PRELIM

Define the cross-ratio $\crossratio{x_1, x_2, x_3, x_4}$ of four points on the complex sphere, as \begin{equation} \crossratio{x_1, x_2, x_3, x_4} = \frac {(x_1 - x_2)(x_3 - x_4)} {(x_1 - x_3)(x_2 - x_4)} \end{equation} The cross-ratio is invariant under permutations of its arguments in the Klein 4-group, that is \begin{equation*} (1),\quad (12)(34),\quad (13)(24),\quad (14)(23) \end{equation*} \begin{equation*} \crossratio{x_1, x_2, x_3, x_4} = \crossratio{x_2, x_1, x_4, x_3} = \crossratio{x_3, x_4, x_1, x_2} = \crossratio{x_4, x_3, x_2, x_1} \end{equation*} Under other permutations it takes 6 different values \begin{equation*} (1),\quad (23),\quad (13),\quad (12),\quad (123),\quad (132) \end{equation*} \begin{equation*} \lambda,\quad\ \frac 1 {\lambda},\quad 1 - \lambda,\quad \frac {\lambda} {\lambda - 1},\quad \frac {\lambda - 1} {\lambda},\quad \frac 1 {1 - \lambda} \end{equation*}

Define a Möbius transformation $L$ as \begin{equation} L(x) = \frac {ax + b} {cx + d} \quad \text{where} \quad ad - bc \ne 0 \end{equation} Then using the Möbius difference formula \begin{equation} \label{eq:Lxy} \frac {L(x) - L(y)} {x - y} = \frac {ad - bc} {(cx + d)(cy + d)} \end{equation} it is easily seen that the cross-ratio of four points is preserved under Möbius transformations \begin{equation} \label{eq:crosspreserved} \crossratio{L(x_1), L(x_2), L(x_3), L(x_4)} = \crossratio{x_1, x_2, x_3, x_4} \end{equation} There exists a unique Möbius transformation mapping any three distinct points $e_1, e_2, e_3$ to $0, \infty, 1$. It is given by \begin{equation} L(x) = \crossratio{x, e_1, e_2, e_3} = \frac {(x - e_1)(e_2 - e_3)} {(x - e_2)(e_1 - e_3)} \end{equation} There exists a unique Möbius transformation mapping any three distinct points $e_1, e_2, e_3$ to any other three distinct points $e_1', e_2', e_3'$. It is given implicitly by \begin{equation} \label{eq:imp} \crossratio{x, e_1, e_2, e_3} = \crossratio{x', e_1', e_2', e_3'} \end{equation} Further if four distinct points $e_1, e_2, e_3, e_4$ have the same cross-ratio as another four distinct points $e_1', e_2', e_3', e_4'$ then they will also mapped to one another by the same formula \eqref{eq:imp}.

Möbius transformations map circles (or straight lines) to circles (or straight lines).

To show this substitute $z = L(x)$ in the equation for a circle $|z - z_0|^2 = r^2$ or straight line $|z - z_0|^2 = |z - z_1|^2$ and use \eqref{eq:Lxy}.

Invariant theory is concerned with finding polynomials in the coefficients of a binary form which are invariant under a group of linear transformations, usually $\SL(2,\C)$. Linear transformations on binary forms $F(x,y)$ are equivalent to Möbius transformations on a polynomial in one unknown $F(x,1)$, and the two notions are often used interchangeably. These idea's are easily generalised to forms on $n$ variables and linear fractional transformations on $n-1$ variables.

There are two main ways of constructing invariants on binary forms. The first is using symmetric expressions in root differences. By applying \eqref{eq:Lxy} it is easy to see that they are invariant. The discriminant of a polynomial is the archetypal example.

The second method is using a generalisation of the Hessian and Jacobian. The $k$-th transvectant of two homogeneous polynomials $F$ and $G$, is defined by (see for example Pg. 16 of [1]) \begin{equation} \label{eq:transvectant} \transvectant{F,G}_k = \frac {(m-k)!(n-k)!} {m!n!} \sum_{i=0}^k (-1)^i {k\choose i} \frac {\partial^k F} { {\partial x}^{k-i} {\partial y}^i } \frac {\partial^k G} { {\partial x}^i {\partial y}^{k-i} } \end{equation} where $m = \degree(F)$ and $n = \degree(G)$.

