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The dual addition formula is obtained as a consequence of the following remarkable algebraic identity, let \begin{equation} \label{eq:Delta} \Delta \space = \space \big[(x_1x_2 + x_1x_3 + x_2x_3) - (e_1e_2 + e_1e_3 + e_2e_3)\big]^2 \space - \space 4 (x_1x_2x_3 - e_1e_2e_3)\big[(x_1+x_2+x_3) - (e_1+e_2+e_3)\big] \end{equation} then \begin{equation} \label{eq:cubicsym} \frac {\prod\limits_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+}\sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \pm\sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \pm\sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^2} \space = \space \Delta \space = \space \frac {\prod\limits_{\textsf{all signs}} \begin{vmatrix} 1 & e_1 & \hphantom{+}\sqrt{(e_1 - x_1)(e_1 - x_2)(e_1 - x_3)} \\ 1 & e_2 & \pm\sqrt{(e_2 - x_1)(e_2 - x_2)(e_2 - x_3)} \\ 1 & e_3 & \pm\sqrt{(e_3 - x_1)(e_3 - x_2)(e_3 - x_3)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^2} \end{equation} This formula is a bit puzzling because when one of the determinants on the left vanishes, one on the right must also vanish. But which one? This question suggests that each of the determinants should be further factorisable. We will show that each determinant can be factored into four further determinants. And that each of these further determinants can then be factored into another four factors, resulting in infinite sequence of products of $4^n$ algebraic factors representing $\Delta$.

Formulation In Terms Of Weierstrass $\wp$ And $\sigma$ Functions

The Frobenius-Stickelberger formula^[1] for $n=2,3$ gives \begin{equation} \label{eq:fs2} \begin{vmatrix} 1 & \wp(u) & \wp^2(u) \\ 1 & \wp(v) & \wp^2(v) \\ 1 & \wp(w) & \wp^2(w) \\ \end{vmatrix} \space = \space \frac {\sigma(u-v)\sigma(u-w)\sigma(v-w)\sigma(u+v)\sigma(u+w)\sigma(v+w)} {\sigma^2(u)\sigma^2(v)\sigma^2(w)} \end{equation} and \begin{equation} \label{eq:fs3} - \tfrac 1 2 \begin{vmatrix} 1 & \wp(u) & \wp'(u) \\ 1 & \wp(v) & \wp'(v) \\ 1 & \wp(w) & \wp'(w) \\ \end{vmatrix} \space = \space \begin{vmatrix} 1 & \wp(u) & \sqrt{(\wp(u)-e_1)(\wp(u)-e_2)(\wp(u)-e_3)} \\ 1 & \wp(v) & \sqrt{(\wp(v)-e_1)(\wp(v)-e_2)(\wp(v)-e_3)} \\ 1 & \wp(w) & \sqrt{(\wp(w)-e_1)(\wp(w)-e_2)(\wp(w)-e_3)} \\ \end{vmatrix} \space = \space \frac {\sigma(u+v+w)\sigma(u-v)\sigma(u-w)\sigma(v-w)} {\sigma^3(u)\sigma^3(v)\sigma^3(w)} \end{equation} Putting $x_1=\wp(u),x_2=\wp(v),x_3=\wp(w)$ in the LHS of \eqref{eq:cubicsym} gives the formula \begin{equation} \label{eq:Delta2} \Delta(u,v,w) \space = \space \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v+w)\sigma(u-v-w)} {\sigma(u)^4\sigma(v)^4\sigma(w)^4} \end{equation} with each of the factors corresponding to one of the determinants in an obvious way.

Factorisation Of RHS Determinants Into $\sigma$ Functions

We now seek to do a factorisation similar to \eqref{eq:fs3} of the determinants on the RHS of \eqref{eq:cubicsym}.

