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Using these definitions for the Weierstrass elliptic functions and these definitions for the Jacobi elliptic functions let us state the following formulae and make some general observations.

Motivation

For the Weierstrass $\wp$ function we have the identity \begin{equation} \label{eq:wp} \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} \enspace = \enspace 2 \cdot \frac {\sigma(z_1+z_2+z_3)\sigma(z_1-z_2)\sigma(z_2-z_3)\sigma(z_3-z_1)} {\sigma^3(z_1)\sigma^3(z_2)\sigma^3(z_3)} \end{equation} For the general even elliptic function $f$ with two poles at $\pm\rho$ with residues $\pm a^{-\tfrac 1 2}$ we have the identity \begin{equation} \label{eq:f} \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} \enspace = \enspace \frac {2\sigma^4(2\rho)} {a^2} \cdot \frac {\sigma(z_1+z_2+z_3+z_4) \prod\limits_{i \lt j}\sigma(z_i-z_j)} {\prod\limits_{i}\sigma^2(z_i - \rho)\sigma^2(z_i + \rho)} \end{equation} For the Jacobi elliptic functions $\cn,\sn,\dn$ with modulus $k^2$ and four poles $\rho_i$ at $\pm iK'$ and $\pm(2K + iK')$ chosen so that $\rho_1+\rho_2+\rho_3+\rho_4=0$ we have the identity \begin{equation} \label{eq:jac} \begin{vmatrix} 1 & \cn(z_1) & \sn(z_1) & \dn(z_1) \\ 1 & \cn(z_2) & \sn(z_2) & \dn(z_2) \\ 1 & \cn(z_3) & \sn(z_3) & \dn(z_3) \\ 1 & \cn(z_4) & \sn(z_4) & \dn(z_4) \\ \end{vmatrix} \enspace = \enspace \frac {4} {k^2} \cdot \frac {\sigma(z_1+z_2+z_3+z_4) \prod\limits_{i \lt j}\sigma(z_i - z_j) \prod\limits_{i \lt j}\sigma(\rho_i - \rho_j)} {\prod\limits_{i,j}\sigma(z_i-\rho_j)} \end{equation}

General Observations

1. The three formulae \eqref{eq:wp}, \eqref{eq:f} and \eqref{eq:jac} are examples of the Extended Frobenius-Stickelberger theorem.
2. When you set $z_1 \space + \space z_2 \space + \space z_3 \enspace [\space + \space z_4] \space = \space 0$ the RHS vanishes and you get the familiar symmetric addition formula.
3. For all these formulae there are infinite sequences of similar formulae generated by adding new basis members one at a time. That is to say you can construct similar formula on an arbitrary number of variables. For example \eqref{eq:f} is the formula obtained by adding $\wp''$ (or equivalently $\wp^2$) to the basis $1,\wp,\wp'$ of \eqref{eq:wp}.
4. The denominators in these formulae are somewhat irrelevant. When cleared on both sides we get an identity involving only $\sigma$ functions.
5. There is an infinite number of such formulae each corresponding to a particular division of the period lattice into various sub-lattices.

Five Algebraic Integrals

Each of the above formulae can be interpreted as the algebraic integral of a certain type of differential equation (Abel sum).

Writing the differential equation like this \begin{equation} \label{eq:diff} \frac {dx_1} {\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)(x_1-e_4)}} \enspace + \enspace \frac {dx_2} {\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)(x_2-e_4)}} \enspace + \enspace \frac {dx_3} {\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)(x_3-e_4)}} \enspace = \enspace 0 \end{equation} several simple determinant style implicit algebraic integrals can be found, with $x_4$ the constant of integration:

$I_1$, $I_2$ and $I_3$ are obtained by applying the Extended Frobenius-Stickelberger formula using the specified basis. $I_4$ and $I_5$ are obtained by symmetry in $I_3$ along with the fact that the first three integrals must be algebraically equivalent. Despite appearances the basis elements are meromorphic elliptic functions with respect to the specified lattice.

Putting $e_1 = 0, \space e_2 = 1, \space e_3 = 1/k^2, \space e_4 = \infty$ in \eqref{eq:diff} gives \begin{equation*} \frac {dx_1} {\sqrt{x_1(1 - x_1)(1 - k^2x_1)}} \enspace + \enspace \frac {dx_2} {\sqrt{x_2(1 - x_2)(1 - k^2x_2)}} \enspace + \enspace \frac {dx_3} {\sqrt{x_3(1 - x_3)(1 - k^2x_3)}} \enspace = \enspace 0 \end{equation*} which we recognise as the differential formula for $x=\sn^2(z)$. So making the following substitutions \begin{equation*} \sqrt{x_i - e_1} = \sn(z_i), \qquad \sqrt{\frac {x_i - e_2} {-e_2}} = \cn(z_i), \qquad \sqrt{\frac {x_i - e_3} {-e_3}} = \dn(z_i), \qquad \sqrt{\frac {x_i - e_4} {-e_4}} = 1 \end{equation*} converts the integrals in Table 1 to addition formulae for the Jacobian Elliptic functions. Therefore when $z_1 + z_2 + z_3 + z_4 = 0$ we have

