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Schwarz-Christoffel Elliptic Integrals

You might wonder if there are any other elliptic integrals similar to the previous section's which are also preserved by Möbius transforms. That is an integral of the form \begin{equation*} \bigint \frac 1 {\sqrt[n] {P(x)}} dx \end{equation*} with associated irreducible curve of genus 1 of the form \begin{equation} \label{eq:curve} y^n = P(x) = K\prod_{i=1}^m {(x-e_i)^{k_i}} \end{equation} where the $e_i$ are distinct and $k_i \gt 0$ and the differential $\displaystyle \frac {dx} {y}$ is holomorphic. It is convenient to assume that the curve has no branch points at infinity because they are just limit cases when some $e_i \rightarrow \infty$. Under those conditions there are just four such polynomials

The elliptic integrals corresponding to these solutions are \begin{equation} \label{eq:integral2} I_2 = \bigint \frac {1} {\sqrt {(x-e_1)(x-e_2)(x-e_3)(x-e_4)}} dx \end{equation} which we investigated in the previous section, and three new integrals \begin{equation} \label{eq:integral3} I_3 = \bigint \frac {1} {\sqrt[3] {(x-e_1)^2(x-e_2)^2(x-e_3)^2}} dx \end{equation} \begin{equation} \label{eq:integral4} I_4 = \bigint \frac {1} {\sqrt[4] {(x-e_1)^2(x-e_2)^3(x-e_3)^3}} dx \end{equation} \begin{equation} \label{eq:integral6} I_6 = \bigint \frac {1} {\sqrt[6] {(x-e_1)^3(x-e_2)^4(x-e_3)^5}} dx \end{equation} We now extend the invariant integral method of the previous section to these new integrals.

Invariant Integrals

Similiar to previously we can construct an integral invariant under Möbius transforms $L$. If $L$ maps $e_1',e_2',e_3'$ to $e_1,e_2,e_3$ then for the integral $I_3$ under the change of variables $x = L(u)$ we get \begin{equation} \label{eq:invariant3} \bigint_{C} {\sqrt[3] \frac {(e_1-e_2)(e_1-e_3)(e_2-e_3)} {(x-e_1)^2(x-e_2)^2(x-e_3)^2}} dx = \bigint_{C\thinspace'} {\sqrt[3] \frac {(e_1'-e_2')(e_1'-e_3')(e_2'-e_3')} {(u-e_1')^2(u-e_2')^2(u-e_3')^2}} du \end{equation} A similar thing can be done for $I_4$ and $I_6$.

We can also write equation \eqref{eq:invariant3} in terms of the coefficients of the associated polynomial \begin{equation} \label{eq:invariantd3} \sqrt[6] {\discrim(a,b,c,d)} \bigint_{C} \frac 1 {\sqrt[3] {(ax^3+bx^2+cx+d)^2}} dx = \sqrt[6] {\discrim(p,q,r,s)} \bigint_{C\thinspace'} \frac 1 {\sqrt[3] {(pu^3+qu^2+ru+s)^2}} du \end{equation} where $\discrim(a,b,c,d)$ is the discriminant of the cubic polynomial \begin{equation} \discrim(a,b,c,d) = -27a^2d^2 + 18abcd - 4ac^3 - 4b^3d + b^2c^2 \end{equation} And similarly for $I_4$ and $I_6$.

We now want to transform the integrals to Weierstrass form. This may seem daunting but once we understand the nature of the elliptic functions uniformising the curves it is very easy.

Tiling Patterns

When the $e_i$ are real the integrals \eqref{eq:integral2}, \eqref{eq:integral3}, \eqref{eq:integral4} and \eqref{eq:integral6} are special cases of the Schwarz-Christoffel mapping. By examining the value of the integrand along the real axis and especially how it changes at the roots we can deduce that it maps the upper half plane to a polygon. The restriction to the real axis is unnecessary and can be weakened to the requirement that the $e_i$ lie on a circle or straight line (exercise for the reader).

For $I_2$ the polygon is a rectangle; for $I_3$ a triangle with internal angles 60, 60, 60; for $I_4$ a triangle with internal angles 90, 45, 45; and for $I_6$ a triangle with internal angles 90, 60, 30.

In the tiling patterns below $e_1, e_2, e_3, e_4$ are represented by green, blue, red and yellow dots respectively. They are the centers of various rotational symmetries. The upper half-plane is light coloured and the lower dark.