Swapping the order of variables under the square root sign for convenience, the determinants on the RHS have the formula \begin{equation} \label{eq:H} H(u,v,w,\delta_1,\delta_2,\delta_3) \space = \space \begin{vmatrix} 1 & e_1 & \delta_1\sqrt{(\wp(u) - e_1)(\wp(v) - e_1)(\wp(w) - e_1)} \\ 1 & e_2 & \delta_2\sqrt{(\wp(u) - e_2)(\wp(v) - e_2)(\wp(w) - e_2)} \\ 1 & e_3 & \delta_3\sqrt{(\wp(u) - e_3)(\wp(v) - e_3)(\wp(w) - e_3)} \\ \end{vmatrix} \end{equation} for suitably chosen $\delta_i$. Using identity \eqref{eq:cubicsym}we can deduce that \begin{equation*} \frac {\imath H(u,v,w,1,1,1) \thinspace \imath H(u,v,w,1,-1,1) \thinspace \imath H(u,v,w,1,1,-1) \thinspace \imath H(u,v,w,1,-1,-1)} {\begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^2} \enspace = \enspace \Delta(u,v,w) \end{equation*} which after rearranging, and noting $H(u,v,w,1,-1,-1) = -H(u,v,w,-1,1,1)$ then using \eqref{eq:Delta2} gives \begin{equation} \label{eq:Hprod} H(u,v,w,1,1,1) H(u,v,w,-1,1,1) H(u,v,w,1,-1,1) H(u,v,w,1,1,-1) \space = \space - \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^2 \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v+w)\sigma(u-v-w)} {\sigma(u)^4\sigma(v)^4\sigma(w)^4} \end{equation} This is true no matter which rule is used to select the signs of the square roots, as long as it is applied consistently over all factors. So we know the factorisation of the product of the determinants, what remains is to determine which factors belong to which determinants.

Handy Preliminary Formulae

Using the DLMF definitions of $\sigma, \omega_i, \eta_i$ we have \begin{equation} \label{eq:periodIdentities} \omega_1+\omega_2+\omega_3 = 0\qquad\qquad \eta_1+\eta_2+\eta_3 = 0\qquad\qquad \eta_2\omega_1 - \eta_1\omega_2 = \eta_3\omega_2 - \eta_2\omega_3 = \eta_1\omega_3 - \eta_3\omega_1 = \tfrac 1 2 \pi \imath \end{equation} which implies \begin{equation} \label{eq:eta} \eta_1 \omega_1 + \eta_2\omega_2 + \eta_3\omega_3 \space = \space -2(\eta_1\omega_2 + \eta_2\omega_3 + \eta_3\omega_1) - \tfrac 3 2 \pi \imath \space = \space -2(\eta_2\omega_1 + \eta_3\omega_2 + \eta_1\omega_3) + \tfrac 3 2 \pi \imath \end{equation} We also have with $i,j,k$ distinct \begin{equation} \label{eq:sigma} \sigma(z+2\omega_i) = - e^{2\eta_i (z+\omega_i)} \sigma(z) \qquad\qquad \sigma(z-\omega_i) = - e^{-2\eta_i z} \sigma(z+\omega_i) \qquad\qquad \sigma(\omega_j - \omega_i) = e^{-2\eta_i\omega_j}\sigma(\omega_k) \end{equation} and therefore \begin{equation} \label{eq:wpwp} \wp(z) - \wp(u) \space = \space - \frac {\sigma(z-u)\sigma(z+u)} {\sigma^2(z)\sigma^2(u)} \qquad\qquad \wp'(z) \space = \space - \frac {\sigma(2z)} {\sigma^4(z)} \qquad\qquad \wp(z) - e_i \space = \space - \frac {\sigma(z-\omega_i)\sigma(z+\omega_i)} {\sigma^2(z)\sigma^2(\omega_i)} \qquad\qquad e_j - e_i \space = \space e^{-2\eta_i\omega_j} \frac {\sigma^2(\omega_k)} {\sigma^2(\omega_i) \sigma^2(\omega_j)} \end{equation} and combining \eqref{eq:eta}, \eqref{eq:sigma} and \eqref{eq:wpwp} gives \begin{equation} \label{eq:eee} \kappa \space = \space \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix} \space = \space (e_1-e_2)(e_2-e_3)(e_3-e_1) \space = \space \frac {e^{\eta_1\omega_1+\eta_2\omega_2+\eta_3\omega_3+\sfrac 3 2 \pi \imath}} {\sigma^2(\omega_1)\sigma^2(\omega_2)\sigma^2(\omega_3)} \space = \space \frac {\sigma(\omega_1 - \omega_2)\sigma(\omega_2 - \omega_3)\sigma(\omega_3 - \omega_1)} {\sigma^3(\omega_1)\sigma^3(\omega_2)\sigma^3(\omega_3)} \end{equation} We also need these sigma duplication formulae \begin{equation} \label{eq:sigmaDup} \sigma(2z) \enspace = \enspace \frac {2\sigma(z)\sigma(z+\omega_1)\sigma(z+\omega_2)\sigma(z+\omega_3)} {\sigma(\omega_1)\sigma(\omega_2)\sigma(\omega_3)} \enspace = \enspace -\frac {2\sigma(z)\sigma(z-\omega_1)\sigma(z-\omega_2)\sigma(z-\omega_3)} {\sigma(\omega_1)\sigma(\omega_2)\sigma(\omega_3)} \end{equation}

