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In this section we investigate how certain elliptic integrals transform under Möbius transformations. Then we use those results to reduce the general integral to Weierstrass form.

We seek to construct a Möbius transformation $x = L(u)$ which transforms the integral below into Weierstrass form.

#fold \bigint_{C} \frac 1 {\sqrt {ax^4 + bx^3 + cx^2 + dx + e}} dx = \bigint_{C'} \frac 1 {\sqrt {4u^3 - g_2u - g_3}} du

Doing so is also equivalent to constructing a birational mapping

x = L(u),\quad y = vL'(u)

between the curves

#stack,width=600,list,aligned,lineSpacing=0.5em y^2 = ax^4 + bx^3 + cx^2 + dx + e v^2 = 4u^3 - g_2u - g_3

and the holomorphic differentials

\frac {dx} {y} = \frac {du} {v}

It is also equivalent to solving the differential equation

f'^2 = af^4 + bf^3 + cf^2 + df + e

using the Weierstrass $\wp$ function

f(z) = L(\wp(z, g_2, g_3))

HISTORICAL NOTE

Invariant Integral

First we determine how the elliptic integral changes when transformed by Möbius transformations.

Consider the integral whose integrand is modelled on the cross-ratio formula:

\bigint_{C} \sqrt {\frac {(e_1 - e_2)(e_3 - e_4)} {(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx

Using the Möbius difference formula we see that it is invariant under Möbius transformations. That is if $L$ is a Möbius transformation mapping $e_1',e_2',e_3',e_4'$ to $e_1,e_2,e_3,e_4$ and the path of integration $C'$ to $C$ then the change of variables $x = L(u)$ gives

#fold \bigint_{C} \sqrt {\frac {(e_1 - e_2)(e_3 - e_4)} {(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C'} \sqrt {\frac {(e_1' - e_2')(e_3' - e_4')} {(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du

The cross-ratio's of the two sets of roots are by necessity equal.

\crossratio{e_1,e_2,e_3,e_4}=\crossratio{e_1',e_2',e_3',e_4'}

Equation invariant remains true if we introduce arbitrary leading coefficients $K$ and $K'$ into the polynomials

#fold \bigint_{C} \sqrt {\frac {K(e_1 - e_2)(e_3 - e_4)} {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C'} \sqrt {\frac {K'(e_1' - e_2')(e_3' - e_4')} {K'(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du

If we let $e_4' \rightarrow \infty$ then invariantK becomes

#fold \bigint_{C} \sqrt {\frac {K(e_1 - e_2)(e_3 - e_4)} {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C'} \sqrt {\frac {K'(e_1' - e_2')} {K'(u - e_1')(u - e_2')(u - e_3')}} du

and the cross-ratio formula crossratio becomes

\crossratio{e_1,e_2,e_3,e_4}=\crossratio{e_1',e_2',e_3',\infty}

We now have the formulae needed to carry out the reduction.

Reduction to Weierstrass Form

Let

#stack,width=800,list,aligned A = K(e_1 - e_2)(e_3 - e_4) B = K(e_3 - e_1)(e_2 - e_4) C = K(e_2 - e_3)(e_1 - e_4)

then it is easily confirmed that

A + B + C = 0

Put

#stack,width=600,list,aligned \epsilon_1 = \tfrac 1 {12} (A - B) \epsilon_2 = \tfrac 1 {12} (C - A) \epsilon_3 = \tfrac 1 {12} (B - C)

so that

\epsilon_1 + \epsilon_2 + \epsilon_3 = 0

and

#stack,width=900,list,aligned K(e_1 - e_2)(e_3 - e_4) = 4(\epsilon_1 - \epsilon_2) K(e_3 - e_1)(e_2 - e_4) = 4(\epsilon_3 - \epsilon_1) K(e_2 - e_3)(e_1 - e_4) = 4(\epsilon_2 - \epsilon_3)

and

\crossratio{e_1,e_2,e_3,e_4} = \crossratio{\epsilon_1,\epsilon_2,\epsilon_3,\infty}

