)a^{-3}z^4 +\bigO(z^5) \end{aligned}

We can similarly compute the power series for $\wp(z,g_2,g_3)$ as follows

\begin{equation*} \wp(z) = z^{-2} + \tfrac 1 {20} g_2 z^2 + \tfrac 1 {28} g_3 z^4 + \tfrac 1 {1200} g_2^2 z^6 + \tfrac 3 {6160} g_2 g_3 z^8 + \bigO(z^{10}) \end{equation*}

Now assuming $f(z)$ has two simple poles at $z_1,z_2 \neq 0$ with residues $\alpha,-\alpha$ we can use the order 2 version of this formula to write $f(z)$ as

\begin{equation*} f(z) = f(0) + \tfrac 1 2 \alpha \left[ \frac {\wp'(z) + \wp'(z_1)} {\wp(z) - \wp(z_1)} -\frac {\wp'(z) + \wp'(z_2)} {\wp(z) - \wp(z_2)} \right] \end{equation*}

Attempting to solve this for the unknown coefficients $\alpha,\wp(z_1),\wp(z_2)$ leads to nonlinear equations. However rewriting it as

\begin{equation*} f(z) = \frac {A_1 \wp'(z) + A_2\wp(z)^2 + A_3\wp(z) + A_4} {\wp(z)^2 + A_5\wp(z) + A_6} \end{equation*}

then moving the denominator to the left hand side of solution, substituting in fseries and pseries, and finally equating the coefficients of the first eight $z^i$ terms, gives six linear equations in the six unknown coefficients $A_1,A_2,A_3,A_4,A_5,A_6$ and a further two linear equations in $g_2,g_3$. This set of eight equations is easily solved to yield:

\begin{align*} A_1 &= -\tfrac 1 2 y_0 \\ A_2 &= x_0 \\ A_3 &= \tfrac 1 {12} \Big[ 3b x_0^2 + 4c x_0 + 3d \Big] \\ A_4 &= \tfrac 1 {144} \Big[ (18ad - 3bc) x_0^2 + (36ae + 9bd - 5c^2) x_0 + 18be - 3cd \Big] \\ A_5 &= -\tfrac 1 6 \Big[ 6a x_0^2 + 3b x_0 + c \Big] \\ A_6 &= \tfrac 1 {144} \Big[ (9b^2 - 24ac) x_0^2 + (6bc - 36ad) x_0 - 36ae + c^2 \Big] \\ g_2 &= \tfrac {1} {12} \Big[ 12ae - 3bd + c^2 \Big] \\ g_3 &= \tfrac {1} {432} \Big[ 72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3 \Big] \end{align*}

In this way $f(z)$ can be expressed as a rational functional of $\wp(z),\wp'(z),a,b,c,d,e$ and $x_0,y_0$. And we have expressed $g_2,g_3$ as polynomials in $a,b,c,d,e$.

Reverse Solution

A similar procedure can be applied to express $\wp(z)$ as a rational function of $f(z),f'(z),a,b,c,d,e$ and $x_0,y_0$.

\begin{equation*} \wp(z) = \frac {B_1 f'(z) + B_2 f(z)^2 + B_3 f(z) + B_4} {f(z)^2 + B_5 f(z) + B_6} \end{equation*}

where

\begin{align*} B_1 &= \tfrac 1 2 y_0 \\ B_2 &= \tfrac 1 {12} \Big[ 6a x_0^2 + 3b x_0 + c \Big] \\ B_3 &= \tfrac 1 {12} \Big[ 3b x_0^2 + 4c x_0 + 3d \Big] \\ B_4 &= \tfrac 1 {12} \Big[ c x_0^2 + 3d x_0 + 6e \Big] \\ B_5 &= -2 x_0 \\ B_6 &= x_0^2 \\ \end{align*}

Is this a general algorithm ?

This algorithm could be applied to any genus 1 curve $F(x,y)=0$ with a known holomorphic differential $\displaystyle dz = \frac {Q(x,y)} {F_y(x,y)} dx = -\frac {Q(x,y)} {F_x(x,y)} dy$.

  1. Choose an arbitrary point on the curve $x_0,y_0$ where $F(x_0,y_0)=0$.
  2. Compute the power series solution for the uniformising elliptic functions $f(z),\space g(z)$ with boundary condition $f(0)=x_0,\space g(0)=y_0$. This is done by repeatedly differentiating the differential equations $f' = F_y(f,g) / Q(f,g)$ and $g' = -F_x(f,g) / Q(f,g)$ derived from the holomorphic differential. The coefficients will be rational functions† of the coefficients of $F$ and $Q$, say $a_i$, and $x_0,y_0$.
  3. Equate $f(z)$ to an appropriate rational function of $\wp(z),\wp'(z)$ with unknown coefficients $A_i$ and rearrange it into a linear equation for the $A_i$
  4. Substitute the power series into this equation and equate the coefficients to zero, giving a set of linear equations for the $A_i$ and a further two relations for $g_2,g_3$. Solve these equations for $A_i$ and $g_2,g_3$
  5. This yields $f(z)$ as a rational function of the coefficients of $a_i$ and $x_0,y_0$ and $\wp(z),\wp'(z)$ and also formulae expressing $g_2,g_3$ as a rational function in the coefficients of $a_i$.

