Lattice-Aligned Elliptic Functions

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In this section we look at order 2 elliptic functions in which both zeros and poles are at half-period lattice points. This is closely related to reducing order 2 functions to special forms and to special cases of various curves.

Nomenclature / Definitions

In this section we use the following nomenclature and definitions

The Legendre curve is $y^2 = ax^3 + bx^2 + cx$.
A Legendre function is any solution $f(z)$ of the differential equation $f'^2 = af^3 + bf^2 + cf$.
The Euler curve is $y^2 = ax^4 + bx^2 + c$.
An Euler function is any solution $f(z)$ of the differential equation $f'^2 = af^4 + bf^2 + c$.
The Edwards curve is $ax^2y^2 + bx^2 + cy^2 + d = 0$.
The Hofstadter curve is $ax^2y + bxy^2 + cx + dy = 0$.
The half-period grid is the set of points $\{0, \omega_1, \omega_2, \omega_3\}$ where $\Omega = [2\omega_1,2\omega_2]$ is the period lattice and $\omega_1 + \omega_2 + \omega_3 = 0$.
A lattice-aligned function is one with all zeroes and poles lying on the half-period grid.
By generic even order 2 elliptic function we mean a solution $f(z)$ of the differential equation $f' = a(f - e_1)(f - e_2)(f - e_3)(f - e_4)$ with $f(0) = e_4$.
The values $\{e_1,e_2,e_3,e_4\}$ are known as the roots of $f$.
By distinct functions we mean functions that differ by more than just a multiplicative constant.
A centre of odd symmetry is a point $z_0$ such that $f(z - z_0)$ is an odd function.
A centre of even symmetry is a point $z_0$ such that $f(z - z_0)$ is an even function.

Summary Of Results

When the zeroes and poles of order 2 elliptic functions align with the half-period grid they gain additional symmetries.

Perfect Squares

When an order 2 elliptic function is a perfect square

It is an even Legendre function →
It has a double zero and a double pole aligned with the half period grid →
It's square root is an Euler function (but with one of its periods doubled).
It's square root is an odd Euler function if it has a zero or pole at the origin, otherwise it is an even Euler function.
A generic even order 2 elliptic function can be transformed to a perfect square by a Möbius transformation which sends one root to zero and one root to infinity →

Odd Symmetry

When an order 2 elliptic function has odd symmetry

It is an Euler function with either a pole or zero at the origin →
It has two simple zeroes at two of the half-period grid points and two simple poles at the other two. →
It has centre's of even symmetry lying halfway between any two poles (or indeed halfway between any two points where $f(z)$ has the same value).
It can be scaled by a multiplicative constant so that it's four roots have the form $\{\pm\alpha, \pm 1/\alpha\}$. When this is done, shifting it by a half-period either negates it, inverts it, or both.
A generic even order 2 elliptic function can be transformed to an odd Euler function with the same period lattice by taking the square root of the product of the two Möbius transforms which sends all four roots to zero or infinity.
The Hofstadter curve is the algebraic relation between two odd Euler functions.

Even-Odd Symmetry

Odd Euler functions become even when shifted by a quarter period, so that the origin lies midway between two poles.
A generic even order 2 elliptic function can be transformed to an even Euler function by a Möbius transformation which sends it's roots to two $\pm$ pairs of roots.
The Edwards curve is the algebraic relation between an even and an odd Euler function.

Formulae For Perfect Squares

There are three distinct lattice aligned, perfect square order 2 elliptic functions with a pole at zero. In terms of sigma products they are given by

\frac {\sigma(z - \omega_1)\sigma(z + \omega_1)} {\sigma^2(z)}, \qquad\qquad \frac {\sigma(z - \omega_2)\sigma(z + \omega_2)} {\sigma^2(z)}, \qquad\qquad \frac {\sigma(z - \omega_3)\sigma(z + \omega_3)} {\sigma^2(z)}

In terms of Weierstrass $\wp$ they may be written

\wp(z) - e_1, \qquad\qquad\wp(z) - e_2, \qquad\qquad\wp(z) - e_3

This follows directly from perfect_sigma and the well known formula

\wp(u) \space - \space \wp(v) \space = \space -\frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)}

In terms of any order 2 elliptic function $f$ they may be written

\frac {f(z) - e_1} {f(z) - e_4}, \qquad\qquad \frac {f(z) - e_2} {f(z) - e_4}, \qquad\qquad \frac {f(z) - e_3} {f(z) - e_4}

Differential Equation For Perfect Squares

Equations square_wp and square_f are simply Möbius transforms moving one root to zero and one to infinity. This is the Möbius transformation that puts their differential equation into Legendre form. That is

g(z) \enspace = \enspace \frac {f(z) - e_1} {f(z) - e_4} \qquad\implies\qquad g'(z)^2 \enspace = \enspace a g^3(z) \enspace + \enspace b g^2(z) \enspace + \enspace c g(z) \qquad\qquad g(0) = \infty

Therefore an order 2 elliptic function is a perfect square if and only if it is a solution of the Legendre differential equation.