Then under the change of variables $x = L(u)$ given implicitly by

\crossratio{x,e_1,e_2,e_3} = \crossratio{u,\epsilon_1,\epsilon_2,\epsilon_3}

we get using invariantcubic and ABCP

\begin{aligned} \bigint_{C} \frac 1 {\sqrt {ax^4 + bx^3 + cx^2 + dx + e}} dx &= \bigint_{C} \frac 1 {\sqrt {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx \\ &= \bigint_{C'} \frac 1 {\sqrt {4(u - \epsilon_1)(u - \epsilon_2)(u - \epsilon_3)}} du \\ &= \bigint_{C'} \frac 1 {\sqrt {4u^3 - g_2u - g_3}} du \end{aligned}

Quartic Invariants

Using ABC, proots and reduction we can obtain expressions for $g_2$ and $g_3$ in terms of the other coefficients.

#stack,width=700,aligned g_2 = -4 (\epsilon_1 \epsilon_2 + \epsilon_2 \epsilon_3 + \epsilon_3 \epsilon_1) = \tfrac {1} {24} (A^2 + B^2 + C^2) = \tfrac {1} {12} (12ae - 3bd + c^2)

and

#stack,width=800,aligned g_3 = 4 \epsilon_1 \epsilon_2 \epsilon_3 = \tfrac {1} {432} (A - B)(B - C)(C - A) = \tfrac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3)

The derivation of the last terms in g2 and g3 is not obvious. The expressions involving $A, B, C$ are symmetric polynomials in $e_1,e_2,e_3,e_4$, a fact which is also not entirely obvious. Therefore by a well known theorem they can be written as polynomials in the coefficients $a,b,c,d,e$. In fact more than that. Because they are symmetric root differences, they are invariants of the quartic polynomial, in the sense of Invariant Theory.

Modular Discriminant

In addition the modular discriminant is defined by

#stack,width=800,aligned \Delta = g_2^3 - 27 g_3^2 = 16 (\epsilon_1 - \epsilon_2)^2 (\epsilon_2 - \epsilon_3)^2 (\epsilon_3 - \epsilon_1)^2 = \tfrac {1} {256} A^2 B^2 C^2 = \tfrac {1} {256} \discrim(a,b,c,d,e)

where $\discrim(a,b,c,d,e)$ is the discriminant of the quartic polynomial.

#flow,indent \discrim(a,b,c,d,e) = 256a^3e^3 - 192a^2bde^2 - 128a^2c^2e^2 + 144a^2cd^2e - 27a^2d^4 + 144ab^2ce^2 - 6ab^2d^2e - 80abc^2de + 18abcd^3 + 16ac^4e - 4ac^3d^2 - 27b^4e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2c^2d^2

Observe that

\discrim(a,b,c,d,e) = \discrim(0,4,0,-g_2,-g_3)

Finally we have achieved our goal of transforming the general integral to Weierstrass form via a Möbius transformation mobius.

Cubic Case

Well not quite. Equation ABC needs an adjustment to deal with the cubic case. Letting $e_4 \rightarrow \infty$ in invariantcubic reveals the appropriate formula is

#stack,width=540,list,aligned A = K(e_1 - e_2) B = K(e_3 - e_1) C = K(e_2 - e_3)

From there the rest of the reduction is the same with $a = 0$ and $e_4 = \infty$ and the $e_4$ terms dropping out of ABCP and reduction.

12th Root Formula

Equation invariant is a pretty formula but its numerator appears to lack symmetry. The explanation for this is simple. The equality of the cross-ratios crossratio implies the following identity

\frac {(e_1 - e_2)(e_3 - e_4)} {(e_1' - e_2')(e_3' - e_4')} = \frac {(e_3 - e_1)(e_2 - e_4)} {(e_3' - e_1')(e_2' - e_4')} = \frac {(e_2 - e_3)(e_1 - e_4)} {(e_2' - e_3')(e_1' - e_4')}

and so invariant also holds for any permutation of the roots in the numerator. It can even be written symmetrically as