Below these two construction methods are illustrated by three examples of invariants which appear later in these notes.

Let $F(x,y)$ be a quadratic form and $f_1, f_2$ be its roots ie. \begin{equation*} F(x,y) = ax^2 + 2bxy + cy^2 = a(x - f_1y)(x - f_2y) \end{equation*} then the discriminant $\discrim(F)$ is an invariant which can be expressed as a determinant, transvectant or symmetric root difference \begin{equation*} ac - b^2 = \begin{vmatrix} a & b \\ b & c \\ \end{vmatrix} = \tfrac 1 2 \transvectant{F, F}_2 = -\tfrac 1 4 a(f_1 - f_2)^2 \end{equation*}

Simultaneous invariants can also be constructed. The resultant of two polynomials is the archetypal example but simpler simultaneous invariants exist. For example let $G(x,y)$ be a cubic form \begin{equation*} G(x,y) = px^3 + 3qx^2y + 3rxy^2 + sy^3 = p(x - g_1y)(x - g_2y)(x - g_3y) \end{equation*} then the following expression is a simultaneous invariant of $F$ and $G$ invariant under simultaneous linear transformations \begin{equation*} a(qs - r^2) + b(qr - ps) + c(pr - q^2) = \begin{vmatrix} a & b & c \\ p & q & r \\ q & r & s \\ \end{vmatrix} = \tfrac 1 2 \transvectant{\transvectant{G, G}_2, F}_2 = - \tfrac 1 {18} ap^2\left[(g_1 - f_1)(g_1 - f_2)(g_2 - g_3)^2 + (g_2 - f_1)(g_2 - f_2)(g_1 - g_3)^2 + (g_3 - f_1)(g_3 - f_2)(g_1 - g_2)^2\right] \end{equation*}

In more complex cases the symmetric root expressions are far less obvious. For example a simultaneous invariant for three quadratics \begin{aligned} F(x,y) &= a_1x^2 + 2b_1xy + c_1y^2 = a_1(x - f_1y)(x - f_2y) \\ G(x,y) &= a_2x^2 + 2b_2xy + c_2y^2 = a_2(x - g_1y)(x - g_2y) \\ H(x,y) &= a_3x^2 + 2b_3xy + c_3y^2 = a_3(x - h_1y)(x - h_2y) \end{aligned} is given by \begin{equation*} a_1b_2c_3 - a_1b_3c_2 - a_2b_1c_3 + a_2b_3c_1 + a_3b_1c_2 - a_3b_2c_1 = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix} = \transvectant{F,\transvectant{G,H}_1}_2 = -\tfrac 1 2 a_1 a_2 a_3\left[(f_1 - g_2)(g_1 - h_2)(h_1 - f_2) + (f_2 - g_1)(g_2 - h_1)(h_2 - f_1)\right] \end{equation*} The right-hand side root expression does not really appear to be symmetric under either $f_1 \leftrightarrow f_2$, $g_1 \leftrightarrow g_2$ or $h_1 \leftrightarrow h_2$, but fully expanding reveals that it is.

In the above formulae binomial coefficients were used in the definitions of the binary forms. Combined with the leading scale factor in \eqref{eq:transvectant} this usually makes for smaller numeric coefficients. However in subsequent sections of these notes that convention becomes a bit awkward and the binomial coefficients are dropped.

Weierstrass's elliptic $\wp$ function is defined by \begin{equation} \label{eq:wpdef} \wp(z) = \frac 1 {z^2} + \sum_{\omega \in \Omega'} \left[ \frac 1 {(z + \omega)^2} - \frac 1 {\omega^2} \right] \end{equation} where \begin{equation} \Omega' = \lbrace 2\omega_1 n + 2\omega_2 m: m,n \in \Z \quad (m,n) \neq (0,0)\rbrace \end{equation} and following historical convention $\omega_1,\omega_2$ denote the half-periods generating the period lattice.

