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Simplest Case

Two even elliptic functions, $f$ and $g$, of order 2 with the same poles must satisfy a simple linear equation \begin{equation} \label{eq:fg1} C_1 + C_2 f(z) + C_3 g(z) = 0 \end{equation} To determine the coefficients $C_1,C_2,C_3$ we can evaluate \eqref{eq:fg1} at two different points $z_1,z_2$ and solve using linear algebra to get \begin{equation} \label{eq:fg1d} \begin{vmatrix} 1 & f(z) & g(z) \\ 1 & f(z_1) & g(z_1) \\ 1 & f(z_2) & g(z_2) \\ \end{vmatrix} \space = \space 0 \end{equation} so that the $C_i$ are given by the cofactors of the first row, explicitly \begin{equation} \label{eq:fg1c} C_1 = \begin{vmatrix} f(z_1) & g(z_1) \\ f(z_2) & g(z_2) \\ \end{vmatrix}, \quad C_2 = - \begin{vmatrix} 1 & g(z_1) \\ 1 & g(z_2) \\ \end{vmatrix}, \quad C_3 = \begin{vmatrix} 1 & f(z_1) \\ 1 & f(z_2) \\ \end{vmatrix} \end{equation} But equation \eqref{eq:fg1} cannot depend on $z_1,z_2$ therefore each $C_i$ must be a constant multiple of some function of $W(z_1,z_2)$. This function can be determined by using the $\sigma$ function representation of $f$ and $g$ \begin{equation} f(z) = f_0 \frac {\sigma(z - a)\sigma(z + a)} {\sigma(z - c)\sigma(z + c)},\quad g(z) = g_0 \frac {\sigma(z - b)\sigma(z + b)} {\sigma(z - c)\sigma(z + c)} \end{equation} and substituting into \eqref{eq:fg1c} and applying the $\sigma$ addition formula yields \begin{equation} C_1 = f_0 g_0 \left[\sigma(c - a)\sigma(c + a) - \sigma(c - b)\sigma(c + b)\right] W(z_1,z_2),\quad C_2 = -g_0 \sigma(c - b)\sigma(c + b) W(z_1,z_2),\quad C_3 = f_0 \sigma(c - a)\sigma(c + a) W(z_1,z_2) \end{equation} where \begin{equation} W(z_1,z_2) = \frac {\sigma(z_1 - z_2)\sigma(z_1 + z_2)} {\sigma(z_1 - c)\sigma(z_1 + c)\sigma(z_2 - c)\sigma(z_2 + c)} \end{equation} and the identity for any $x,y,z,a,b,c$ \begin{equation} \label{eq:fg1i} \begin{vmatrix} \sigma(x - a)\sigma(x + a) & \sigma(x - b)\sigma(x + b) & \sigma(x - c)\sigma(x + c) \\ \sigma(y - a)\sigma(y + a) & \sigma(y - b)\sigma(y + b) & \sigma(y - c)\sigma(y + c) \\ \sigma(z - a)\sigma(z + a) & \sigma(z - b)\sigma(z + b) & \sigma(z - c)\sigma(z + c) \\ \end{vmatrix} \space = \space 0 \end{equation}

Next Case

Two even elliptic functions, $f$ and $g$, of order 2 must satisfy a simple equation \begin{equation} \label{eq:fg2} C_1 + C_2 f(z) + C_3 g(z) + C_4 f(z)g(z)= 0 \end{equation} \begin{equation} W(z_1,z_2,z_3) = \frac {\sigma(z_1 + z_2 + z_3)\sigma(z_1 - z_2)\sigma(z_1 - z_3)\sigma(z_2 - z_3)} {\prod\limits_{i=1}^{3}\sigma(z_i - c)\sigma(z_i + c)\sigma(z_i - d)\sigma(z_i + d)} \end{equation} and the identity for any $w,x,y,z,a,b,c,d$ \begin{equation} \label{eq:fg2i} \begin{vmatrix} \sigma(w,a,b) & \sigma(w,a,d) & \sigma(w,b,c) & \sigma(w,c,d) \\ \sigma(x,a,b) & \sigma(x,a,d) & \sigma(x,b,c) & \sigma(x,c,d) \\ \sigma(y,a,b) & \sigma(y,a,d) & \sigma(y,b,c) & \sigma(y,c,d) \\ \sigma(z,a,b) & \sigma(z,a,d) & \sigma(z,b,c) & \sigma(z,c,d) \\ \end{vmatrix} \space = \space 0 \end{equation}

