Lattice-Aligned Elliptic Functions

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.

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

Perfect Squares

Odd Symmetry

Even-Odd Symmetry

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 \begin{equation} \label{eq:perfect_sigma} \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)} \end{equation} In terms of Weierstrass $\wp$ they may be written

\begin{equation} \label{eq:square_wp} \wp(z) - e_1, \qquad\qquad\wp(z) - e_2, \qquad\qquad\wp(z) - e_3 \end{equation}

This follows directly from \eqref{eq:perfect_sigma} and the well known formula \begin{equation*} \wp(u) \space - \space \wp(v) \space = \space -\frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)} \end{equation*}

In terms of any order 2 elliptic function $f$ they may be written \begin{equation} \label{eq:square_f} \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} \end{equation}

Equations \eqref{eq:square_wp} and \eqref{eq: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 \begin{equation} \label{eq:legendre-form} 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 \end{equation} Therefore an order 2 elliptic function is a perfect square if and only if it is a solution of the Legendre differential equation.

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

\begin{equation} \label{eq:odd_sigma} \phi(z) \enspace = \enspace \frac {\sigma(z) \thinspace \sigma(z - \omega_1)} {\sigma(z+\omega_2) \thinspace \sigma(z+\omega_3)} \end{equation}

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 \begin{equation*} \phi(z) \enspace = \enspace \frac {\sigma(z) \thinspace \sigma(z - \mu)} {\sigma(z-\rho_1) \thinspace \sigma(z-\rho_2)} \end{equation*} 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 \begin{equation*} \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] \end{equation*} 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 \eqref{eq: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 \begin{equation*} \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) \end{equation*}

If instead $\phi$ has a simple pole at zero, it's reciprocal must have form \eqref{eq: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 \eqref{eq:odd_sigma}.

Translating \eqref{eq:odd_sigma} by half-periods and utilising the pseudo-periodicity formula for $\sigma$ gives

\begin{equation} \label{eq:odd_shift} \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)} \end{equation}

The first is equivalent to odd symmetry \begin{equation*} \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) \end{equation*} The second can be deduced using the pseudo periodicity formula for $\sigma(z)$ \begin{equation*} \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)} \end{equation*} 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 $\phi(z)$.

Equations \eqref{eq:odd_shift} suggests scaling $\phi$ by multiplicative constant as follows \begin{equation} \label{eq:odd_std} \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)} \end{equation} so that \begin{equation} \label{eq:even_odd_psi} \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)} \end{equation} 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.

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

\begin{equation} \label{eq:odd_from_odd} \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 \end{equation}

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 \eqref{eq: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 \eqref{eq: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

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

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 \eqref{eq:odd_from_odd}.

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

\begin{equation} \label{eq:euler_form} \phi'(z)^2 \enspace = \enspace a \phi^4(z) \enspace + \enspace b \phi^2(z) \enspace + \enspace c \end{equation}

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 \begin{equation*} \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 \end{equation*} 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 \eqref{eq:euler_form} to obtain \begin{equation*} \phi'' \space = \space \phi \cdot \left(2a \phi^2 + b\right) \end{equation*} and then differentiate $\phi^{[2n]}$ formula to obtain \begin{equation*} \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) \end{equation*} and then differentiate the $\phi^{[2n+1]}$ formula to obtain \begin{equation*} \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) \end{equation*} Therefore the polynomials $P_n$ and $Q_n$ are given iteratively by \begin{equation*} 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) \end{equation*}

The Weierstrass invariants for this quartic are \begin{equation} \label{eq:euler_invariants} 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 \end{equation}

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

\begin{equation*} 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 \end{equation*} Determine $a$ by substituting the Laurent expansion for $\phi$ at $z=\omega_2$ into \eqref{eq:euler_form}. Determine $b$ by differentiating \eqref{eq:euler_form} twice, then evaluating it at $z=0$. Determine $c$ by evaluating \eqref{eq:euler_form} at $z=0$.

The four roots of \eqref{eq:euler_form} say $\pm\alpha,\pm\beta$ are given by \begin{equation*} \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 \end{equation*} Determine $\alpha$ and $\beta$ by evaluating $\phi$ at the centres of even symmetry $\tfrac 1 2\omega_1 + \omega_i$.