The $\wp$ function satisfies the following differential equation \begin{equation} {\wp'(z)}^{2} = 4 \wp(z)^3 - g_2 \wp(z) -g_3, \qquad \wp(0) = \infty \end{equation} where \begin{equation} \label{eq:g2def} g_2 = 60 \sum_{w \in \Omega'} \frac 1 {\omega^4}, \qquad g_3 = 140 \sum_{w \in \Omega'} \frac 1 {\omega^6} \end{equation} The modular discriminant $\Delta$, Kleins absolute invariant $J$, and $j$-invariant are defined by \begin{equation} \label{eq:jdef} \Delta = g_2^3 - 27 g_3^2,\qquad J = g_2^3 / \Delta, \qquad j = 1728 J \end{equation} The $\wp$ function can be thought of as the inverse of the elliptic integral \begin{equation} \wp^{-1}(z) = \bigint_{-\infty}^{z} \frac 1 {\sqrt {4x^3 - g_2 x - g_3}} dx \end{equation}

The half periods are given by the formula \begin{equation} \omega_i = \bigint_{-\infty}^{e_i} \frac 1 {\sqrt {4x^3 - g_2 x - g_3}} dx\qquad i=1,2,3 \end{equation} where the $e_i$ are the roots of the cubic $4x^3 - g_2x - g_3 = 0$

General elliptic functions are analytic functions doubly periodic and meromorphic in the whole of the complex plane. The order of an elliptic function is the number of times it takes each value (properly counted) in a fundamental region. A fundamental region is a period parallelogram or more generally any reasonable region which contains exactly one point modulo the period lattice.

Two basic properties of general elliptic functions can be obtained by integrating around the boundary of a fundamental region.

The first basic property of an elliptic function $f$ is that the sum of residues in a fundamental region is zero. This property ensures that if $R$ is a rational function with the same principal parts as $f$ in a fundamental region then the following infinite sum converges everywhere except at the poles \begin{equation} f(x) - f(y) = \sum_{\omega \in \Omega} \left[ R(x + \omega) - R(y + \omega) \right] \end{equation} where \begin{equation} \Omega = \lbrace 2\omega_1 n + 2\omega_2 m: m,n \in \Z \rbrace \end{equation} This equation is really just a simple generalisation of the defining equation \eqref{eq:wpdef} for the Weierstrass $\wp$ function. Put $R(x) = \frac 1 {x^2}$ and let $y \rightarrow 0$.

The second basic property of an elliptic function $f$ is that the sum of zeroes is equal to the sum of poles, modulo $\Omega$, in a fundamental region. This property ensures that if $R$ is a rational function with the same zeroes and poles as $f$ in a fundamental region then the following infinite product converges everywhere except at the zeroes and poles \begin{equation} \frac {f(x)} {f(y)} = \prod_{\omega \in \Omega} \frac {R(x + \omega)} {R(y + \omega)} \end{equation} This property is easily seen to extend to all values. If $f$ is of order $n$ and $z_1 \ldots z_n$ are the $n$ properly counted, distinct solutions of $f(z_i) = k$ for any $k$, then $z_1 + \ldots z_n$ is equal to the sum of the locations of the poles of $f$, modulo $\Omega$. It is convenient to call this point $\centre(f)$. This point often plays an important role, for example every elliptic function of order 2 has even symmetry about its centre.

A third basic property is that every elliptic function $f(z)$ can be written as a rational function of $\wp(z)$ and $\wp'(z)$ and it is quite easy to make an explicit formula.

For example if $f$ has three non-zero simple poles $a$, $b$ and $c$ with respective residues $\alpha$, $\beta$ and $\gamma$ with $\alpha + \beta + \gamma = 0$ then

\begin{equation} \label{eq:fwp2} f(z) = f(0) + \tfrac 1 2 \alpha \frac {\wp'(z) + \wp'(a)} {\wp(z) - \wp(a)} + \tfrac 1 2 \beta \frac {\wp'(z) + \wp'(b)} {\wp(z) - \wp(b)} + \tfrac 1 2 \gamma \frac {\wp'(z) + \wp'(c)} {\wp(z) - \wp(c)} \end{equation}

This formula follows from the fact that if \begin{equation*} g(z,a) = \tfrac 1 2 \frac {\wp'(z) + \wp'(a)} {\wp(z) - \wp(a)} \end{equation*} then $g(z,a) = -\frac 1 z + \bigO(z)$ and $g(z,a) = \frac 1 {z-a} + \bigO(z-a)$ and $g$ is a second order elliptic function with a simple pole at $0$ with residue $-1$ and a simple pole at $a$ with residue $1$. Therefore \begin{equation*} h(z) = \alpha g(z,a) + \beta g(z,b) + \gamma g(z,c) \end{equation*} is an order 4 elliptic function with simple poles at $a,b,c,0$ with residues $\alpha,\beta,\gamma,-(\alpha+\beta+\gamma)$ and $h(0)=0$. The result follows when $\alpha+\beta+\gamma=0$.