The elliptic functions $f$ which invert these integrals have additional rotational symmetries of the form \begin{equation*} f(\zeta (z - z_i) + z_i) = f(z) \end{equation*} where $f(z_i) = e_i$ and $\zeta$ is a $n/(n - k_i)$ th root of unity. With respect to this group they take each value once in a fundamental region. In the tiling patterns above a fundamental region is one dark plus one light triangle (or parallelogram). The fundamental region is isomorphic to a sphere.

Prototype Curves

The Weierstrass elliptic function has periods with 6-fold rotational symmetry when $g_2 = 0$ and 4-fold rotational symmetry when $g_3 = 0$. As a result $\wp(\zeta z) = \zeta^{-2} \wp(z)$ where $\zeta$ is a suitable root of unity. And if you differentiate it sufficient times you get an elliptic function with rotational symmetry.

For example by differentiating $\wp'\thinspace^2 = 4\wp^3 + 1$ and then eliminating $\wp$ we get $2\wp''\thinspace^3 = 27(\wp'\thinspace^2 - 1)^2$, which is a special case prototype of $I_3$. \begin{equation} \label{eq:prototype3} y^3 = \tfrac {27} {2} (x^2 - 1)^2 \end{equation} Another way of expressing this is that the birational map $x=v,\space y=6u^2$ maps curve \eqref{eq:prototype3} to the Weierstrass curve \begin{equation*} v^2 = 4u^3 + 1 \end{equation*} and the holomorphic differentials similarly \begin{equation*} \frac {dx} y = \frac {du} v \end{equation*} That is to say the change of variables $x=\sqrt {4u^3 + 1}$ transforms the prototype case of $I_3$ to Weierstrass form \begin{equation} \label{eq:protoint3} \bigint \frac 1 {\sqrt[3]{\tfrac {27} {2} (x^2 - 1)^2}} dx = \bigint \frac 1 {\sqrt{4u^3 + 1}} du \end{equation} A similar prototype curve can be obtained for $I_4$ and $I_6$, see the table below. For $I_4$ the change of variables $x=6u^2 - \tfrac 1 2$ transforms the prototype to Weierstrass form \begin{equation} \label{eq:protoint4} \bigint \frac 1 {\sqrt[4]{\tfrac {16} {3} (x - 1)^2(2x + 1)^3}} dx = \bigint \frac 1 {\sqrt{4u^3 - u}} du \end{equation} and for $I_6$ the change of variables $x=120u^3-12$ transforms the prototype to Weierstrass form \begin{equation} \label{eq:protoint6} \bigint \frac 1 {\sqrt[6]{\tfrac {1944} {5} (x - 18)^3(x + 12)^4}} dx = \bigint \frac 1 {\sqrt{4u^3 - 1}} du \end{equation} The results are summarised in the table below.

Reduction to Weierstrass Form

We now have sufficient formulae to carry out the reduction of $I_3$ to Weierstrass form. In equation \eqref{eq:invariant3} add a factor $K'$ to the numerator and denominator of the right hand side and let $e_3' \rightarrow \infty$. Let L be the Möbius transform mapping $1,-1,\infty$ to $e_1,e_2,e_3$. Under the change of variables $x = L(t)$ and then $t=\sqrt {4u^3 + 1}$ we have using \eqref{eq:protoint3} \begin{equation} \bigint {\sqrt[3] \frac {(e_1-e_2)(e_1-e_3)(e_2-e_3)} {(x-e_1)^2(x-e_2)^2(x-e_3)^2}} dx = \sqrt[3] {K'(e_1'-e_2')} \bigint \frac 1 {\sqrt[3] {K'(t-e_1')^2(t-e_2')^2}} dt = \sqrt[3] {27} \bigint {\frac 1 {\sqrt[3] {\tfrac {27} {2} (t^2 - 1)^2}}} dt = 3 \bigint \frac 1 {\sqrt{4u^3 + 1}} du \end{equation}

Rolling all that together under a single (implicit) change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{\sqrt {4u^3 + 1}, 1,-1,\infty}$

Similarly for $I_4$ under the change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{6u^2-\tfrac 1 2, 1, -\tfrac 1 2, \infty}$

And for $I_6$ under the change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{120u^3-12, 12, -18, \infty}$

Computation of $g_2$ and $g_3$ Invariants

From these equations we can finally calculate the $g_2$ and $g_3$ invariants for the curves in terms of their coefficients. For example to calculate $g_3$ for $I_3$ \begin{equation} g_3\left(\bigint {\frac 1 {\sqrt[3] {(ax^3+bx^2+cx+d)^2}}} dx\right) = \left[\frac {3} {\sqrt[6]{\discrim(a,b,c,d)}}\right]^{-6} g_3\left(\bigint \frac 1 {\sqrt{4u^3 + 1}} du\right) = - \frac 1 {3^6} \discrim(a,b,c,d) \end{equation} We can now be put the integrals into the same form as the previous section.