#fold \bigint_{C} \frac {\sqrt[12] {(e_1 - e_2)^2(e_1 - e_3)^2(e_1 - e_4)^2(e_2 - e_3)^2(e_2 - e_4)^2(e_3 - e_4)^2}} {\sqrt{(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C'} \frac {\sqrt[12] {(e_1' - e_2')^2(e_1' - e_3')^2(e_1' - e_4')^2(e_2' - e_3')^2(e_2' - e_4')^2(e_3' - e_4')^2}} {\sqrt{(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du

The symmetry in this formula allows it to be written purely in terms of the coefficients of the polynomials. If the roots of two quartic polynomials can be mapped to one another by a Möbius transform $L$ then under the change of variables $x = L(u)$ we get

#fold \sqrt[12] {\discrim(a,b,c,d,e)}\bigint_{C} \frac 1 {\sqrt{ax^4 + bx^3 + cx^2 + dx + e}} dx = \sqrt[12] {\discrim(p,q,r,s,t)} \bigint_{C'} \frac 1 {\sqrt{pu^4 + qu^3 + ru^2 + su + t}} du

Formula invariantdeltaK remains valid when one or both of the polynomials are cubic ie. $a=0$ or $p=0$. In such cases the "missing" 4th root used to define the Möbius transform is $\infty$.

FUN FACT

Jacobian and Legendre Curves

We can now apply the reduction formulae to compute the $g_2$ and $g_3$ invariants for other standard curves.

The curve associated with Jacobi's elliptic sine function is

y^2 = (1-x^2)(1-k^2x^2)

Put $K = k^2$, $e_1=1$, $e_2=-1$, $e_3=1/k$, $e_4= -1/k$ then using ABC

A=4k,\quad B=(k-1)^2,\quad C=-(k+1)^2

A more symmetric version of this curve is

y^2 = (x^2-\alpha^2)(x^2-\alpha^{-2})

Put $K = 1$, $e_1=\alpha$, $e_2=-\alpha$, $e_3=1/\alpha$, $e_4= -1/\alpha$ then using ABC

A=4,\quad B=(\alpha-1/\alpha)^2,\quad C=-(\alpha+1/\alpha)^2

The Legendre curve is

y^2 = x(x-1)(x-\lambda)

Put $K = 1,\space e_1=0,\space e_2=\lambda,\space e_3=1,\space e_4= \infty$ then using ABC3

A=-\lambda,\quad B=1,\quad C=\lambda - 1

In summary

Table 1
Curve $A,\space B,\space C$ $g_2$ $g_3$ $\Delta$
${y^2 = K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}$ ABC $\frac {1} {24} (A^2 + B^2 + C^2)$ $\frac {1} {432} (A - B)(B - C)(C - A)$ $\frac {1} {256} A^2 B^2 C^2$
$y^2 = ax^4 + bx^3 + cx^2 + dx + e$ ${\frac {1} {12} (12ae - 3bd + c^2)}$ ${\frac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3)}$ $\tfrac 1 {256} \discrim(a,b,c,d,e)\quad$ see discriminant
$y^2 = 4x^3 - g_2x - g_3$ $g_2$ $g_3$ $g_2^3 - 27g_3^2$
${y^2 = 4(x - e_1)(x - e_2)(x - e_3) \space \dagger}$ ABCP ${-4 (e_1 e_2 + e_2 e_3 + e_3 e_1)}$ ${4 e_1 e_2 e_3}$ ${16 (e_1 - e_2)^2 (e_2 - e_3)^2 (e_3 - e_1)^2}$
$y^2 = (1 - x^2)(1 - k^2x^2)$ ABCjacobi $\frac {1} {12} (k^4 + 14k^2 + 1)$ $\frac {1} {216} (k^2 + 1)(k^2 + 6k + 1)(k^2 - 6k + 1)$ $\frac {1} {16} k^2(k - 1)^4(k + 1)^4$
$y^2 = (x^2 - \alpha^2)(x^2 - \alpha^{-2})$ ABCalpha $\frac {1} {12} (\alpha^4 + 14 + \alpha^{-4})$ $\frac {1} {216} (\alpha^2 + \alpha^{-2})(\alpha^4 - 34 + \alpha^{-4})$ $\frac {1} {16} \left(\alpha^2 - \alpha^{-2}\right)^4$
$y^2 = x(x - 1)(x - \lambda)$ ABClegendre $\frac {1} {12} (\lambda^2 - \lambda + 1)$ $\frac {1} {432} (\lambda - 2)(2\lambda - 1)(\lambda + 1)$ $\frac {1} {256} \lambda^2(\lambda - 1)^2$