$\approx \textsf{Cayley} \approx$ $I_1$ is a symmetric root difference expression invariant under Möbius transforms. Therefore when rationalised the resulting polynomial must be a simultaneous invariant (in the sense of classical Invariant Theory) of the two quartic polynomials $R(x) = (x - e_1)(x - e_2)(x - e_3)(x - e_4)$ and $S(x) = (x - x_1)(x - x_2)(x - x_3)(x - x_4)$. We have \begin{equation} \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & x_1^2 & \hphantom{+} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \pm \sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \enspace = \enspace \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^4 \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \end{equation} where the simultaneous invariant $\mathfrak{R}$ can be expressed in terms of scaled transvectants \begin{equation} \mathfrak{R}(R,S) \enspace = \enspace 64 \Big[ \mathfrak{S}(R,S)^2 \enspace - \enspace \resultant(R,S) \Big] \qquad\qquad \mathfrak{S}(R,S) \enspace = \enspace 12 \thinspace \transvectant{\transvectant{R,R}_2,\transvectant{S,S}_2}_4 \enspace - \enspace \transvectant{R,S}_4^2 \enspace - \enspace \transvectant{R,R}_4 \transvectant{S,S}_4 \end{equation} Because $\mathfrak{R}$ is symmetric in $R$ and $S$ it has $x \leftrightarrow e$ symmetry. Like many invariants in the classical theory it has a simple interpretation in terms of the roots of the two quartics $R$ and $S$, in this case being that it vanishes if either $R$ or $S$ is a perfect square.

There are "trivial" zeroes of the form $x_i=x_j$ for which $I_1$ vanishes and these correspond to the Vandermonde determinant. There are also "non-trivial" zeroes of the form $e_1=e_2$ and $e_3 = e_4$ for which $I_1$ vanishes and these correspond to the invariant $\mathfrak{R}$. These zeroes also provide the key to understanding the $\sigma$ factorisations of $I_1 \dots I_5$ when expressed in terms of $f$.

For $I_2$ we have \begin{equation} \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+}\sqrt{(x_1-e_1)(x_1-e_2)} & \hphantom{+}\sqrt{(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & \pm\sqrt{(x_2-e_1)(x_2-e_2)} & \pm\sqrt{(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & \pm\sqrt{(x_3-e_1)(x_3-e_2)} & \pm\sqrt{(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & \pm\sqrt{(x_4-e_1)(x_4-e_2)} & \pm\sqrt{(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \enspace = \enspace 2^{32} (e_1-e_2)^{16}(e_3-e_4)^{16} \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{16} \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4)^{4} \end{equation} For $I_3$ we have \begin{equation} \prod_{\textsf{all signs}} \begin{vmatrix} \sqrt{x_1-e_1} & \phantom{+}\sqrt{x_2-e_1} & \phantom{+}\sqrt{x_3-e_1} & \phantom{+}\sqrt{x_4-e_1} \\ \sqrt{x_1-e_2} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_3-e_2} & \pm\sqrt{x_4-e_2} \\ \sqrt{x_1-e_3} & \pm\sqrt{x_2-e_3} & \pm\sqrt{x_3-e_3} & \pm\sqrt{x_4-e_3} \\ \sqrt{x_1-e_4} & \pm\sqrt{x_2-e_4} & \pm\sqrt{x_3-e_4} & \pm\sqrt{x_4-e_4} \\ \end{vmatrix} \enspace = \enspace 2^{256} \begin{vmatrix} 1 & e_1 & e_1^2 & e_1^3 \\ 1 & e_2 & e_2^2 & e_2^3 \\ 1 & e_3 & e_3^2 & e_3^3 \\ 1 & e_4 & e_4^2 & e_4^3 \\ \end{vmatrix}^{64} \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{64} \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4)^{16} \end{equation}

NOTES

Schwarz-Christoffel Examples ?