Representation Formula

Starting from the formula \begin{equation} \label{eq:gf2} g(z) = \frac {C_1 + C_2 f(z) + C_3 f(z)^2 + C_4 f'(z)} {C_5 + C_6 f(z) + C_7 f(z)^2} \end{equation} rearrange it to \begin{equation} \label{eq:gf} C_1 + C_2 f(z) + C_3 f(z)^2 + C_4 f'(z) + C_5 g(z) + C_6 f(z) g(z) + C_7 f(z)^2 g(z) = 0 \end{equation} then evaluate it at 6 distinct complex numbers $z_1 \ldots z_6$ and solve for $C_i$ using linear algebra to get \begin{equation} \label{eq:gf1} \begin{vmatrix} 1 & f(z) & f(z)^2 & f'(z) & g(z) & f(z)g(z) & f(z)^2 g(z) \\ 1 & f(z_1) & f(z_1)^2 & f'(z_1) & g(z_1) & f(z_1)g(z_1) & f(z_1)^2 g(z_1) \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1 & f(z_6) & f(z_6)^2 & f'(z_6)^2 & g(z_6) & f(z_6)g(z_6) & f(z_6)^2 g(z_6) \\ \end{vmatrix} \space = \space 0 \end{equation} where $C_i$ is the cofactor of the first row and $i$-th column of the left hand side of \eqref{eq:gf1}. For example \begin{equation*} C_1 = \begin{vmatrix} f(z_1) & f(z_1)^2 & f'(z_1) & g(z_1) & f(z_1)g(z_1) & f(z_1)^2 g(z_1) \\ f(z_2) & f(z_2)^2 & f'(z_2) & g(z_2) & f(z_2)g(z_2) & f(z_2)^2 g(z_2) \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ f(z_6) & f(z_6)^2 & f'(z_6) & g(z_6) & f(z_6)g(z_6) & f(z_6)^2 g(z_6) \\ \end{vmatrix} \end{equation*} Letting $z_i \rightarrow 0$ we get the $C_i$ expressed in terms of the first 7 coefficients in the power series expansion of $f$ and $g$ at 0 \begin{equation*} C_1 = \begin{vmatrix} h_2(0) & h_3(0) & \ldots & h_7(0) \\ h_2'(0) & h_3'(0) & \ldots & h_7'(0) \\ \vdots & \vdots & \ddots & \vdots \\ h_2^{[5]}(0) & h_3^{[5]}(0) & \ldots & h_7^{[5]}(0) \\ \end{vmatrix} \quad C_2 = \begin{vmatrix} h_1(0) & h_3(0) & \ldots & h_7(0) \\ h_1'(0) & h_3'(0) & \ldots & h_7'(0) \\ \vdots & \vdots & \ddots & \vdots \\ h_1^{[5]}(0) & h_3^{[5]}(0) & \ldots & h_7^{[5]}(0) \\ \end{vmatrix} \quad \text{etc.} \end{equation*} where $h_1 = 1,\thinspace h_2 = f,\thinspace h_3 = f^2,\thinspace h_4 = f',\thinspace h_5 = g,\thinspace h_6 = fg,\thinspace h_7 = f^2g$.

Let $a_1,a_2$ be the poles of $f$ and $a_3,a_4$ be the poles of $g$. Then applying EFS formula to the polar divisor $2a_1 + 2a_2 + a_3 + a_4$ gives

Question

Hence for $n=3$ if $\sigma(z,u) = \sigma(z - u_1)\sigma(z - u_2)\sigma(z - u_3)$ where $u_1+u_2+u_3=0$ etc. then \begin{equation} \label{eq:efs4} \begin{vmatrix} \sigma(z_1,u) & \sigma(z_1,v) & \sigma(z_1,w) \\ \sigma(z_2,u) & \sigma(z_2,v) & \sigma(z_2,w) \\ \sigma(z_3,u) & \sigma(z_3,v) & \sigma(z_3,w) \\ \end{vmatrix} \space = \space K \sigma(z_1 + z_2 + z_3) \sigma(z_1-z_2)\sigma(z_1-z_3)\sigma(z_2-z_3) \end{equation} where $K$ is independent of the $z_i$, what is $K$?
If $\{u_i\} \equiv \{v_i\}$ then the LHS vanishes.
Observe $\frac d {dz} \sigma(z,u) = \zeta(z,u)\sigma(z,u)$ where $\zeta(z,u) = \zeta(z-u_1) + \zeta(z-u_2) +\zeta(z-u_3)$
So taking the limit as $z_3 \rightarrow z_2$ then $z_2 \rightarrow z_1$ then $z_1 = z$ gives \begin{equation} \label{eq:efs5} \begin{vmatrix} 1 & 1 & 1 \\ \zeta(z,u) & \zeta(z,v) & \zeta(z,w) \\ \zeta(z,u)^2 + \wp(z,u) & \zeta(z,v)^2 + \wp(z,v) & \zeta(z,w)^2 + \wp(z,w) \\ \end{vmatrix} \space = \space K \frac {\sigma(3z)} {\sigma(z,u)\sigma(z,v)\sigma(z,w)} \end{equation} Note $\zeta(0,u)^2 + \wp(0,u) = 0$ is a well known identity when $u_1+u_2+u_3=0$ so both sides vanish at $z=0$.