$\dagger \space e_1 + e_2 + e_3 = 0$

The $j$ invariants are given by

#stack,width=950,aligned,lineSpacing=1em j = 1728 {g_2^3 \over g_2^3 - 27g_3^2} = 32 {(A^2 + B^2 + C^2)^3 \over A^2 B^2 C^2} = 16 {(k^4 + 14k^2 + 1)^3 \over k^2(k^2 - 1)^4} = 16 {(\alpha^8 + 14\alpha^4 + 1)^3 \over \alpha^4(\alpha^4 - 1)^4} = 256 {(\lambda^2 - \lambda + 1)^3 \over \lambda^2(\lambda - 1)^2}

Special Lattices

Table 2
Lattice $\lambda \space = \space \crossratio{e_1,e_2,e_3,e_4}$ $g_2, \space g_3, \space \Delta$ $J$
Degenerate $\lambda \space = \space 0,1,\infty$ $\Delta \space = \space 0$ $J \space = \space \infty$
Hexagonal $\lambda \space = \space \frac {1 \pm \imath \sqrt{3}} {2}$ $g_2 \space = \space 0$ $J \space = \space 0$
Square $\lambda \space = \space -1,\tfrac 1 2,2$ $g_3 \space = \space 0$ $J \space = \space 1$
Rectangular $\lambda \in \R$ $J \in \R \enspace \textsf{and} \enspace J \space \ge \space 1$

Lattice is rectangular iff the four roots are distinct and lie on a circle or straight line.

Dihedral Symmetry

The $\lambda$ version of the $j$-invariant formula factorises as follows

#fold {(\delta^2 - \delta + 1)^3 \over \delta^2(\delta - 1)^2} \enspace - \enspace {(\lambda^2 - \lambda + 1)^3 \over \lambda^2(\lambda - 1)^2} \enspace = \enspace \frac {(\lambda - \delta)(1 - \lambda \delta)(1 - \lambda - \delta)(1 - \delta + \lambda \delta)(1 - \lambda + \lambda \delta)(\lambda + \delta - \lambda \delta)} {\lambda^2 (\lambda - 1)^2 \thinspace \delta^2 (\delta - 1)^2}

Therefore, as expected, the $j$-invariants are equal when

\delta \enspace = \enspace \lambda,\quad \frac 1 \lambda, \quad 1-\lambda, \quad \frac 1 {1-\lambda}, \quad \frac {\lambda-1} \lambda, \quad \frac \lambda {\lambda-1}

When interpreted as rotations on the sphere via a suitably positioned stereographic projection they form the dihedral symmetry group of order 6.

Octahedral Symmetry

The $\alpha$ version of the $j$-invariant formula factorises as follows

#fold,marked {(\beta^8 + 14\beta^4 + 1)^3 \over \beta^4(\beta^4 - 1)^4} \enspace - \enspace {(\alpha^8 + 14\alpha^4 + 1)^3 \over \alpha^4(\alpha^4 - 1)^4} \enspace = \enspace $ \frac {\prod\limits_{i=0}^3 \left(\alpha - \zeta^i \beta \right) \cdot \prod\limits_{i=0}^3 \left(1 - \zeta^i \alpha\beta\right) \cdot \prod\limits_{i=0}^3 \prod\limits_{j=0}^3 \left(1 - \zeta^i\alpha - \zeta^j\beta - \zeta^{i+j} \alpha\beta\right)} {\vphantom{\prod\limits_{i=0}^0} \alpha^4 (1 - \alpha^4)^4 \thinspace \beta^4 (1 - \beta^4)^4}

where $\zeta$ is a primitive fourth root of unity. Therefore the $j$-invariants are equal when

\beta \space = \space \zeta^i \alpha, \quad \frac 1 {\zeta^i \alpha}, \quad \zeta^j \frac {1 - \zeta^i\alpha} {1 + \zeta^i\alpha} \qquad \textsf{where} \qquad i=0\ldots 3, \enspace j=0 \ldots 3

When interpreted as rotations on the sphere via stereographic projection they form the (chiral) octahedral symmetry group of order 24.