Another simple way to obtain this formula is to apply the addition formula for the Weierstrass Zeta function to the formula \begin{equation*} f(z) = f(0) + \alpha \zeta(z-a) + \beta \zeta(z-b)+ \gamma \zeta(z-c) + \alpha \zeta(a) + \beta \zeta(b)+ \gamma \zeta(c) \end{equation*}

And if $f$ has three non-zero simple zeroes $r,s,t$ with $r + s + t = a + b + c$ then

\begin{equation} \label{eq:fwp4} f(z) = f(0) \space \left. \begin{vmatrix} 1 & \wp(a) & \wp'(a) \\ 1 & \wp(b) & \wp'(b) \\ 1 & \wp(c) & \wp'(c) \end{vmatrix} \space \begin{vmatrix} 1 & \wp(z) & \wp'(z) & \wp(z)^2 \\ 1 & \wp(r) & \wp'(r) & \wp(r)^2 \\ 1 & \wp(s) & \wp'(s) & \wp(s)^2 \\ 1 & \wp(t) & \wp'(t) & \wp(t)^2 \end{vmatrix} \space \middle/ \space \begin{vmatrix} 1 & \wp(r) & \wp'(r) \\ 1 & \wp(s) & \wp'(s) \\ 1 & \wp(t) & \wp'(t) \end{vmatrix} \space \begin{vmatrix} 1 & \wp(z) & \wp'(z) & \wp(z)^2 \\ 1 & \wp(a) & \wp'(a) & \wp(a)^2 \\ 1 & \wp(b) & \wp'(b) & \wp(b)^2 \\ 1 & \wp(c) & \wp'(c) & \wp(c)^2 \end{vmatrix} \right. \end{equation}

This formula follows from the fact that the numerator is an order 4 elliptic function with four simple zeroes at $r,s,t,-(r+s+t)$ and a quadruple pole at zero and similarly the denominator has four simple zeroes at $a,b,c,-(a+b+c)$ and a quadruple pole at zero. The quadruple poles cancel, as do the zeros at $-(r+s+t) = -(a+b+c)$ and the result follows.

Another way to obtain this formula is by applying the Frobenius-Stickelberger addition formula to the representation of $f$ using the Weierstrass Sigma function \begin{equation*} f(z) = f(0) \frac {\sigma(z-r)\sigma(z-s)\sigma(z-t)\thinspace\sigma(a)\sigma(b)\sigma(c)} {\sigma(z-a)\sigma(z-b)\sigma(z-c)\thinspace\sigma(r)\sigma(s)\sigma(t)} \end{equation*}

This property is not confined to the Weierstrass $\wp$ function. Any two elliptic functions $f(z)$ and $g(z)$, with the same periods, satisfy an irreducible algebraic equation of genus 1, (or in degenerate cases genus 0). If the relation is genus 1 they also generate the entire field of elliptic functions.

It is also possible to give an explicit formula for this algebraic relation. From the order of the elliptic functions we can deduce the general form of the relation. For example if $f(z)$ and $g(z)$ are elliptic functions of order 2 then \begin{equation} \label{eq:fg1} C_1 + C_2 f(z) + C_3 f(z)^2 + C_4 g(z) + C_5 f(z)g(z) + C_6 f(z)^2 g(z) + C_7 g(z)^2 + C_8 f(z)g(z)^2 + C_9 f(z)^2 g(z)^2 = 0 \end{equation} If we evaluate $f$ and $g$ at 8 distinct complex numbers $z_1 \ldots z_8$ then we can solve \eqref{eq:fg1} for $C_i$ using linear algebra to get

\begin{equation} \label{eq:f2g2} \begin{vmatrix} 1 & f(z) & f(z)^2 & g(z) & f(z)g(z) & f(z)^2 g(z) & g(z)^2 & f(z)g(z)^2 & f(z)^2 g(z)^2 \\ 1 & f(z_1) & f(z_1)^2 & g(z_1) & f(z_1)g(z_1) & f(z_1)^2 g(z_1) & g(z_1)^2 & f(z_1)g(z_1)^2 & f(z_1)^2 g(z_1)^2 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1 & f(z_8) & f(z_8)^2 & g(z_8) & f(z_8)g(z_8) & f(z_8)^2 g(z_8) & g(z_8)^2 & f(z_8)g(z_8)^2 & f(z_8)^2 g(z_8)^2 \\ \end{vmatrix} \space = \space 0 \end{equation}