Sigma Product For Order 2 Elliptic Function With Odd Symmetry

Every odd order 2 elliptic function $\phi$ with a zero at the origin has (up to a multiplicative constant) the form

          \phi(z) \enspace = \enspace \frac {\sigma(z) \thinspace \sigma(z - \omega_1)} {\sigma(z+\omega_2) \thinspace \sigma(z+\omega_3)}
        

Since it is an odd function of order two it must have a simple zero at $z=0$ and have a $\sigma$ product of the form

\phi(z) \enspace = \enspace \frac {\sigma(z) \thinspace \sigma(z - \mu)} {\sigma(z-\rho_1) \thinspace \sigma(z-\rho_2)}

But because of odd symmetry both $\mu$ and $-\mu$ must be zeroes of $\phi$, and since there can only be two zeroes in total, we must have

\mu \enspace \equiv \enspace -\mu \mod \left[2\omega_1,2\omega_2\right] \qquad\qquad \textsf{and} \qquad\qquad \mu \enspace \not\equiv \enspace 0 \mod \left[2\omega_1,2\omega_2\right]

which implies $\mu$ is one of the half-periods $\omega_1,\omega_2,\omega_3$ where $\omega_1 + \omega_2 + \omega_3 = 0$. The two poles $\rho_1, \rho_2$ cannot be a distinct $\pm$ pair, because then sum of zeroes would not equal the sum of poles. So they must also satisfy a similar condition and be the other two half-periods. The plus signs are required in the denominator to ensure that the sum of the zeroes equals the sum of the poles.

To show that odd_sigma is an odd function, use the pseudo-periodicity formula $\sigma(z+\omega_i) = -e^{2\eta_i z}\sigma(z - \omega_i)$ to give

\phi(-z) \enspace = \enspace \frac {\sigma(z) \thinspace \sigma(z + \omega_1)} {\sigma(z - \omega_2) \thinspace \sigma(z - \omega_3)} \enspace = \enspace \frac {\sigma(z) \thinspace (-1) e^{2\eta_1 z} \sigma(z - \omega_1)} {(-1) e^{-2\eta_2 z}\sigma(z + \omega_2) \thinspace (-1) e^{-2\eta_3 z} \sigma(z + \omega_3)} \enspace = \enspace (-1)^3 e^{2(\eta_1+\eta_2+\eta_3)z} \phi(z) \enspace = \enspace -\phi(z)

If instead $\phi$ has a simple pole at zero, it's reciprocal must have form odd_sigma. Therefore, up to a multiplicative constant and taking reciprocals, there are just three odd order 2 elliptic functions on the period lattice $\left[2\omega_1,2\omega_2\right]$. They are the three functions obtained by permuting $\omega_1,\omega_2,\omega_3$ in odd_sigma.

Locations Of Centre's Of Odd Symmetry

Translating odd_sigma by half-periods and utilising the pseudo-periodicity formula for $\sigma$ gives

        \phi(z + \omega_1) \space = \space -\phi(z) \qquad\qquad \textsf{and} \qquad\qquad
        \phi(z + \omega_2) \space = \space e^{2\eta_2\omega_3} \frac 1 {\phi(z)} \qquad\qquad \textsf{and} \qquad\qquad
        \phi(z + \omega_3) \space = \space e^{2\eta_3\omega_2} \frac 1 {\phi(z)}
      

The first is equivalent to odd symmetry

\phi(z + \omega_1) \enspace = \enspace \frac {\sigma(z + \omega_1) \sigma(z)} {\sigma(z - \omega_3) \sigma(z - \omega_2)} \enspace = \enspace \phi(-z) \enspace = \enspace -\phi(z)

The second can be deduced using the pseudo periodicity formula for $\sigma(z)$

\phi(z + \omega_2) \enspace = \enspace \frac {\sigma(z + \omega_2) \sigma(z + \omega_3 + 2\omega_2)} {\sigma(z + 2\omega_2) \sigma(z - \omega_1)} \enspace = \enspace \frac {\sigma(z + \omega_2) (-1)e^{2\eta_2(z+\omega_3+\omega_2)} \sigma(z + \omega_3)} {(-1) e^{2\eta_2 (z + \omega_2)}\sigma(z) \sigma(z - \omega_1)} \enspace = \enspace e^{2\eta_2\omega_3} \frac 1 {\phi(z)}

and the third one follows by swapping $2 \leftrightarrow 3$ in the above formula.

Therefore there are three other centres of odd symmetry lying on the half-period grid, relative to $\phi(z)$.

NOTES

Equations odd_shift suggests scaling $\phi$ by multiplicative constant as follows

\phi(z) \enspace = \enspace e^{-\eta_2\omega_3} \cdot \frac {\sigma(z) \thinspace \sigma(z - \omega_1)} {\sigma(z+\omega_2) \thinspace \sigma(z+\omega_3)}

so that

\phi(-z) \space = \space - \phi(z) \qquad\qquad\qquad \phi(z + \omega_1) \space = \space - \phi(z) \qquad\qquad\qquad \phi(z + \omega_2) \space = \space \frac 1 {\phi(z)} \qquad\qquad\qquad \phi(z + \omega_3) \space = \space - \frac 1 {\phi(z)}

However this simple idea is complicated by the fact that such a scaling factor is only uniquely determined up to an arbitrary fourth root of unity.

Determining Other Two Distinct Odd Elliptic Functions From Given One

If $\phi$ is an odd order 2 elliptic function on the lattice $[2\omega_1, 2\omega_2]$, with zeroes at $0, \omega_1$, then the three distinct odd order 2 elliptic functions $\phi_i$ with zeroes at $0, \omega_i$ are given by the formula

        \phi_i(z) \enspace = \enspace \phi(z) \space + \space \residue(\phi,\omega_i) \cdot \frac {\phi'(z) - \phi'(0)} {\phi(z)} \qquad \textsf{for} \qquad i=1,2,3
      

Because $\phi(z)$ has simple zeroes at $0, \omega_1$ it has simple poles at $\omega_2, \omega_3$. This implies $\phi'(z) - \phi'(0)$ is even with double zero at $0$ and two double poles at $\omega_2, \omega_3$. This in turn implies $\left[\phi'(z) - \phi'(0)\right]/\phi(z)$ is odd with simple zero at $0$ and three simple poles at $\omega_1, \omega_2, \omega_3$.