Extract The Square Roots

We now seek to show $\sqrt{(\wp(u)-e_1)(\wp(v)-e_1)(\wp(w)-e_1)}$ is a meromorphic function and give it a definition, unambiguous with regard to sign. Define $h_i(z)$ so that it's residue is $+1$ at $z=0$ \begin{equation} h_i(z) \space = \space \sqrt{\wp(z) - e_i} \space = \space -e^{\eta_i z} \frac {\sigma(z - \omega_i)} {\sigma(z)\sigma(\omega_i)} \space = \space e^{-\eta_i z} \frac {\sigma(z + \omega_i)} {\sigma(z)\sigma(\omega_i)} \qquad\qquad\qquad h_1(z)h_2(z)h_3(z) \space = \space -\tfrac 1 2 \wp'(z) \end{equation} We get the relations with $i,j,k$ distinct \begin{equation} \label{eq:hIdentities} h_i(z) = \frac 1 z \space + \space \bigO(z) \qquad\qquad h_i(-z) = -h_i(z) \qquad\qquad h_i(z + 2\omega_i) = h_i(z) \qquad\qquad h_i(z + 2\omega_j) = -h_i(z) \qquad\qquad h_i'(z) = -h_j(z)h_k(z) \end{equation} Now we can explicitly define the square roots in \eqref{eq:H}. Define \begin{equation} \label{eq:H2} H(u,v,w,\delta_1,\delta_2,\delta_3) \enspace = \enspace \begin{vmatrix} 1 & e_1 & \delta_1\thinspace h_1(u) h_1(v) h_1(w) \\ 1 & e_2 & \delta_2\thinspace h_2(u) h_2(v) h_2(w) \\ 1 & e_3 & \delta_3\thinspace h_3(u) h_3(v) h_3(w) \\ \end{vmatrix} \end{equation} From this we can see \begin{equation} \label{eq:Hformulae} \begin{aligned} H(-u,v,w,1,1,1) \space &= \space - H(u,v,w,1,1,1) \\ H(u+2\omega_1,v,w,1,1,1) \space &= \space H(u,v,w,1,-1,-1) \\ H(u+2\omega_2,v,w,1,1,1) \space &= \space H(u,v,w,-1,1,-1) \\ H(u+2\omega_3,v,w,1,1,1) \space &= \space H(u,v,w,-1,-1,1) \\ \end{aligned} \end{equation} and similar identities for the $v$ and $w$ variables. Therefore in the arguments $u,v,w$ $H$ is an odd function and further $H$ is an elliptic function of order 4 over the doubled period lattice $2\Omega=[4\omega_1,4\omega_2]$.