If $D$ is the determinant on the left hand side of \eqref{eq:f2g2}, then the $C_i$ are the 9 cofactors of the first row of $D$. While these cofactors are functions of the $z_i$ they must be all constant multiples of some single function of the $z_i$ because the algebraic relation itself must be independent of the $z_i$.

This function can be determined as follows. Assume $f$ has two simple poles at $a_1,a_2$ and $g$ two simple poles at $a_3,a_4$ where all the $a_i$ are distinct. The first row of each of the 9 cofactors is a basis for the 8 dimensional vector space of elliptic functions with poles at the $a_i$ of order two or less. Applying the Extended Frobenius-Stickelberger formula to each cofactor gives a constant $K_i$ times a product of sigma functions, explicitly \begin{equation*} D = \left[ K_1 + K_2 f(z) + K_3 f(z)^2 + K_4 g(z) + \ldots + K_9 f(z)^2 g(z)^2\right] \frac {\sigma\left(z_1 + \ldots + z_8 - 2(a_1 + a_2 + a_3 + a_4)\right) \space \prod\limits_{j=1}^8 \space \prod\limits_{i = 1}^{j - 1} \sigma(z_i-z_j)} {\prod\limits_{i=1}^8 \prod\limits_{j=1}^4 \sigma(z_i - a_j)^2} \end{equation*} From this it is seen that \eqref{eq:f2g2} only "fails" when the argument of one of the $\sigma$-functions is zero. That is the only non-trivial failure is when $\sum z_i \equiv 2 \sum a_j \mod \Omega$.

A notable special case occurs when all $z_i \rightarrow a$. Putting $h_1(z) = f(z),\space h_2(z) = f(z)^2 \ldots h_8(z) = f(z)^2 g(z)^2$ we get \begin{equation*} \begin{vmatrix} 1 & f(z) & f(z)^2 & \ldots & f(z)^2 g(z)^2 \\ 1 & h_1(a) & h_2(a) & \ldots & h_8(a) \\ 0 & h_1'(a) & h_2'(a) & \dots & h_8'(a) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & h_1^{[7]}(a) & h_2^{[7]}(a) & \ldots & h_8^{[7]}(a) \\ \end{vmatrix} \space = \space 0 \end{equation*} This formula expresses the coefficients $C_i$ in terms of the first 8 coefficients in the power series expansion of $f$ and $g$ at $a$. It is, of course, also the formula you get if substitute the power series expansions of $f$ and $g$ into \eqref{eq:fg1} and solve for $C_i$. The factor function in this case is \begin{equation*} \frac {\sigma\left(8a - 2(a_1 + a_2 + a_3 + a_4)\right)} {\prod\limits_{j=1}^4 \sigma(a - a_j)^{16}} \end{equation*} and the formula only "fails" when $a \equiv \tfrac 1 4 \sum a_j \mod \tfrac 1 8 \Omega$.

A birational mapping is a rational mapping whose inverse is also a rational mapping.

For example the curve $x^3 + y^3 = 1$ is mapped to the curve $\displaystyle 6v^2 = u^3 - 2$ by the birational mapping \begin{equation*} x = \frac {v - 1} {v + 1},\qquad y = \frac {u} {v + 1} \end{equation*} the inverse mapping is \begin{equation*} u = \frac {2y} {1 - x},\qquad v = \frac {1 + x} {1 - x} \end{equation*}

We can compute the genus of a curve by counting it's branch points $b_i$ and then applying Eulers graph formula $\chi = V - E + F = 2 - 2g$. On a surface which covers the complex sphere $n$ times run a simple curve through all $m$ branch points and all points conjugate to branch points. This curve divides each layer into two faces, each face is bounded by $m$ edges, and each layer has $m$ vertices (each of which is shared with $b_i - 1$ other layers). Therefore $V = mn - \sum (b_i - 1)$, $E = mn$ and $F = 2n$, so $\chi = 2n - \sum (b_i - 1)$ and \begin{equation} g = \tfrac 1 2 \sum_1^m {(b_i-1)} - n + 1 \end{equation}

For example the curve $y^6 = (x - a)^3(x - b)^4(x - c)^5$ has genus 1.