From the power series expansion $\phi(z) = \residue(\phi,\omega_2)(z-\omega_2)^{-1} + \bigO(z-\omega_2)$ we can deduce $\left[\phi'(z) - \phi(0)\right]/\phi(z) = -1.(z-\omega_2)^{-1} + \bigO(z-\omega_2)$. This implies $\phi(z) + \residue(\phi,\omega_2).\left[\phi'(z) - \phi(0)\right]/\phi(z)$ is odd with simple zero at 0 and two simple poles at $\omega_1, \omega_3$ because the pole at $\omega_2$ is cancelled out.

It then follows that odd_from_odd is true for $i=2$ and a similar power series expansion shows it is true for $i=3$. And since $\residue(\phi,\omega_1)=0$ it is true for $i=1$ as well.

The most well known odd order 2 elliptic function is the Jacobi elliptic function $\sn(z)$. Applying odd_from_odd to this function gives the other two distinct odd order 2 elliptic functions on the lattice $[4K, 2iK']$, with a zero at $z=0$, as

\sn(z) \enspace \pm \enspace \frac {\cn(z)\dn(z) \space - \space 1} {k\sn(z)}

Put $\phi(z) = \sn(z)$ so then $\omega_1 = 2K$ and $\omega_2 = iK'$.
Also $\phi'(z) = \cn(z)\dn(z)$ and $\phi'(0) = \cn(0)\dn(0) = 1$.
And $\residue(\phi, \omega_2) = \residue(\sn, iK') = 1/k$ so $\residue(\phi, \omega_3) = -1/k$.
The result follows by substituting these values into odd_from_odd.

Differential Equation For Odd Order 2 Elliptic Functions

While every order 2 elliptic function has four centre's of even symmetry, it has in general has no centre of odd symmetry.

A necessary and sufficient condition for a non-trivial order 2 elliptic function $\phi$ to have odd symmetry is that it satisfies a differential equation of the form

\phi'(z)^2 \enspace = \enspace a \phi^4(z) \enspace + \enspace b \phi^2(z) \enspace + \enspace c

To show necessity note that since $\phi$ is order 2 it satisfies a differential equation of the form $\phi'^2 = A \phi^4 + B \phi^3 + C \phi^2 + D \phi + E$. By mapping $z \rightarrow -z$ we see that since $\phi(-z)=-\phi(z)$ it also satisfies the differential equation $\phi'^2 = A \phi^4 - B \phi^3 + C \phi^2 - D \phi + E$. As $\phi$ is not a constant, the only way this can happen is if $B = D = 0$.

To show sufficiency, under the boundary condition $\phi(0)=0$, use induction to get polynomials $P_n$ and $Q_n$ such that

\phi^{[2n]}(z) \space = \space \phi(z) \cdot P_n\left(\phi^2(z)\right) \qquad \textsf{and} \qquad \phi^{[2n+1]}(z) = \phi'(z) \cdot Q_n\left(\phi^2(z)\right) \qquad \textsf{for} \qquad n \ge 0

This implies that when $\phi(0) = 0$, all the even derivatives $\phi^{[2n]}(0)$ vanish and that therefore $\phi(z)$ is an odd function.

In more detail the inductive steps are compute $\phi''$ by differentiating euler_form to obtain

\phi'' \space = \space \phi \cdot \left(2a \phi^2 + b\right)

and then differentiate $\phi^{[2n]}$ formula to obtain

\phi^{[2n+1]} \space = \space \frac {d} {dz} \phi^{[2n]} \space = \space \phi' \cdot P_n(\phi^2) \space + \space 2 \phi^2 \phi' P_n'(\phi^2) \space = \space \phi' \cdot \left[ P_n(\phi^2) \space + \space 2 \phi^2 P_n'(\phi^2) \right] \space = \space \phi' \cdot Q_n(\phi^2)

and then differentiate the $\phi^{[2n+1]}$ formula to obtain

\phi^{[2n+2]} \space = \space \frac {d} {dz} \phi^{[2n+1]} \space = \space \phi'' Q_n(\phi^2) \space + \space 2f \phi'^{2} Q_n'(\phi^2) \space = \space \phi \cdot \left[\left(2a \phi^2+b\right)Q_n(\phi^2) \space + \space 2 \left(a \phi^4 + b \phi^2 + c\right) Q_n'(\phi^2) \right] \space = \space \phi \cdot P_{n+1}(\phi^2)

Therefore the polynomials $P_n$ and $Q_n$ are given iteratively by

P_0(x) = 1 \qquad\qquad Q_n(x) = P_n(x) \space + \space x P'_n(x) \qquad\qquad P_{n+1}(x) = (2ax+b)Q_n(x)\space + \space (ax^2+bx+c)Q'_n(x)

The Weierstrass invariants for this quartic are

g_2 \enspace = \enspace ac \space + \space \tfrac 1 {12} b^2 \qquad\qquad\qquad g_3 \enspace = \enspace \tfrac 1 6 abc \space - \space \tfrac 1 {216} b^3 \qquad\qquad\qquad \Delta \enspace = \enspace \tfrac 1 {16} ac\left(b^2 \space - \space 4ac\right)^2

NOTES

If $\phi$ has zeroes at $0$ and $\omega_1$ the coefficients $a,b,c$ are given by

a \space = \space \tfrac 1 {12} \left[ \frac {\phi^{[5]}(0)} {\phi'(0)^3} \space - \space \frac {\phi'''(0)^2} {\phi'(0)^4} \right] \space = \space \frac 1 {\residue(\phi,\omega_2)^2} \space = \space \frac 1 {\residue(\phi,\omega_3)^2}, \qquad b \space = \space \frac {\phi'''(0)} {\phi'(0)}, \qquad c \space = \space \phi'(0)^2

Determine $a$ by substituting the Laurent expansion for $\phi$ at $z=\omega_2$ into euler_form. Determine $b$ by differentiating euler_form twice, then evaluating it at $z=0$. Determine $c$ by evaluating euler_form at $z=0$.