OctahedronDiagram

DISCUSSION

Invariants of the holomorphic differential

Just to emphasize the point. Because of reduction, g2 and g3 we have calculated $\sum {\frac 1 {\omega^4}}$ and $\sum {\frac 1 {\omega^6}}$ where $\omega$ ranges over all the non-zero periods of the differential $\frac {dx} {y}$. Explicitly let

y^2 = ax^4 + bx^3 + cx^2 + dx + e

and let $\omega_C$ be the period of the differential associated with closed path $C$

\omega_C = \bigint_{C} \frac {dx} {y}

then

\begin{aligned} 60 \sum_{C} {\frac 1 {\omega_C^4}} &= \tfrac {1} {12} (12ae - 3bd + c^2) \\\\ 140 \sum_{C} {\frac 1 {\omega_C^6}} &= \tfrac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3) \end{aligned}

where $C$ ranges over all non-trival closed paths in the homology group of the curve.

Curve as a differential equation

To solve the differential equation below for $f$ in terms of the Weierstrass $\wp$ function:

f'(z)^2 = a f(z)^4 + b f(z)^3 + c f(z)^2 + d f(z) + e

Let $e_1,e_2,e_3,e_4,\epsilon_1,\epsilon_2,\epsilon_3,g_2,g_3$ be defined as in ABC thru g3. Assume a boundary condition of $f(0) = e_4$. Then $f(z)$ is given implicitly by the cross ratio formula

\crossratio{f(z),e_1,e_2,e_3} = \crossratio{\wp(z,g_2,g_3),\epsilon_1,\epsilon_2,\epsilon_3}

In fact solution, and the invariance of cross-ratio's under the Möbius transforms, implies for any $w,x,y,z$

\crossratio{f(w),f(x),f(y),f(z)} = \crossratio{\wp(w),\wp(x),\wp(y),\wp(z)} = \frac {\sigma(w-x)\sigma(w+x)\sigma(y-z)\sigma(y+z)} {\sigma(w-y)\sigma(w+y)\sigma(x-z)\sigma(x+z)}

The general solution of the fde is given by

f(z) = L(\wp(z+z_0,g_2,g_3))

where $z_0$ is effectively the constant of integration. For the boundary condition $f(0) = x_0$ there are in general two solutions corresponding to the two possible values of $f'(0) = y_0$. It might appear from solution that this solution would explicitly involve the roots $e_1,e_2,e_3,\epsilon_1,\epsilon_2,\epsilon_3$. But that is not the case because, for a fixed value of $x_0$ and $y_0$, such a solution must be unchanged when the roots are permuted, and therefore only involves the rational functions of the coefficients $a,b,c,d,e$ along with $x_0, y_0$. We could attempt to find this solution by expanding gensolution using the addition formula for $\wp(z+z_0)$. But using this power series method is more direct.

Addition Formula In 4 Variables

Similar to the well known symmetric addition formula for $\wp$ there is a symmetric addition formula for the general elliptic function $f$ of order 2.

If $z_1 + z_2 + z_3 + z_4 \equiv 2(a_1 + a_2) \mod \Omega$ where $a_1$ and $a_2$ are the location of the poles of $f$, then:

\begin{vmatrix} 1 & f(z_1) & f(z_1)^2 & f'(z_1) \\ 1 & f(z_2) & f(z_2)^2 & f'(z_2) \\ 1 & f(z_3) & f(z_3)^2 & f'(z_3) \\ 1 & f(z_4) & f(z_4)^2 & f'(z_4) \\ \end{vmatrix} \space = \space 0

EXAMPLE

To get a symmetric 3 variable formula put $z_4 = 2(a_1 + a_2)$.