An Addition Formula for $h_i$

The functions $h_1,h_2,h_3$ are equivalent, after scaling, to the Jacobian elliptic functions $\ns,\cs,\ds$ with $\displaystyle k^2 = \frac {e_3-e_1} {e_2-e_1}$. From this observation, it is easy to determine this addition formulae for them

$\sigma$ Factorisation

With respect to the doubled period lattice $2\Omega$ the expression $\Delta(u,v,w)$, considered as a function of $u$ has a set of 16 zeroes at $u = \pm v \pm w - 2\omega_i$ for $i=0,1,2,3$ (putting $\omega_0=0$ for notational convenience). Therefore each of the four $H$ factors in \eqref{eq:Hprod}, considered as a function of $u$, must have four zeroes from this set. Since $H$ has odd symmetry it's set of 4 zeroes must also have odd symmetry, that is they must be invariant under $(u,v,w) \mapsto (\pm u, \pm v \pm w)$. So if $u+v+w$ is a factor of $H$ so is $u-v+w$, $u+v-w$ and $u-v-w$. Therefore the set of zeroes are partitioned between the $H$ factors as in this table

Similarly $\Delta$ has 16 poles consisting of 4 poles of order 4 at $u = 2\omega_i$ for $i=0,1,2,3$. Therefore $H(u,v,w,+1,+1,+1)$ can be expressed as a $\sigma$ product, over the period lattice $\Omega$, as follows

$\sigma$ Factorisations For Other Sign Variants

Under the substitution $u \mapsto u+2\omega_1$ equation \eqref{eq:H3} becomes

Other sign variants are similar.

Symmetric Product Form

Although equations \eqref{eq:H3} and \eqref{eq:H4} are symmetric in $u,v,w$ the righthand sides do not appear so. The righthand side symmetry can be made clear like this

CROSS CHECK

$H$ in terms of $\Delta$

We have

Proportional Rows

Show that if $u+v+w=0$ then \begin{equation*} \frac {\begin{vmatrix} \wp(u) & \wp'(u) \\ \wp(v) & \wp'(v) \\ \end{vmatrix}} {\begin{vmatrix} 1 & \wp'(u) \\ 1 & \wp'(v) \\ \end{vmatrix}} \enspace = \enspace \frac {\begin{vmatrix} e_1 & h_1(u)h_1(v)h_1(w) \\ e_2 & h_2(u)h_2(v)h_2(w) \\ \end{vmatrix}} {\begin{vmatrix} 1 & h_1(u)h_1(v)h_1(w) \\ 1 & h_2(u)h_2(v)h_2(w) \\ \end{vmatrix}} \end{equation*}

Another Form

Using CAS it is easy to confirm the following algebraic identity \begin{equation} \label{eq:sqrt3Alg} \prod_{\textsf{all signs}} \begin{vmatrix} \sqrt{x_1-e_1} & \hphantom{+}\sqrt{x_2-e_1} & \hphantom{+}\sqrt{x_3-e_1} \\ \sqrt{x_1-e_2} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_3-e_2} \\ \sqrt{x_1-e_3} & \pm\sqrt{x_2-e_3} & \pm\sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 256 \cdot \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^4 \cdot \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^4 \end{equation} The sigma product associated with the determinant on the left is

Taking the product of \eqref{eq:sqrt3} over the 16 possible combinations of substitutions $z_2 \mapsto z_2 + 2\omega_i$ and $z_3 \mapsto z_3 + 2\omega_j$ where $i,j \in \{0,1,2,3\}$ and utilising the sigma duplication formula etc. must eventually lead to \begin{equation} \label{eq:sqrt3prod} \prod_{\textsf{all signs}} \begin{vmatrix} h_1(z_1) & \hphantom{+}h_1(z_2) & \hphantom{+}h_1(z_3) \\ h_2(z_1) & \pm h_2(z_2) & \pm h_2(z_3) \\ h_3(z_1) & \pm h_3(z_2) & \pm h_3(z_3) \\ \end{vmatrix} \enspace = \enspace 256 \cdot \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^4 \cdot \frac {\prod\limits_{i \gt j} \sigma^4(z_i - z_j)\sigma^4(z_i + z_j)} {\prod\limits_i \sigma^{16}(z_i)} \end{equation} which is the elliptic function equivalent of \eqref{eq:sqrt3Alg} - I haven't fully computed it to verify this fact. However we can easily see that there will be $6 \cdot 16 / 4 = 24$ applications of the $\sigma$ duplication formula. Each application will consume one power of 2 and half a power of the $e$-determinant. Therefore the leading constant will be $4^{16} \cdot 2^{-24} = 256$. And the power on the $e$-determinant will be $16 - \tfrac 1 2 \cdot 24 = 4$. So the leading constant term is verified.