The four roots of euler_form say $\pm\alpha,\pm\beta$ are given by

\alpha \space = \space \phi(\tfrac 1 2 \omega_1), \qquad -\alpha \space = \space \phi(\tfrac 1 2 \omega_1 + \omega_1), \qquad \beta \space = \space \phi(\tfrac 1 2 \omega_1 + \omega_2), \qquad -\beta \space = \space \phi(\tfrac 1 2 \omega_1 + \omega_3), \qquad

Determine $\alpha$ and $\beta$ by evaluating $\phi$ at the centres of even symmetry $\tfrac 1 2\omega_1 + \omega_i$.

Differential Equations For The Three Distinct Odd Order 2 Elliptic Functions

For the period lattice defined by invariants $g_2$ and $g_3$ with $g_2^3 - 27g_3^2 \ne 0$, the three distinct odd order 2 elliptic functions have differential equations of the form euler_form where

b^3 \space - \space 9 g_2 b \space + \space 54 g_3 \space = \space 0 \qquad \textsf{and} \qquad ac \space = \space g_2 \space - \space \tfrac 1 {12} b^2

This follows by solving the invariant equations euler_invariants for $b$ and $ac$.

This cubic in $b$ always has three distinct roots because it's discriminant is a multiple of $g_2^3 - 27g_3^2$. It's three roots correspond to the three distinct odd functions. If given one solution $a,b,c$ the other two solutions $a',b',c'$ are given by

b' \space = \space -\tfrac 1 2 b \space \pm \space 3\sqrt{ac} \qquad \textsf{and} \qquad \sqrt{a'c'} \space = \space \tfrac 1 4 b \space \pm \space \tfrac 1 2 \sqrt{ac}

This follows by solving $\displaystyle \frac {(b'^3 - 9 g_2 b' + 54 g_3) \space - \space (b^3 - 9 g_2 b + 54 g_3)} {b' - b} \space = \space 0$ for $b'$ etc.

CROSS CHECK

\begin{aligned} a'c' + \tfrac 1 {12}b'^2 \space &= \space \left(\tfrac 1 {16} b^2 + \tfrac 1 4 b\sqrt{ac} + \tfrac 1 4 ac\right) \space + \space \tfrac 1 {12} \left(\tfrac 1 4 b^2 - 3b\sqrt{ac} + 9ac\right) \\ &= \space ac \space + \space \tfrac 1 {12} b^2 \quad \checkmark \\\\ \tfrac 1 6 a'b'c' - \tfrac 1 {216} b'^3 \space &= \space \tfrac 1 6 \left(\tfrac 1 {16} b^2 + \tfrac 1 4 b\sqrt{ac} + \tfrac 1 4 ac\right)\left(-\tfrac 1 2 b + 3\sqrt{ac} \right) \space - \space \tfrac 1 {216} \left( -\tfrac 1 8 b^3 + \tfrac 9 4 b^2\sqrt{ac} - \tfrac {27} 2 bac + 27ac\sqrt{ac}\right) \\ &= \space \left(-\tfrac 1 {192} + \tfrac 1 {1728} \right) b^3 \space + \space \left( \tfrac 3 {96} - \tfrac 1 {48} - \tfrac 1 {96} \right)b^2\sqrt{ac} \space + \space \left(\tfrac 3 {24} - \tfrac 1 {48} + \tfrac 1 {16} \right)abc \space + \space \left(\tfrac 1 8 - \tfrac 1 {8} \right)ac\sqrt{ac} \\ &= \space \tfrac 1 6 abc\space - \space \tfrac 1 {216} b^3 \quad \checkmark \end{aligned}

Now every odd function with a zero at $z=0$ is just a constant multiple of the function $\phi(z,a,b)$ where $\phi'^2=a\phi^4+b\phi^2+a$ and $\phi(0)=0$. And the two other distinct odd functions on the same lattice are

\phi\left(z,\space \tfrac 1 2 a \space + \space \tfrac 1 4 b, \space 3a \space - \space \tfrac 1 2 b\right) \qquad \textsf{and} \qquad \phi\left(z,\space -\tfrac 1 2 a \space + \space \tfrac 1 4 b, \space -3a \space - \space \tfrac 1 2 b\right)

Symmetrical Roots Elliptic Curve

The Euler curve with the highest degree of root-symmetry is

y^2 \enspace = \enspace (x - \alpha)(x - \alpha^{-1})(x + \alpha)(x + \alpha^{-1}) \enspace = \enspace x^4 \space - \space (\alpha^2 + \alpha^{-2}) x^2 \space + \space 1

with invariants

g_2(\alpha) = \tfrac 1 {12}\left(\alpha^4\space + \space 14 + \space \alpha^{-4}\right) \qquad\qquad g_3(\alpha) = \tfrac 1 {216}\left(\alpha^2\space + \space \alpha^{-2}\right)\left(\alpha^4 \space - \space 34 + \alpha^{-4}\right) \qquad\qquad \Delta(\alpha) = \tfrac 1 {16} \left(\alpha^2 - \alpha^{-2}\right)^4

and cross-ratio

\lambda(\alpha) \space = \space \crossratio{\alpha,\alpha^{-1},-\alpha,-\alpha^{-1}} \space = \space - \tfrac 1 4 {\left({\alpha - \alpha^{-1}}\right)^{2}}