Addition Formula For The Integral

The natural way to express the above addition formula in terms of integrals is to say that if $R$ is a polynomial, of degree 3 or 4, and

\bigint_{l_1}^u \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_1}^v \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_2}^w \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_2}^x \frac {dt} {\sqrt{R(t)}} \space \equiv \space 0 \mod \Omega

then $u,v,w,x$ satisfy the algebraic relation

\begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & x & x^2 & \sqrt{R(x)} \\ \end{vmatrix} \space = \space 0

The congruence relation, modulo the period lattice $\Omega$, is to take into account different paths of integration. There is also some complexity in the interpretation of the lower bounds. The congruence is only true when $l_1$ and $l_2$ are a pair of conjugate points. That is points which have the same complex $t$ value but correspond to two different points (or a branch point) on the underlying Riemann surface where the integration takes place.

As we will see below, many of the standard addition formula for elliptic integrals and functions are an expansion or specialisation of this formula.

ALTERNATIVE EXPRESSION

Level 1 Addition Formula In 3 Variables For Degree 3

Assume $R(t) = K(t-a)(t-b)(t-c)$.

We can obtain a 3 variable addition formula by letting $x \rightarrow \infty$ in addvar4 giving

\begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 0 & 0 & 1 & 0 \\ \end{vmatrix} \space = \space - \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \\ \end{vmatrix} \space = \space (v-w)\sqrt{\smash[b]{R(u)}} \space + \space (w-u)\sqrt{\smash[b]{R(v)}} \space + \space (u-v)\sqrt{\smash[b]{R(w)}} \space = \space 0

In order to distinguish this from the closely related formulae below, call it the level 1 formula.

Level 2 Addition Formula In 3 Variables For Degree 3

We can partialy rationalise the level 1 formula by multiplying by a conjugate expression to give

\begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \\ \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \\ \end{vmatrix} \space = \space (w-u)(w-v)\left(K\left[(u + v + w) - (a + b + c)\right]\left(u - v\right)^2 - \left(\sqrt{\smash[b]{R(u)}} - \sqrt{\smash[b]{R(v)}}\right)^2\right) \space = \space 0

Call this the level 2 addition formula. It can be solved for $w$ to give the non-trivial solution

w = (a + b + c) + \frac 1 K \left[\frac {\sqrt{R(u)} - \sqrt{R(v)}} {u - v}\right]^2 - (u + v)

When $K=4$ and $a+b+c=0$ this is the standard addition formula for the Weierstrass integral.

Level 3 Addition Formula in 3 Variables For Degree 3

We can repeat this process once more to produce level 3 addition formulae.

\begin{aligned} \left. \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & -\sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & -\sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \end{vmatrix} \space \middle/ \space \begin{vmatrix} 1 & u & u^2 \\ 1 & v & v^2 \\ 1 & w & w^2 \\ \end{vmatrix}^2 \right. \\\\ \space = \space K^2 \Big[\big((uv + uw + vw) - (ab + ac + bc)\big)^2 - 4\big((u + v + w) - (a + b + c)\big)(uvw - abc)\Big] \space = \space 0 \end{aligned}

This is an example of Abel's classic addition theorem for algebraic integrals. Rather mysteriously, in this case it is equivalent to the discriminant expression

\discriminant_{t}\big( R(t) - K(t-u)(t-v)(t-w) \big) \space = \space 0

FUN FACT

Differential Equation Equivalent

The three variable addition formulae can also be interpreted as the solution of a different equation. For example the integral of this differential equation

\frac {dx} {\sqrt{(x-a)(x-b)(x-c)}} = \frac {dy} {\sqrt{(y-a)(y-b)(y-c)}}

is equation deg3lev3. Explicitly

\left[(xy + xC + yC) - (ab + ac + bc)\right]^2 - 4\left[(x + y + C) - (a + b + c)\right]\left[xyC - abc\right] \space = \space 0

where $C$ is the constant of integration.