Let $g_3=- 3^{-6} \discrim(a,b,c,d)$, then under the change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{\sqrt {1 - {4t^3}/{g_3}}, 1,-1,\infty}$ \begin{equation} \bigint {\frac 1 {\sqrt[3] {(ax^3+bx^2+cx+d)^2}}} dx = \bigint \frac 1 {\sqrt{4t^3 - g_3}} dt \end{equation}

For $I_4$, let $g_2=2^{-6}(aq^2 - bpq + cp^2) (b^2 - 4ac)$, then under the change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{6t^2/g_2-\tfrac 1 2, 1, -\tfrac 1 2, \infty}$ \begin{equation} \bigint {\frac 1 {\sqrt[4] {(px + q)^2(ax^2 + bx + c)^3}}} dx = \bigint \frac 1 {\sqrt{4t^3 - g_2 t}} dt \end{equation}

For $I_6$, let $g_3=2^{-4}3^{-6}(ad - bc)(af - be)^2(cf - de)^3$, then under the change of variables $\crossratio{x,e_1,e_2,e_3} = \crossratio{120t^3/g_3-12, 12, -18, \infty}$ \begin{equation} \bigint {\frac 1 {\sqrt[6] {(ax + b)^3(cx + d)^4(ex + f)^5}}} dx = \bigint \frac 1 {\sqrt{4t^3 - g_3}} dt \end{equation}

The invariants for all curves are summarised in the following table.

Differential equations

To solve the differential equation below for $f$ in terms of the Weierstrass $\wp'$ function: \begin{equation} \label{eq:fde} f'(z)^3 = \left(a f(z)^3 + b f(z)^2 + c f(z) + d\right)^2 \end{equation} Let $e_1,e_2,e_3$ be the roots of the polynomial and $g_3$ be as defined in the table above for $I_3$. Assume a boundary condition of $f(0) = e_3$. Then $f(z)$ is given implicitly by the cross ratio formula \begin{equation} \label{eq:solution} \crossratio{f(z),e_1,e_2,e_3} = \crossratio{\wp'(z,0,g_3),1,-1,\infty} \end{equation} and for any $w,x,y,z$ the two cross-ratio's and sigma ratio are equal

and $\zeta$ is a primitive cube root of unity. Similar formulae hold for $I_4$ and $I_6$. This equation leads to special addition formulae for the sigma function.

Connection to the Beta and Hypergeometic functions

The integrals $I_3$, $I_4$ and $I_6$ are closely connected to cases $B(\tfrac 1 3, \tfrac 1 3)$, $B(\tfrac 1 2, \tfrac 1 4)$ and $B(\tfrac 1 2, \tfrac 1 3)$ of the Beta, Incomplete Beta and Hypergeometric functions. Under the change of variables $t=\crossratio{x,a,c,b}$ with $0 \lt \alpha,\beta,\gamma \lt 1$ and $\alpha+\beta+\gamma=1$ we have \begin{equation} \label{eq:beta} B-A = \bigint_{a}^{b} \frac 1 {(x-a)^{1-\alpha}(x-b)^{1-\beta}(x-c)^{1-\gamma}} dx = \frac 1 {(a-b)^{\gamma}(b-c)^{\alpha}(c-a)^{\beta}} \bigint_{0}^{1} t^{\alpha-1}(1-t)^{\beta-1} dt = \frac {\Gamma(\alpha)\Gamma(\beta)\Gamma(\gamma)} {\pi\thinspace(a-b)^{\gamma}(b-c)^{\alpha}(c-a)^{\beta}} \sin(\pi\gamma) \end{equation} This integral maps the interior of the circle through the points $a,b,c$ to a triangle, say $A,B,C$, with internal angles $\pi\alpha,\pi\beta,\pi\gamma$. The lengths of the sides of the triangle are therefore \begin{equation} |A-B| = |d| \sin(\pi\gamma), \qquad |B-C| = |d| \sin(\pi\alpha), \qquad |C-A| = |d| \sin(\pi\beta) \end{equation} where \begin{equation} d = \frac {\Gamma(\alpha)\Gamma(\beta)\Gamma(\gamma)} {\pi\thinspace (a-b)^{\gamma}(b-c)^{\alpha}(c-a)^{\beta}} \end{equation} and $|d|$ is the diameter (by the law of sines) of the triangle's circumscribing circle.