There are 24 values of $\alpha$ for which curve_sym has the same $j$-invariant. If $\zeta$ is a primitive fourth root of unity they are given by this 24th degree polynomial with the remarkably simple factorisation

j(\beta) \space - \space j(\alpha) \enspace = \enspace \frac {\prod\limits_{i=0}^3 \left(\zeta^i \alpha - \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} 2^4 \thinspace 3^6 \thinspace \alpha^4\beta^4 \thinspace \left(1 - \alpha^4\right)^4\left(1 - \beta^4\right)^4} \enspace = \enspace 0

This formula is the factorisation of

j(\beta) \space - \space j(\alpha) \space = \space 27 \cdot \frac {g_2(\alpha)^3 g_3(\beta)^2 \space - \space g_2(\beta)^3 g_3(\alpha)^2} {\Delta(\alpha)\Delta{\beta}}

It's simplicity is explained by computing this equation using the cross-ratio invariant $\lambda$ instead of the $g_2,g_3$ invariants. This gives six degree 4 polynomials, arising from permuting the arguments of $\lambda$

\lambda(\beta) \space = \space \lambda(\alpha), \quad \frac 1 {\lambda(\alpha)}, \quad 1 - \lambda(\alpha), \quad \frac 1 {1 - \lambda(\alpha)}, \quad \frac {\lambda(\alpha)} {\lambda(\alpha) - 1} \quad \textsf{or} \quad \frac {\lambda(\alpha) - 1} {\lambda(\alpha)}

Each of these polynomials can be further factorised because of the four symmetries in $\lambda(\alpha)$

\lambda(\alpha) \space = \space \lambda(-\alpha) \space = \space \lambda(1 / \alpha) \space = \space \lambda(-1 / \alpha)

giving the factorisation fac_sym into 24 degree 1 polynomials.

From this it follows that the 24 values are

\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

and the transformations which carry them onto one another are the Möbius transforms of the octahedral group.

Symmetrical Roots Elliptic Function

Let $\phi$ be the elliptic function with differential equation

\phi'^2 \enspace = \enspace \phi^4 \enspace - \enspace (\alpha^2 + \alpha^{-2}) \thinspace \phi^2 \enspace + \enspace 1

and boundary condition $\phi(0) = 0$, and for definiteness $\phi'(0) = 1$. We have the basic identities

\phi(z,\thinspace \alpha) \space = \space -\phi(-z, \thinspace \alpha) \space = \space \phi(z, \thinspace -\alpha) \space = \space \phi(z, \thinspace 1 /\alpha) \space = \space -\i\phi(\i z, \thinspace \i \alpha)

We also have the three distinct odd functions given by

        \phi(z, \space \alpha), \qquad \phi\left(\frac {2\i\alpha z} {1 - \alpha^2}, \space \frac {1-\alpha} {1 + \alpha} \right), \qquad
        \phi\left(\frac {2\alpha z} {1 + \alpha^2}, \space \frac {1 - \i\alpha} {1 + \i\alpha} \right)
      

We want to construct a function $\phi(\mu z, \beta)$ which has the same periods as $\phi(z, \alpha)$. Taking the value of $\beta$ from beta_values with $i=j=0$ and equating the $g_2$ and $g_3$ invariants gives a formula for $\mu$

\mu \enspace = \enspace \sqrt[4]{\frac {g_2(\beta)} {g_2(\alpha)}} \enspace = \enspace \sqrt[6]{\frac {g_3(\beta)} {g_3(\alpha)}} \enspace = \enspace \pm \frac {2\i\alpha} {1 - \alpha^2} \qquad \textsf{where} \qquad\beta \space = \space \frac {1-\alpha} {1 + \alpha}

Both the $g_2$ and $g_3$ formulae are required to determine which are the correct two roots.

Another Version Of $\sn$ Two Odd Companions

Using euler_sym we can express $\phi$ in terms of $\sn$ as

\phi(z, \space \alpha) \space = \space {\alpha} \sn(\frac {z} {\alpha}, \space \alpha^2)

From this we deduce these another set of Jacobian elliptic function formulae for the three distinct odd elliptic functions on the lattice $\left[4K,2\i K'\right]$

\sn\left(z, \space k\right), \quad\qquad \sn\left(\frac {2 \i k} {\big(1 - \sqrt{k}\big)^2} \thinspace z, \space \left(\frac {1 - \sqrt{k}}{1 + \sqrt{k}}\right)^2 \right), \qquad\qquad \sn\left(\frac {2 k} {\big(1 - \i\sqrt{k}\big)^2} \thinspace z, \space \left(\frac {1 - \i\sqrt{k}}{1 + \i\sqrt{k}}\right)^2 \right)

This follows by converting equation sym_3 into a formula for $\sn$ using identity sym_sn.

\phi(\alpha z, \alpha) \space = \space \alpha \sn\left(z, \alpha^2\right) \qquad\qquad\textsf{and}\qquad\qquad \phi(\mu \alpha z, \beta) \space = \space \beta \sn\left(\frac {\mu \alpha z} {\beta}, \beta^2\right) \space = \space \beta \sn\left(\frac {2\i\alpha^2 z} {1 - \alpha^2} \cdot \frac {1+\alpha} {1-\alpha}, \left(\frac {1-\alpha} {1+\alpha}\right)^2\right)

Because of the identity

\sn(z, \space \frac 1 k) \space = \space k \sn(\frac z k, k)

using the opposite square root sign will still give the same function (up to a constant multiplier).