HISTORICAL NOTE

Addition Formulae In 3 Variables For Degree 4

Assume $R(t) = K(t-a)(t-b)(t-c)(t-d)$.

By letting $x$ take some specific value in equation addvar4 we can get an addition formula for degree 4 in the 3 variables. However, without extra requirements, these formulae are not rational in the coefficients of $R$. For example $x = \infty, 0\space \text{or}\space d$ gives

\begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 0 & 0 & 1 & \sqrt{K} \\ \end{vmatrix} \space = \space 0, \qquad \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & 0 & 0 & \sqrt{Kabcd} \\ \end{vmatrix} \space = \space 0 \qquad \text{or} \qquad \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & d & d^2 & 0 \\ \end{vmatrix} \space = \space 0

The corresponding level 2 addition formulae are not rational unless $\sqrt{K}, \sqrt{Kabcd} \space \text{or} \space d$, respectively, are rational. For this reason it's better to focus on the 4 variable addition formula in the degree 4 case.

Level 2 Addition Formula In 4 Variables For Degree 4

Similar to above, computation of the 4 variable level 2 addition formula gives

#fold,marked \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & x & x^2 & \sqrt{R(x)} \\ \end{vmatrix} \space \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & x & x^2 & -\sqrt{R(x)} \\ \end{vmatrix} \frac 1 {(u-x)(v-x)(w-x)} \space = $ \space K(u - v)^2(v - w)^2(w - u)^2\big[(u + v + w + x) - (a + b + c + d)\big] \space + \space (u-x)S(u,v,w) \space + (v-x)S(v,u,w) \space + \space (w-x)S(w,u,v) \space = \space 0

where

S(u,v,w) = (u - v)(u - w)\left(\sqrt{\smash[b]{R(v)}} - \sqrt{\smash[b]{R(w)}}\right)^2

solving for $x$ gives

x = \frac {K(u - v)^2(v - w)^2(w - u)^2\big[(u + v + w) - (a + b + c + d)\big] \space + \space u S(u,v,w) \space + \space v S(v,u,w) \space + \space w S(w,u,v)} {-K(u - v)^2(v - w)^2(w - u)^2 \space + \space S(u,v,w) \space + \space S(v,u,w) \space + \space S(w,u,v)}

There are many ways of expressing this formula, see Quartic Addition Formulae.

Level 4 Addition Formula In 4 Variables For Degree 4

Computation of the 4 variable level 4 addition formula, using CAS, gives

\prod_{\text{all signs}} \begin{vmatrix} 1 & u & u^2 & \hphantom{\pm}\sqrt{R(u)} \\ 1 & v & v^2 & \pm\sqrt{R(v)} \\ 1 & w & w^2 & \pm\sqrt{R(w)} \\ 1 & x & x^2 & \pm\sqrt{R(x)} \\ \end{vmatrix} \space = \space \begin{vmatrix} 1 & u & u^2 & u^3 \\ 1 & v & v^2 & v^3 \\ 1 & w & w^2 & w^3 \\ 1 & x & x^2 & x^3 \\ \end{vmatrix}^4 \space K^4 \space P(u,v,w,x,a,b,c,d) \space = \space 0

where $P$ is a polynomial with several hundred terms.

For reasons which are not at all obvious it has the following symmetry

\begin{equation} P(u,v,w,x,a,b,c,d) \space = \space P(a,b,c,d,u,v,w,x) \end{equation}

ANOTHER FUN FACT

References

[1] Euler, Leonhard (1761) Observationes de comparatione arcuum curvarum irrectificibilium Euler Archive - All Works. 252.

[2] Barrios, Jose (2009) A Brief History of Elliptic Integral Addition Theorems Rose-Hulman Undergraduate Mathematics Journal: Vol. 10 : Iss. 2 , Article 2.

[3] Trotti, Manuela (2016) Applications of elliptic functions to solve differential equations [Laurea magistrale], Università di Bologna, Corso di Studio in Matematica [LM-DM270]