† Actually there are isolated special values of $x_0,y_0$ where the coefficients of the power series are not rational functions of the $a_i$. This happens when there are multiple points on the Riemann surface of $F$ with same coordinates $x_0,y_0$. For example on curve curve there is one such case. When $a \ne 0$ there are two points with coordinates $\infty,\infty$

Special Case 1

When $x_0 \rightarrow \infty$ we have $y_0/x_0^2 \rightarrow \sqrt{a}$, one of the poles say $z_2 \rightarrow 0$ and solution collapses to

\begin{equation*} f(z) = \tfrac 1 2 \alpha \frac {\wp'(z) + \wp'(z_1)} {\wp(z) - \wp(z_1)} + \text{const.} = \frac {\pm 24\sqrt{a}\thinspace\wp'(z) - 12b\thinspace\wp(z) - 6ad + bc} {48a\thinspace\wp(z) + 8ac - 3b^2} \end{equation*}

So there are two solutions with $f(0)=\infty$ and $f'(0)=\infty$ and the solution is no longer rational.

Special Case 2

When $R(x_0)=0$ we have $y_0=0$ and solution collapses to the Möbius transform computed in a previous section.

\begin{equation*} f(z) = L(\wp(z)) = x_0 + \frac {B_1} {\wp(z) + B_2} \end{equation*}

where

\begin{align*} B_1 &= \tfrac 1 {4} \Big[4 a x_0^3 + 3 b x_0^2 + 2 c x_0 + d\Big] \\ B_2 &= -\tfrac 1 {12} \Big[6 a x_0^2 + 3 b x_0 + c\Big] \\ \end{align*}

The reduction is not obvious but follows from the fact that the numerator and denominator of solution have a common factor when $R_0=0$ because

\begin{equation*} \resultant(A_2 x^2 + A_3 x + A_4, x^2 + A_5 x + A_6) = \tfrac 1 {64} R_0 (R_0 R_3^2 - R_1^2 R_4) = 0 \end{equation*}

Rational Points

If $a,b,c,d,e$ are rational and $(x_0, y_0)$ is a rational point on curve curve then solution has rational coefficients. It puts the rational points on curve in one to one correspondence with the rational points on the corresponding Weierstrass curve. In particular the point $(x_0, y_0)$ corresponds to the point $(\infty, \infty)$ on the Weierstrass curve.

Using solution we can compute the birational map of the Jacobi curve

\begin{equation*} y^2 = (1-x^2)(1-k^2x^2) \end{equation*}

to the Weierstrass curve

\begin{equation*} v^2 = 4u^3 - g_2u - g_3 \end{equation*}

We obtain

\begin{equation*} g_2=\tfrac 1 {12}(k^4 + 14k^2 + 1),\qquad g_3=\tfrac 1 {216}(k^6 - 33k^4 - 33k^2 + 1) \end{equation*}

and for eight obvious points on jacobi

Table 1
$x_0$ $y_0$ $x$ $y$
$\pm 1$ $0$ $\displaystyle x = \pm \frac {12u + k^2 - 5} {12u - 5k^2 + 1}$ $\displaystyle y = \pm \frac {72(1 - k^2)v} {(12u - 5k^2 + 1)^2}$
$\pm 1/k$ $0$ $\displaystyle x = \pm \frac {12u - 5k^2 + 1} {k(12u + k^2 - 5)}$ $\displaystyle y = \pm \frac {-72(1 - k^2)v} {k(12u + k^2 - 5)^2}$
$0$ $\pm 1$ $\displaystyle x = \pm \frac {-72v} {(12u + k^2 + 6k + 1)(12u + k^2 - 6k + 1)}$ $\displaystyle y = \pm \frac {(12u + k^2 - 5)(12u - 5k^2 + 1)} {(12u + k^2 + 6k + 1)(12u + k^2 - 6k + 1)}$
$\infty$ $\infty$ $\displaystyle x = \pm \frac {72v} {24k(6u - k^2 - 1)}$ $\displaystyle y = \pm \frac {(12u + k^2 - 5)(12u - 5k^2 + 1)} {24k(6u - k^2 - 1)}$

Addition Formula

When we use this algorithm to compute a rational mapping of a curve onto itself we are also computing the addition formulae for the uniformising elliptic functions. For example in solution substitute the coefficients for the Weierstrass curve $a=0,b=4,c=0,d=-g_2,e=-g_3$. Then $f(z)=\wp(z+z_0)$ where $\wp(z_0)=x_0$ and $\wp'(z_0)=y_0$ and we have

\begin{equation*} \wp(z+z_0) = \frac {-\tfrac 1 2 \wp'(z_0)\wp'(z) + \wp(z_0)\wp(z)^2 + \wp(z_0)^2\wp(z) - \tfrac 1 4 g_2\wp(z) + \tfrac 1 4 g_2\wp(z_0) -\tfrac 1 2 g_3} {\wp(z)^2 - 2 \wp(z_0)\wp(z) + \wp(z_0)^2} \end{equation*}

which is equivalent to the standard addition formula

\begin{equation*} \wp(x+y) = \tfrac 1 4 \left[\frac {\wp'(x) - \wp'(y)} {\wp(x) - \wp(y)}\right]^2 - \wp(x) - \wp(y) \end{equation*}