Preliminaries

Define the cross-ratio $\crossratio{x_1, x_2, x_3, x_4}$ of four points on the complex sphere, as

The cross-ratio is invariant under permutations of its arguments in the Klein 4-group, that is

The obvious identities between these formulae eg. $\lambda + (1 - \lambda) = 1$ are equivalent to the cross-ratio identity formula

it is easily seen that the cross-ratio of four points is preserved under Möbius transformations

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

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

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

The equation for a circle / straight line is

az\bar{z} + bz + \bar{b}\bar{z} + c = 0

where $a,c$ are real and $b$ is complex and $|b|^2 \gt ac$. This is because when $a \ne 0$ it is the equation of a circle $|z + \bar{b}/a|^2 = |b|^2/a^2 - c/a$ and when $a = 0$ it is the equation of a straight line.

Substituting in a rotation and scale$z \mapsto dz$ where $d \ne 0$ is complex yields another equation of the same form

ad\bar{d}z\bar{z} + bdz + \bar{b}\bar{d}\bar{z} + c = 0
Substituting in a translation $z \mapsto z + e$ where $e$ is complex also yields another equation of the same form

#flow,marked az\bar{z} $+ \left(b + a\bar{e}\right)z $+ \left(\bar{b} + ae\right)\bar{z} $+ \left(c + be + \bar{b}\bar{e} + ae\bar{e}\right) = 0
Similarly an inversion $z \mapsto 1/z$ also yields another equation of the same form namely

cz\bar{z} + \bar{b}z + b\bar{z} + a = 0

Since every Möbius transform can be reduced to a sequence of rotations, scales, translations and inversions the result follows.

This last condition is equivalent to Ptolemy's Theorem if the four vertices of a proper quadrilateral lie on a circle then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides.

Let $e_1,e_2,e_3,e_4$ be four distinct points in the complex plane. Put $A = (e_1 - e_2)(e_3 - e_4)$, $B = (e_1 - e_3)(e_4 - e_2)$, $C = (e_1 - e_4)(e_2 - e_3)$ then $A + B + C = 0$ and $\crossratio{e_1,e_2,e_3,e_4} = -A / B$. The condition that the cross-ratio be real is then $A\bar{B} - \bar{A}B = 0$. Using the fact that $C = -(A + B)$ we can compute the identity

#flow,marked \left(|A| + |B| + |C|\right) \cdot \left(|A| + |B| - |C|\right) \cdot $ \left(|A| - |B| + |C|\right) \cdot $ \left(|A| - |B| - |C|\right) \space = $ \space \left(A\bar{B}\space - \space \bar{A}B\right)^2

Therefore if the cross-ratio is real and the four points are distinct, either $|A|+|B|-|C|=0$, $|A|-|B|+|C|=0$ or $|A|-|B|-|C|=0$. Which factor vanishes is determined by the cyclic order of the four points around the circle.

If we join the four points with straight lines to form a quadrilateral the products of the lengths of the opposite sides and diagonals are $|A|,|B|,|C|$ because $|A| = |e_1 - e_2| \cdot |e_3 - e_4|$ etc. If the cyclic order of the points around the circle is $e_1,e_2,e_3,e_4$ then $|A|$ and $|C|$ are the products of the lengths of opposite sides and $|B|$ is the product of the lengths of the diagonals and $|A| - |B| + |C| = 0$ which is Ptolemy's theorem.

When $a\overline{c} - \overline{a}c \ne 0$ the equation $L(z) = \overline{L(z)}$ is a circle

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 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])

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

then the discriminant $\discrim(F)$ is an invariant which can be expressed as a determinant, transvectant or symmetric root difference

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

then the following expression is a simultaneous invariant of $F$ and $G$ invariant under simultaneous linear transformations

In more complex cases the symmetric root expressions are far less obvious. For example a simultaneous invariant for three quadratics

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 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.

and following historical convention $\omega_1,\omega_2$ denote the half-periods generating the period lattice.

The modular discriminant $\Delta$, Kleins absolute invariant $J$, and $j$-invariant are defined by

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

This equation is really just a simple generalisation of the defining equation 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

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.

It also follows from this property that a non-constant elliptic function $f(z)$ has the same number of zeroes and poles, in a fundamental region, otherwise $f(z + \epsilon)$ for suitable small non-zero $\epsilon$ would be a non-constant elliptic function whose sum of zeroes does not equal it's sum of poles.

And from this that an elliptic function $f(z)$ with no poles must be constant, otherwise $f(z) - f(0)$ would be a non-constant elliptic function with no poles but at least one zero.

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

#flow,indent,marked 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)} $

This formula follows from the fact that if

g(z,a) = \tfrac 1 2 \frac {\wp'(z) + \wp'(a)} {\wp(z) - \wp(a)}

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

h(z) = \alpha g(z,a) + \beta g(z,b) + \gamma g(z,c)

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

#flow,indent 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)

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

#fold,marked f(z) = f(0) \space \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 \cdot \space $ \begin{vmatrix} 1 & \wp(r) & \wp'(r) \\ 1 & \wp(s) & \wp'(s) \\ 1 & \wp(t) & \wp'(t) \end{vmatrix}^{-1} \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}^{-1}

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.

This formula can be extended to an arbitrary number of poles and zeroes by using the monomial basis sequence $1,\wp,\wp',\wp^2,\wp\wp',\wp^3,\wp^2\wp',\wp^4 \dots$

Another way to obtain this formula is by applying the Frobenius-Stickelberger addition formula to the representation of $f$ using the Weierstrass Sigma function

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)}

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.

One way to see that any pair of elliptic functions satisfy an algebraic relation is by considering the set of monomials $f(z)^i g(z)^j$ where $i,j \le n$. This set of monomials has size $(n+1)^2$. By considering the poles of these monomials it can be seen they are members of a vector space of dimension $\bigO(n)$. Therefore if $n$ is chosen large enough the number of monomials will exceed the dimension of this vector space implying there is a linear relation between them. That linear relation is an algebraic relation between $f$ and $g$.

If we evaluate $f$ and $g$ at 8 distinct complex numbers $z_1 \ldots z_8$ then we can solve fg1 for $C_i$ using linear algebra to get

\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

If $D$ is the determinant on the left hand side of 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

#fold,marked 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] \cdot $ \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}

From this it is seen that 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{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

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 fg1 and solve for $C_i$. The factor function in this case is

\frac {\sigma\left(8a - 2(a_1 + a_2 + a_3 + a_4)\right)} {\prod\limits_{j=1}^4 \sigma(a - a_j)^{16}}

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

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

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

g = \tfrac 1 2 (1 + 1 + 1 + 2 + 2 + 5) - 6 + 1 = 1

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

\frac 1 y \frac {dx} {dt} = \frac {2t + \bigO(t^2)} {t + \bigO(t^2)} = 2 + \bigO(t)

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

\frac 1 y \frac {dx} {dt} = \frac {4t^3 + \bigO(t^4)} {t^3 + \bigO(t^4)} = 4 + \bigO(t)

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

\frac 1 y \frac {dx} {dt} = \frac {-t^{-2} + \bigO(1)} {t^{-2} + \bigO(t^{-1})} = -1 + \bigO(t)

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

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

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.

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

The holomorphic differential 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