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 \eqref{eq:euler_form} where

\begin{equation} \label{eq:euler_form_three} 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 \end{equation}

This follows by solving the invariant equations \eqref{eq: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

\begin{equation} \label{eq:euler_form_from_other} 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} \end{equation}

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.

\begin{eqnarray*} 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) \\ & = & ac \space + \space \tfrac 1 {12} b^2 \quad \checkmark \\\\ \tfrac 1 6 a'b'c' - \tfrac 1 {216} b'^3 & = & \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) \\ & = & \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} \\ & = & \tfrac 1 6 abc\space - \space \tfrac 1 {216} b^3 \quad \checkmark \end{eqnarray*}

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 \begin{equation} \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) \end{equation}

The Euler curve with the highest degree of root-symmetry is \begin{equation} \label{eq:curve_sym} 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 \end{equation} with invariants \begin{equation} 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 \end{equation} and cross-ratio \begin{equation} \lambda(\alpha) \space = \space \crossratio{\alpha,\alpha^{-1},-\alpha,-\alpha^{-1}} \space = \space - \tfrac 1 4 {\left({\alpha - \alpha^{-1}}\right)^{2}} \end{equation} There are 24 values of $\alpha$ for which \eqref{eq: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

\begin{equation} \label{eq:fac_sym} 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 \end{equation}

This formula is the factorisation of \begin{equation*} 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}} \end{equation*} 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 $\lambda$ \begin{equation*} \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)} \end{equation*} Each of these polynomials can be further factorised because of the four symmetries in $\lambda(\alpha)$ \begin{equation*} \lambda(\alpha) \space = \space \lambda(-\alpha) \space = \space \lambda(1 / \alpha) \space = \space \lambda(-1 / \alpha) \end{equation*} giving the factorisation \eqref{eq:fac_sym} into 24 degree 1 polynomials.

From this it follows that the 24 values are \begin{equation} \label{eq:beta_values} \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 \end{equation} and the transformations which carry them onto one another are the Möbius transforms of the octahedral group.

Let $\phi$ be the elliptic function with differential equation \begin{equation} \label{eq:euler_sym} \phi'^2 \enspace = \enspace \phi^4 \enspace - \enspace (\alpha^2 + \alpha^{-2}) \thinspace \phi^2 \enspace + \enspace 1 \end{equation} and boundary condition $\phi(0) = 0$, and for definiteness $\phi'(0) = 1$. We have the basic identities \begin{equation} \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 -\imath\phi(\imath z, \thinspace \imath \alpha) \end{equation} We also have the three distinct odd functions given by

\begin{equation} \label{eq:sym_3} \phi(z, \space \alpha), \qquad \phi\left(\frac {2\imath\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 - \imath\alpha} {1 + \imath\alpha} \right) \end{equation}

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 \eqref{eq:beta_values} with $i=j=0$ and equating the $g_2$ and $g_3$ invariants gives a formula for $\mu$ \begin{equation*} \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\imath\alpha} {1 - \alpha^2} \qquad \textsf{where} \qquad\beta \space = \space \frac {1-\alpha} {1 + \alpha} \end{equation*} Both the $g_2$ and $g_3$ formulae are required to determine which are the correct two roots.

Using \eqref{eq:euler_sym} we can express $\phi$ in terms of $\sn$ as \begin{equation} \label{eq:sym_sn} \phi(z, \space \alpha) \space = \space {\alpha} \sn(\frac {z} {\alpha}, \space \alpha^2) \end{equation} From this we deduce these another set of Jacobian elliptic function formulae for the three distinct odd elliptic functions on the lattice $\left[4K,2\imath K'\right]$

\begin{equation} \label{eq:sn_other} \sn\left(z, \space k\right), \quad\qquad \sn\left(\frac {2 \imath 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 - \imath\sqrt{k}\big)^2} \thinspace z, \space \left(\frac {1 - \imath\sqrt{k}}{1 + \imath\sqrt{k}}\right)^2 \right) \end{equation}

This follows by converting equation \eqref{eq:sym_3} into a formula for $\sn$ using identity \eqref{eq:sym_sn}. \begin{equation*} \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\imath\alpha^2 z} {1 - \alpha^2} \cdot \frac {1+\alpha} {1-\alpha}, \left(\frac {1-\alpha} {1+\alpha}\right)^2\right) \end{equation*} Because of the identity \begin{equation*} \sn(z, \space \frac 1 k) \space = \space k \sn(\frac z k, k) \end{equation*} using the opposite square root sign will still give the same function (up to a constant multiplier). \begin{equation*} \left(1 - \sqrt{k}\right)^2 \sn\left(\frac {2 \imath 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 \imath k} {\left(1 + \sqrt{k}\right)^2} \thinspace z, \space \left(\frac {1 + \sqrt{k}}{1 - \sqrt{k}}\right)^2 \right) \end{equation*}