\left(1 - \sqrt{k}\right)^2 \sn\left(\frac {2 \i k} {\left(1 - \sqrt{k}\right)^2} \thinspace z, \space \left(\frac {1 - \sqrt{k}}{1 + \sqrt{k}}\right)^2 \right) \enspace = \enspace \left(1 + \sqrt{k}\right)^2 \sn\left(\frac {2 \i k} {\left(1 + \sqrt{k}\right)^2} \thinspace z, \space \left(\frac {1 + \sqrt{k}}{1 - \sqrt{k}}\right)^2 \right)

Formulae For The Three Odd Order 2 Elliptic Functions

There are three distinct odd order 2 elliptic functions with a zero at the origin. They can be expressed in terms of the Weierstrass $\sigma$ function as

\frac {\sigma(z) \thinspace \sigma(z - \omega_1)} {\sigma(z + \omega_2) \thinspace \sigma(z + \omega_3)}, \qquad \frac {\sigma(z) \thinspace \sigma(z - \omega_2)} {\sigma(z + \omega_1) \thinspace \sigma(z + \omega_3)}, \qquad \frac {\sigma(z) \thinspace \sigma(z - \omega_3)} {\sigma(z + \omega_1) \thinspace \sigma(z + \omega_2)}

and in terms of the Weierstrass $\wp$ function as

\frac {\wp(z) - e_1} {\wp'(z)}, \qquad \frac {\wp(z) - e_2} {\wp'(z)}, \qquad \frac {\wp(z) - e_3} {\wp'(z)}

and in terms of the Weierstrass $h$ functions as

\frac {h_1(z)} {h_2(z)h_3(z)}, \qquad \frac {h_2(z)} {h_1(z)h_3(z)}, \qquad \frac {h_3(z)} {h_1(z)h_2(z)}

and in terms of the Weierstrass $\zeta$ function as

\eta_1 \space + \space \zeta(z + \omega_2) \space - \space \zeta(z - \omega_3), \qquad \eta_2 \space + \space \zeta(z + \omega_3) \space - \space \zeta(z - \omega_1), \qquad \eta_3 \space + \space \zeta(z + \omega_1) \space - \space \zeta(z - \omega_2)

and in terms of the Jacobian elliptic functions on the lattice $\left[4K,2iK'\right]$ as

\sn(z), \qquad \sn(z) \space \pm \space \frac {\cn(z)\dn(z) - 1} {k\sn(z)}

and in terms of the Jacobian elliptic functions on the lattice $\left[2K,2iK'\right]$ as

\frac {\sn(z)} {\cn(z)\dn(z)}, \qquad \frac {\sn(z)\cn(z)} {\dn(z)}, \qquad \frac {\sn(z)\dn(z)} {\cn(z)}

and in terms of any even order 2 elliptic function $f$ with roots $e_1,e_2,e_3,e_4$ and $f(0) = e_4$ as

\frac {\left(f(z) - e_1\right) \left(f(z) - e_4\right)} {f'(z)}, \qquad \frac {\left(f(z) - e_2\right) \left(f(z) - e_4\right)} {f'(z)}, \qquad \frac {\left(f(z) - e_3\right) \left(f(z) - e_4\right)} {f'(z)}

Vector Space Spanned By Odd Order 2 Elliptic Functions

The three odd functions with a zero at the origin are linearly dependent. Two of them, along with the unit function form a basis for the 3-dimensional vector space all elliptic functions with at most simple poles at $\omega_1,\omega_2,\omega_3$. This is confirmed by the identity

\begin{vmatrix} 1 & \large\frac {\wp(z_1) \space - \space e_1} {\wp'(z_1)} & \large\frac {\wp(z_1) \space - \space e_2} {\wp'(z_1)} \\ 1 & \large\frac {\wp(z_2) \space - \space e_1} {\wp'(z_2)} & \large\frac {\wp(z_2) \space - \space e_2} {\wp'(z_2)} \\ 1 & \large\frac {\wp(z_3) \space - \space e_1} {\wp'(z_3)} & \large\frac {\wp(z_3) \space - \space e_2} {\wp'(z_3)} \\ \end{vmatrix} \enspace = \enspace \frac {(e_2 - e_1)} {\wp'(z_1)\wp'(z_2)\wp'(z_3)} \cdot \begin{vmatrix} 1 & \wp(z_1) & \wp'(z_1) \\ 1 & \wp(z_2) & \wp'(z_2) \\ 1 & \wp(z_3) & \wp'(z_3) \\ \end{vmatrix}

The three odd functions with a pole at the origin are linearly independent. The three of them, along with the unit function form a basis for the 4-dimensional vector space all elliptic functions with at most simple poles at $0, \omega_1,\omega_2,\omega_3$. This is confirmed by the identity