Computation of periods

So for the $I_3$ tiling pattern, using \eqref{eq:beta} with $\alpha=\beta=\gamma=\tfrac 1 3$, the 6 minimal periods are given by \begin{equation} \omega = (B-A)(1 + \zeta)\zeta^{k} = \frac {\Gamma\left(\tfrac 1 3\right)^3} {\pi\thinspace\Delta^{1/6}}\tfrac 3 2\zeta^{k + \tfrac 1 2} \end{equation} where $\Delta$ is the discriminant of the cubic polynomial, $\zeta$ is the 6th root of unity and $k=0,1,2,3,4,5$. For the $I_4$ tiling pattern with $\alpha=\tfrac 1 2,\beta=\gamma=\tfrac 1 4$, the 4 minimal periods are given by \begin{equation} \omega = (B-A)2i^{k} = \frac {\Gamma\left(\tfrac 1 2\right)\Gamma\left(\tfrac 1 4\right)^2} {\pi\thinspace\Delta_1^{1/4}\Delta_2^{1/4}}\sqrt{2}\thinspace i^{k} \end{equation} where $\Delta_1$ is the of the discriminant of the quadratic and $\Delta_2$ is the resultant of the quadratic and linear polynomials and $k=0,1,2,3$. For the $I_6$ tiling pattern with $\alpha=\tfrac 1 2,\beta=\tfrac 1 6,\gamma=\tfrac 1 3$, the 6 minimal periods are given by \begin{equation} \omega = (B-A)2\zeta^{k} = \frac {\Gamma\left(\tfrac 1 2\right)\Gamma\left(\tfrac 1 3\right)\Gamma\left(\tfrac 1 6\right)} {\pi\thinspace\Delta_1^{1/6}\Delta_2^{1/3}\Delta_3^{1/2}}\sqrt{3}\thinspace\zeta^{k} \end{equation} where $\Delta_1,\Delta_2,\Delta_3$ are the resultants of the linear polynomials and $k=0,1,2,3,4,5$.

Alternate computation of periods

Combining the change of variables used in \eqref{eq:protoint4} and \eqref{eq:beta}, $\displaystyle t = \frac 1 {4u^2}$, allows us to compute a half period of the lemniscatic Weierstrass elliptic integral \begin{equation} \tfrac 1 2 \omega = \bigint_{\sfrac 1 2}^{\infty} \frac 1 {\sqrt{4u^3 - u}} du = 2^{-3/2} \bigint_0^1 t^{-3/4} (1-t)^{-1/2} dt = 2^{-3/2} \frac {\Gamma\left(\tfrac 1 4\right) \Gamma\left(\tfrac 1 2\right)} {\Gamma\left(\tfrac 3 4\right)} = \frac {\Gamma\left(\tfrac 1 4\right)^2} {4\sqrt{\pi}} \end{equation} which in turn allows us to compute the sum of the inverse fourth powers of the Gaussian integers \begin{equation} \sum_{(n,m) \ne (0,0)} \frac 1 {(n + mi)^4} = \tfrac 1 {60} g_2(1,i) = \tfrac 1 {60} \omega^{-4} = \frac {4\thinspace\pi^2} {15\thinspace\Gamma\left(\tfrac 1 4\right)^{8}} \end{equation} Combining the change of variables used in \eqref{eq:protoint6} and \eqref{eq:beta}, $\displaystyle t = \frac 1 {4u^3}$, allows us to compute a half period of the equianharmonic Weierstrass elliptic integral \begin{equation} \tfrac 1 2 \omega = \bigint_{4^{-1/3}}^{\infty} \frac 1 {\sqrt{4u^3 - 1}} du = 2^{-2/3}3^{-1} \bigint_0^1 t^{-5/6} (1-t)^{-1/2} dt = 2^{-2/3}3^{-1} \frac {\Gamma\left(\tfrac 1 6\right) \Gamma\left(\tfrac 1 2\right)} {\Gamma\left(\tfrac 2 3\right)} = \frac {\Gamma\left(\tfrac 1 3\right)^3} {4\pi} \end{equation} which in turn allows us to compute the sum of the inverse sixth powers of the Eisenstein integers \begin{equation} \sum_{(n,m) \ne (0,0)} \frac 1 {(n + m\zeta)^6} = \tfrac 1 {140} g_3(1,\zeta) = \tfrac 1 {140}\omega^{-6} = \frac {16\thinspace\pi^6}{35\thinspace\Gamma\left(\tfrac 1 3\right)^{18}} \end{equation} where $\zeta=e^{2\pi i/3}$.