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 \begin{equation} \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)} \end{equation} and in terms of the Weierstrass $\wp$ function as \begin{equation} \label{eq:p_odd} \frac {\wp(z) - e_1} {\wp'(z)}, \qquad \frac {\wp(z) - e_2} {\wp'(z)}, \qquad \frac {\wp(z) - e_3} {\wp'(z)} \end{equation} and in terms of the Weierstrass $h$ functions as \begin{equation} \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)} \end{equation} and in terms of the Weierstrass $\zeta$ function as \begin{equation} \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) \end{equation} and in terms of the Jacobian elliptic functions on the lattice $\left[4K,2iK'\right]$ as \begin{equation} \sn(z), \qquad \sn(z) \space \pm \space \frac {\cn(z)\dn(z) - 1} {k\sn(z)} \end{equation} and in terms of the Jacobian elliptic functions on the lattice $\left[2K,2iK'\right]$ as \begin{equation} \frac {\sn(z)} {\cn(z)\dn(z)}, \qquad \frac {\sn(z)\cn(z)} {\dn(z)}, \qquad \frac {\sn(z)\dn(z)} {\cn(z)} \end{equation} 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 \begin{equation} \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)} \end{equation}

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{equation} \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} \end{equation}

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{equation} \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} \end{equation}

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

\begin{equation} \label{eq:odd} ax^2y \enspace + \enspace bxy^2 \enspace + \enspace cx \enspace + \enspace dy \enspace = \enspace 0 \end{equation}

Every pair of order 2 elliptic functions satisfies an algebraic relation in the form of the biquadratic curve \begin{equation*} \sum_{0 \le i,j \le 2} a_{ij} x^i y^j \enspace = \enspace 0 \end{equation*} 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 \begin{equation*} 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 \end{equation*} and \begin{equation*} a_{21}x^2y \enspace + \enspace a_{12}xy^2 \enspace + \enspace a_{10}x \enspace + \enspace a_{01}y \enspace = \enspace 0 \end{equation*} 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 \begin{equation*} 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 \end{equation*} 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 \eqref{eq:odd} with all four coefficients non-zero.

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

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

\begin{equation} \label{eq:odd-normalised} \phi(z) \enspace = \enspace e^{\eta_j\omega_k + \sfrac 1 2 n\pi \imath} \cdot \frac {\sigma(z)\sigma(z + \omega_j + \omega_k)} {\sigma(z + \omega_j)\sigma(z + \omega_k)} \end{equation}

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

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

\begin{equation} \label{eq:normalised_euler_form} \phi'(z)^2 \enspace = \enspace a\phi^4(z) \enspace + \enspace b \phi^2(z) \enspace + \enspace a \end{equation}

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 \eqref{eq:euler_form} and clearing denominators gives \begin{equation*} \phi'(z)^2 \enspace = \enspace a \enspace + \enspace b \phi(z)^2 \enspace + \enspace c \phi(z)^4 \end{equation*} and the only way this can happen is if $a = c$.

Equation \eqref{eq: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 \begin{equation} \crossratio{\alpha,-\alpha,1/\alpha,\phi(z)} \enspace = \enspace \crossratio{e_1,e_2,e_3,f(z)} \end{equation} where $\alpha$ is found by solving the quadratic equation obtained by equating the cross-ratio's of the roots \begin{equation} \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} \end{equation}

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 \begin{equation} 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)} \end{equation} and in terms of the Weierstrass $\wp$ function as \begin{equation} \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)} } \end{equation} and in terms of the Weierstrass $h$ functions as \begin{equation} 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)} \end{equation} 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 \begin{equation} \label{eq:norm_f} \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)} } \end{equation} The expressions under the square root signs in \eqref{eq: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).

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

\begin{equation} \label{eq:reduced_hors} \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)} \end{equation}

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 \begin{equation} 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)} \end{equation} then substituting into \eqref{eq:reduced_hors} gives \begin{equation*} \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} \end{equation*} Now note that equation \eqref{eq: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 $(x,y) \rightarrow (-x,-y)$.

Lattice Aligned Order 2 Elliptic Functions

Nomenclature / Definitions

Perfect Squares

Odd Symmetry

Even-Odd Symmetry