\begin{vmatrix} 1 & \large\frac {\wp'(z_1)} {\wp(z_1) \space - \space e_1} & \large\frac {\wp'(z_1)} {\wp(z_1) \space - \space e_2} & \large\frac {\wp'(z_1)} {\wp(z_1) \space - \space e_3} \\ 1 & \large\frac {\wp'(z_2)} {\wp(z_2) \space - \space e_1} & \large\frac {\wp'(z_2)} {\wp(z_2) \space - \space e_2} & \large\frac {\wp'(z_2)} {\wp(z_2) \space - \space e_3} \\ 1 & \large\frac {\wp'(z_3)} {\wp(z_3) \space - \space e_1} & \large\frac {\wp'(z_3)} {\wp(z_3) \space - \space e_2} & \large\frac {\wp'(z_3)} {\wp(z_3) \space - \space e_3} \\ 1 & \large\frac {\wp'(z_4)} {\wp(z_4) \space - \space e_1} & \large\frac {\wp'(z_4)} {\wp(z_4) \space - \space e_2} & \large\frac {\wp'(z_4)} {\wp(z_4) \space - \space e_3} \\ \end{vmatrix} \enspace = \enspace 64 \cdot \frac {(e_1 - e_2)(e_2 - e_3)(e_3 - e_1)} {\wp'(z_1)\wp'(z_2)\wp'(z_3)\wp'(z_4)} \cdot \begin{vmatrix} 1 & \wp(z_1) & \wp(z_1)^2 & \wp'(z_1) \\ 1 & \wp(z_2) & \wp(z_2)^2 & \wp'(z_2) \\ 1 & \wp(z_3) & \wp(z_3)^2 & \wp'(z_3) \\ 1 & \wp(z_4) & \wp(z_4)^2 & \wp'(z_4) \\ \end{vmatrix}

Algebraic Relation Between Pair Of Odd Order 2 Elliptic Functions

Any pair of distinct odd order 2 elliptic functions satisfy an algebraic relation of the form

ax^2y \enspace + \enspace bxy^2 \enspace + \enspace cx \enspace + \enspace dy \enspace = \enspace 0

Every pair of order 2 elliptic functions satisfies an algebraic relation in the form of the biquadratic curve

\sum_{0 \le i,j \le 2} a_{ij} x^i y^j \enspace = \enspace 0

By looking at this curve under the mapping $(x,y) \rightarrow (-x,-y)$ we see that if $x$ and $y$ are both odd functions they must satisfy this relation with all odd terms zero and with all even terms zero. That is

a_{22}x^2y^2 \enspace + \enspace a_{20}x^2 \enspace + \enspace a_{11}xy \enspace + \enspace a_{02}y^2 \enspace + \enspace a_{00} = \enspace 0

and

a_{21}x^2y \enspace + \enspace a_{12}xy^2 \enspace + \enspace a_{10}x \enspace + \enspace a_{01}y \enspace = \enspace 0

By examining behaviour near the zeroes and poles of the functions it can be seen that in the former we must have $a_{22} = a_{00} = 0$. Then, since the two functions are not multiples of one another, all the even coefficients must be zero. The discriminant of the second curve with respect to $y$ is

D(x) \space = \space a_{21}^2 x^4 \space + \space 2 \left(a_{21}a_{01} - 4 a_{10} a_{12} \right) x^2 \space + \space a_{01}^2

From examination of $D$ we can see that if any of the coefficents are zero it will not have four distinct roots. Thus every distinct pair of odd order 2 elliptic functions satisfies a relation of the form odd with all four coefficients non-zero.

This curve is a special case of both the biquadratic curve and the cubic curve.

Definition Of Normalised Odd Function

An odd order 2 elliptic function $\phi$ is normalised if translating it's argument by a half-period negates and/or inverts it ie.

\phi(z+\omega_i) \space = \space -\phi(z), \qquad \phi(z+\omega_j) \space = \space \frac 1 {\phi(z)}, \qquad \phi(z + \omega_k) \space = \space -\frac 1 {\phi(z)}

The normalisation factor is only determined up to an arbitrary fourth root of unity. Therefore the general formula for a normalised odd function $\phi$ is

        \phi(z) \enspace = \enspace e^{\eta_j\omega_k + \sfrac 1 2 n\pi \i} \cdot \frac {\sigma(z)\sigma(z + \omega_j + \omega_k)} {\sigma(z + \omega_j)\sigma(z + \omega_k)}
      

where $j,k$ are distinct and $n=0,1,2,3$. Since $\eta_j\omega_k - \eta_k\omega_j = \pm \sfrac 1 2 \pi \i$ this formula is symmetric in $j,k$.

Differential Equation For Normalised Odd Order 2 Elliptic Functions

A normalised odd order 2 elliptic function $\phi$ has a differential equation of the form

\phi'(z)^2 \enspace = \enspace a\phi^4(z) \enspace + \enspace b \phi^2(z) \enspace + \enspace a

An odd order 2 elliptic function $\phi(z)$ is normalised if $\phi(z + \omega_i) = 1 / \phi(z)$ for some $i \in \{1,2,3\}$. Substituting this into euler_form and clearing denominators gives

\phi'(z)^2 \enspace = \enspace a \enspace + \enspace b \phi(z)^2 \enspace + \enspace c \phi(z)^4

and the only way this can happen is if $a = c$.

Equation normalised_euler_form is equivalent to the differential equation having four roots of the form $\alpha, -\alpha, 1 / \alpha, -1 / \alpha$.