On the $x$-sphere it has 3 branch points of order 2 at $a$, 2 of order 3 at $b$, 1 of order 6 at $c$ and no branch points at infinity. So the genus is \begin{equation*} g = \tfrac 1 2 (1 + 1 + 1 + 2 + 2 + 5) - 6 + 1 = 1 \end{equation*}

Holomorphic differentials, a.k.a Abelian differentials of the first kind, are regular differential 1-forms on a Riemann surface. In simple terms this means that their power series expansion in a local uniformising variable has no poles.

For example on the curve $y^4 = (x - a)^2(x - b)^3(x - c)^3$ the differential $\displaystyle \frac {dx} {y}$ is holomorphic.

The local power series expansion at $x=a$ is given by $x = a + t^2 + \bigO(t^3)$ and $y =t + \bigO(t^2)$ so the power series for the differential is \begin{equation} \frac 1 y \frac {dx} {dt} = \frac {2t + \bigO(t^2)} {t + \bigO(t^2)} = 2 + \bigO(t) \end{equation} The local power series expansion at $x=b$ is given by $x = b + t^4 + \bigO(t^5)$ and $y =t^3 + \bigO(t^4)$ so \begin{equation} \frac 1 y \frac {dx} {dt} = \frac {4t^3 + \bigO(t^4)} {t^3 + \bigO(t^4)} = 4 + \bigO(t) \end{equation} and similar at $x=c$. The local power series at $x = \infty$ is given by $x = t^{-1} + \bigO(1)$ and $y = t^{-2} + \bigO(t^{-1})$ so \begin{equation} \frac 1 y \frac {dx} {dt} = \frac {-t^{-2} + \bigO(1)} {t^{-2} + \bigO(t^{-1})} = -1 + \bigO(t) \end{equation} and the differential has no poles and is therefore holomorphic.

On a surface of genus $g$ there are $g$ linearly independent holomorphic differentials. The generic curve $F(x, y) = 0$ of degree $m$ in $x$ and $n$ in $y$ has genus $g = (m - 1)(n - 1)$. The holomorphic differentials are given by \begin{equation} \label{eq:holodiff} dz = \frac {Q(x,y)} {F_y(x,y)} dx = -\frac {Q(x,y)} {F_x(x,y)} dy \end{equation} where $\degree_x(Q) \leq m-2$ and $\degree_y(Q) \leq n-2$.

If $F$ has singular points it will have genus less than $g$ and the space of holomorphic differentials will by a subspace of that given above. In particular if it has genus 1 the curve will have $k = mn - m - n$ singular points and there will be a single holomorphic differential \begin{equation} dz = \begin{vmatrix} 1 & x & y & \cdots & x^{m-2}y^{n-2} \\ 1 & x_1 & y_1 & \cdots & x_1^{m-2}y_1^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_k & y_k & \cdots & x_k^{m-2}y_k^{n-2} \end{vmatrix} \frac {dx} {F_y(x,y)} \end{equation} where the $(x_i,y_i)$ are the coordinates of the singular points. In the generic case the singular points are the points at which both partial derivatives of $F$ vanish simultaneously ie. \begin{equation} F(x_i,y_i) = F_x(x_i,y_i) = F_y(x_i,y_i) = 0 \end{equation}

The uniformisation theorem implies that genus 1 curves can be nicely parametrised by a pair of elliptic functions in the same way the circle $x^2 + y^2 = 1$ is parameterised by trignometric functions \begin{equation*} x = \cos \theta \qquad y = \sin \theta \end{equation*} The holomorphic differential \eqref{eq:holodiff} on a curve of genus 1 given by $F(x,y) = 0$ is effectively a pair of simultaneous differential equations for the parameterising elliptic functions \begin{equation} Q(x,y) \frac {dx} {dz} = F_y(x,y) \qquad Q(x,y) \frac {dy} {dz} = -F_x(x,y) \end{equation}

Preliminaries