\begin{equation*} \frac {dx_1} {\sqrt[3]{(x_1-e_1)^2(x_1-e_2)^2(x_1-e_3)^2}} \enspace + \enspace \frac {dx_2} {\sqrt[3]{(x_2-e_1)^2(x_2-e_2)^2(x_2-e_3)^2}} \enspace = \enspace 0 \qquad\qquad \implies \qquad\qquad \begin{vmatrix} \sqrt[3]{x_1-e_1} & \sqrt[3]{x_1-e_2} & \sqrt[3]{x_1-e_3} \\ \sqrt[3]{x_2-e_1} & \sqrt[3]{x_2-e_2} & \sqrt[3]{x_2-e_3} \\ \sqrt[3]{x_3-e_1} & \sqrt[3]{x_3-e_2} & \sqrt[3]{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation*} where $x_3$ is the constant of integration. \begin{equation*} \frac {dx_1} {\sqrt[4]{(x_1-e_1)^2(x_1-e_2)^3(x_1-e_3)^3}} \enspace + \enspace \frac {dx_2} {\sqrt[4]{(x_2-e_1)^2(x_2-e_2)^3(x_2-e_3)^3}} \enspace = \enspace 0 \qquad\qquad \implies \qquad\qquad \begin{vmatrix} \sqrt{x_1-e_1} & \sqrt[4]{x_1-e_2} & \sqrt[4]{x_1-e_3} \\ \sqrt{x_2-e_1} & \sqrt[4]{x_2-e_2} & \sqrt[4]{x_2-e_3} \\ \sqrt{x_3-e_1} & \sqrt[4]{x_3-e_2} & \sqrt[4]{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation*} \begin{equation*} \frac {dx_1} {\sqrt[6]{(x_1-e_1)^3(x_1-e_2)^4(x_1-e_3)^5}} \enspace + \enspace \frac {dx_2} {\sqrt[6]{(x_2-e_1)^3(x_2-e_2)^4(x_2-e_3)^5}} \enspace = \enspace 0 \qquad\qquad \implies \qquad\qquad \begin{vmatrix} \sqrt{x_1-e_1} & \sqrt[3]{x_1-e_2} & \sqrt[6]{x_1-e_3} \\ \sqrt{x_2-e_1} & \sqrt[3]{x_2-e_2} & \sqrt[6]{x_2-e_3} \\ \sqrt{x_3-e_1} & \sqrt[3]{x_3-e_2} & \sqrt[6]{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation*}

Explanation For $x \leftrightarrow e$ Symmetry ???

\begin{equation*} \begin{vmatrix} \sqrt{f(u_1) - f(v_1)} & \sqrt{f(u_1) - f(v_2)} & \sqrt{f(u_1) - f(v_3)} & \sqrt{f(u_1) - f(v_4)} \\ \sqrt{f(u_2) - f(v_1)} & \sqrt{f(u_2) - f(v_2)} & \sqrt{f(u_2) - f(v_3)} & \sqrt{f(u_2) - f(v_4)} \\ \sqrt{f(u_3) - f(v_1)} & \sqrt{f(u_3) - f(v_2)} & \sqrt{f(u_3) - f(v_3)} & \sqrt{f(u_3) - f(v_4)} \\ \sqrt{f(u_4) - f(v_1)} & \sqrt{f(u_4) - f(v_2)} & \sqrt{f(u_4) - f(v_3)} & \sqrt{f(u_4) - f(v_4)} \\ \end{vmatrix} \enspace = \enspace const. \frac {\sigma\big(\tfrac 1 2 (u_1+u_2+u_3+u_4) - \tfrac 1 2 (v_1+v_2+v_3+v_4)\big) \prod\limits_{i \lt j} \sigma\big(\tfrac 1 2 (u_i-u_j)\big) \prod\limits_{i \lt j} \sigma\big(\tfrac 1 2 (v_i-v_j)\big)} {\sqrt{ \prod \sigma(u_i-\rho)\sigma(u_i+\rho) \thinspace \prod \sigma(v_i-\rho)\sigma(v_i+\rho)} } \end{equation*} therefore \begin{equation*} u_1+u_2+u_3+u_4 \space = \space v_1+v_2+v_3+v_4 \qquad\qquad \implies \qquad\qquad \mathfrak{R}(f(u_1),f(u_2),f(u_3),f(u_4),f(v_1),f(v_2),f(v_3),f(v_4)) \space = \space 0 \end{equation*} Actually it can't have this form because $f$ is even and therefore we actually have $u_1\pm u_2\pm u_3\pm u_4\pm v_1\pm v_2\pm v_3\pm v_4 = 0$ which would imply $\mathfrak{R}$ was (skew-)symmetric in all eight arguments - which it isn't. To avoid this issue you could have $u_1\pm u_2\pm u_3\pm u_4=0$ and $v_1\pm v_2\pm v_3\pm v_4 = 0$ but that seems unlikely.

References

[1] F.G. Frobenius and L. Stickelberger Zur Theorie der elliptischen Functionen J. reine angew. Math 83 (1877), 175–179