Transforming General Order 2 Elliptic To Normalised Euler Form With A Möbius Transform

An arbitrary even order 2 elliptic function $f$ with roots $e_1,e_2,e_3,e_4$ can be put into normalised Euler form by a Möbius transformation which maps them to four roots in the form $\alpha,-\alpha,1/\alpha,-1/\alpha$. Such a transformation is given implicitly by the cross-ratio formula

\crossratio{\alpha,-\alpha,1/\alpha,\phi(z)} \enspace = \enspace \crossratio{e_1,e_2,e_3,f(z)}

where $\alpha$ is found by solving the quadratic equation obtained by equating the cross-ratio's of the roots

\crossratio{\alpha,-\alpha,1/\alpha,-1/\alpha} \enspace = \enspace - \left[ \frac {2\alpha} {1 - \alpha^2} \right]^2 \enspace = \enspace \crossratio{e_1,e_2,e_3,e_4}

For Normalised Odd Order 2 Elliptic Functions

There are three distinct normalised odd order 2 elliptic functions with a zero at the origin. They can be expressed in terms of the Weierstrass $\sigma$ function as

e^{-\eta_2\omega_3} \cdot \frac {\sigma(z) \thinspace \sigma(z - \omega_1)} {\sigma(z + \omega_2) \thinspace \sigma(z + \omega_3)}, \qquad\qquad e^{-\eta_3\omega_1} \cdot \frac {\sigma(z) \thinspace \sigma(z - \omega_2)} {\sigma(z + \omega_1) \thinspace \sigma(z + \omega_3)}, \qquad\qquad e^{-\eta_1\omega_2} \cdot \frac {\sigma(z) \thinspace \sigma(z - \omega_3)} {\sigma(z + \omega_1) \thinspace \sigma(z + \omega_2)}

and in terms of the Weierstrass $\wp$ function as

\sqrt{ \frac{(\wp(z) - e_1)(e_2 - e_3)} {(\wp(z) - e_2)(\wp(z) - e_3)} }, \qquad\qquad \sqrt{ \frac{(\wp(z) - e_2)(e_3 - e_1)} {(\wp(z) - e_3)(\wp(z) - e_1)} }, \qquad\qquad \sqrt{ \frac{(\wp(z) - e_3)(e_1 - e_2)} {(\wp(z) - e_1)(\wp(z) - e_2)} }

and in terms of the Weierstrass $h$ functions as

h_3(\omega_2) \cdot \frac {h_1(z)} {h_2(z)h_3(z)}, \qquad\qquad h_1(\omega_3) \cdot \frac {h_2(z)} {h_1(z)h_3(z)}, \qquad\qquad h_2(\omega_1) \cdot \frac {h_3(z)} {h_1(z)h_2(z)}

and in terms of any even order 2 elliptic function $f$ with roots $e_1,e_2,e_3,e_4$ and $f(0) = e_4$ as

\sqrt{\frac{(f(z) - e_1)(f(z) - e_4)(e_2 - e_3)} {(f(z) - e_2)(f(z) - e_3)(e_1 - e_4)} }, \qquad\qquad \sqrt{\frac{(f(z) - e_2)(f(z) - e_4)(e_3 - e_1)} {(f(z) - e_3)(f(z) - e_1)(e_2 - e_4)} }, \qquad\qquad \sqrt{\frac{(f(z) - e_3)(f(z) - e_4)(e_1 - e_2)} {(f(z) - e_1)(f(z) - e_2)(e_3 - e_4)} }

The expressions under the square root signs in norm_f are weight zero symmetric root differences. Such expression are invariant under Möbius transforms and therefore give the same value for every possible $f$ (with the same periods).

Algebraic Relation Between Pair Of Normalised Odd Order 2 Elliptic Functions

Any pair of distinct odd order 2 elliptic functions, after scaling by suitable normalisation factors, satisfy an algebraic relation of the form

        \left(y \space + \space \frac 1 y\right) \enspace = \enspace \sqrt{\lambda} \left(x \space + \space \frac 1 x\right)
        \qquad \textsf{where} \qquad \lambda \space = \space \crossratio{e_1,e_2,e_3,e_4} \space = \space \frac {(e_1 - e_2)(e_3 - e_4)} {(e_1 - e_3)(e_2 - e_4)}
      

Using $h$ function formulae to get rid of ambiguous square roots, scale and possibly invert the two odd functions so that they have the following form

x \space = \space h_3(\omega_2) \cdot \frac {h_1(z)}{h_2(z)h_3(z)} \qquad\qquad y \space = \space h_3(\omega_1) \cdot \frac {h_2(z)}{h_1(z)h_3(z)} \qquad\qquad \sqrt{\lambda} \space = \space \frac {h_3(\omega_2)} {h_3(\omega_1)}

then substituting into reduced_hors gives

\begin{aligned} h_3(\omega_1) \left(\frac {h_3(\omega_1)h_2} {h_1 h_3} \space + \space \frac {h_1 h_3} {h_3(\omega_1) h_2}\right) \enspace &= \enspace h_3(\omega_2) \left(\frac {h_3(\omega_2)h_1} {h_2 h_3} \space + \space \frac {h_2 h_3} {h_3(\omega_2) h_1}\right) \\\\ h_3(\omega_1)^2 h_2^2 \space + \space h_1^2 h_3^2 \enspace &= \enspace h_3(\omega_2)^2 h_1^2 \space + \space h_2^2 h_3^2 \\\\ (e_1 - e_3)(\wp - e_2) \space + \space (\wp - e_1)(\wp - e_3) \enspace &= \enspace (e_2 - e_3)(\wp - e_1) \space + \space (\wp - e_2)(\wp - e_3) \\\\ \wp^2 \space - \space 2e_3\wp \space + \space e_1e_3 + e_2e_3 - e_1e_2 \enspace &= \enspace \wp^2 \space - \space 2e_3\wp \space + \space e_1e_3 + e_2e_3 - e_1e_2 \qquad \checkmark \end{aligned}

Now note that equation reduced_hors is invariant under $x \rightarrow 1/x$ and $y \rightarrow 1/y$, so any inversions can be undone without affecting the formula.

This is the only cubic curve invariant under inversions in both $x$ and $y$ and double negations $(x,y) \rightarrow (-x,-y)$.