Level 1 Addition Formulae

The level 1 addition formulae can be computed from the birational mapping to the Weierstrass curve as follows:

1. Note that one row of the addition formula should be $\begin{vmatrix} 1 & u & v \end{vmatrix}$ where $u$ and $v$ are the variables for the Weierstrass curve.
2. Substitute in the values for $u$ and $v$ using the birational mapping in table 3, and substitute the value for $y$ from the curves equation.
3. Apply a general Möbius transform to $x$ to move the first root to $e_1$, the second to $e_2$, and $\infty$ to $e_3$.
4. Finally simplify the determinant using row and column operations.

Note these root expressions are rational functions of $x$ and $y$. For example in $I_3$ (ignoring $K$) we have \begin{equation*} \sqrt[3]{(x-e_1)(x-e_2)(x-e_3)} \space = \space \frac {(x-e_1)(x-e_2)(x-e_3)} {y} \end{equation*} and in $I_4$ we have \begin{equation*} \sqrt{(x-e_2)/(x-e_3)} \space = \space \frac {(x-e_1)(x-e_2)^2(x-e_3)} {y^2}, \qquad\qquad \sqrt[4]{(x-e_1)^2(x-e_2)/(x-e_3)^3} \space = \space \frac {(x-e_1)(x-e_2)} {y} \end{equation*} and in $I_6$ we have \begin{equation*} \sqrt{(x-e_1)/(x-e_3)} \space = \space \frac {(x-e_1)^2(x-e_2)^2(x-e_3)^2} {y^3}, \qquad\qquad \sqrt[3]{(x-e_2)/(x-e_3)} \space = \space \frac {(x-e_1)^2(x-e_2)^3(x-e_3)^3} {y^4} \end{equation*}

Three Variable Level 2 Addition Formulae

Therefore applying a change of variables listed above to these formulae for the Weierstrass curve \begin{equation*} \bigint_{\infty}^{u_1} \frac {du} {\sqrt{4u^3 - 1}} \enspace + \enspace \bigint_{\infty}^{u_2} \frac {du} {\sqrt{4u^3 - 1}} \enspace + \enspace \bigint_{\infty}^{u_3} \frac {du} {\sqrt{4u^3 - 1}} \enspace = \enspace 0 \end{equation*} gives \begin{equation*} \bigint_{e_3}^{x_1} \frac {dx} {\sqrt[6]{(x-e_1)^3(x-e_2)^4(x-e_3)^5}} \enspace + \enspace \bigint_{e_3}^{x_2} \frac {dx} {\sqrt[6]{(x-e_1)^3(x-e_2)^4(x-e_3)^5}} \enspace + \enspace \bigint_{e_3}^{x_3} \frac {dx} {\sqrt[6]{(x-e_1)^3(x-e_2)^4(x-e_3)^5}} \enspace = \enspace 0 \end{equation*} and because the point $(e_3,0)$ on the curve corresponds to the point $(\infty,\infty)$ on the Weierstrass curve, then \begin{equation*} \begin{vmatrix} 1 & \sqrt{(x_1 - e_1)/(x_1 - e_3)} & \sqrt[3]{(x_1 - e_2)/(x_1 - e_3)} \\ 1 & \sqrt{(x_2 - e_1)/(x_2 - e_3)} & \sqrt[3]{(x_2 - e_2)/(x_2 - e_3)} \\ 1 & \sqrt{(x_3 - e_1)/(x_3 - e_3)} & \sqrt[3]{(x_3 - e_2)/(x_3 - e_3)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation*} Using methods similar to those for the quartic curve this can be solved for $x_3$ to give \begin{equation*} x_3 \enspace = \enspace \frac 1 {x_1x_2} \prod_{i=0}^5 \genfrac {}{}{1.5pt}{0} {\begin{vmatrix} 1 & \sqrt{(x_1 - e_1)/(x_1 - e_3)} & \sqrt[3]{(x_1 - e_2)/(x_1 - e_3)} \\ 1 & \sqrt{(x_2 - e_1)/(x_2 - e_3)} & \sqrt[3]{(x_2 - e_2)/(x_2 - e_3)} \\ 1 & \zeta^{3i} \sqrt{e_1/e_3} & \zeta^{2i} \sqrt[3]{e_2/e_3} \\ \end{vmatrix}} {\begin{vmatrix} 1 & \sqrt{(x_1 - e_1)/(x_1 - e_3)} & \sqrt[3]{(x_1 - e_2)/(x_1 - e_3)} \\ 1 & \sqrt{(x_2 - e_1)/(x_2 - e_3)} & \sqrt[3]{(x_2 - e_2)/(x_2 - e_3)} \\ 1 & \zeta^{3i} & \zeta^{2i} \\ \end{vmatrix}} \end{equation*} where $\zeta$ is a primitive sixth root of unity.