Yet Another Form

The functions $h_1,h_2,h_3$ are roughly equivalent to the Jacobian elliptic functions $\sn,\cn,\dn$. Along with the constant function they span the vector space of elliptic functions over the doubled period lattice $2\Omega$ with a pole of order at most one at $0,2\omega_1,2\omega_2,2\omega_3$. Therefore we can apply the extended Frobenius-Stickelberger formula to give

Multiplying by $z_4$ then letting $z_4 \rightarrow 0$ in \eqref{eq:sqrt4} gives

ALTERNATE DERIVATION

Under the substitution $(z_1,z_2,z_3) \mapsto (z_1+2\omega_1,z_2+2\omega_2,z_3+2\omega_3)$ the expression $\sigma(\frac {z_1+z_2+z_3} 2)$ is preserved. There are 6 similar substitutions.

Under the substitution $(z_1,z_2,z_3) \mapsto (z_1+2\omega_1,z_2+2\omega_1,z_3)$ the expression $\sigma(\frac {z_1+z_2+z_3} 2)$ is preserved modulo an exponential factor. There are 9 similar substitutions.

Along with the identity substitution that gives 16 substitutions which preserve $\sigma(\frac {z_1+z_2+z_3} 2)$. If we look at the LHS we see that these substitutions flip the signs on the $h_i$ terms in such a way that there is always an even number of minus signs in each row and an even number in each column. So what does the algebraic product of these 16 determinants look like ??? Wait that's not quite right. What we need to do is take the product over a representative from each of the 16 equivalence classes!

Presumably a careful evaluation of the product of variants of \eqref{eq:sqrt4reduced} arising from the 256 substitutions $(z_1,z_2,z_3) \mapsto (z_1+2\omega_i,\pm z_2+2\omega_j,\pm z_3+2\omega_k)$ will yield a formula something like \begin{equation} \prod_{\textsf{all signs}} \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & \hphantom{+}\sqrt{x_1-e_1} & \pm\sqrt{x_2-e_1} & \pm\sqrt{x_3-e_1} \\ 1 & \pm\sqrt{x_1-e_2} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_3-e_2} \\ 1 & \pm\sqrt{x_1-e_3} & \pm\sqrt{x_2-e_3} & \pm\sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 2^{320} \Delta^{16} \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^{32} \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{32} \end{equation}

Correct Algebraic Solution

What we need to do is chose the variants of \eqref{eq:sqrt4reduced} that arise from the 16 substitutions $(z_1,z_2,z_3) \mapsto (z_1 + 2\omega_i, \thinspace\pm z_2, \thinspace\pm z_3)$ for $i=0,1,2,3$. After a little simplification \begin{equation*} \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & h_1(z_1) & h_2(z_1) & h_3(z_1) \\ \pm 1 & h_1(z_2) & h_2(z_2) & h_3(z_2) \\ \pm 1 & h_1(z_3) & h_2(z_3) & h_3(z_3) \\ \end{vmatrix}, \qquad \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & -h_1(z_1) & h_2(z_1) & h_3(z_1) \\ \pm 1 & h_1(z_2) & h_2(z_2) & h_3(z_2) \\ \pm 1 & h_1(z_3) & h_2(z_3) & h_3(z_3) \\ \end{vmatrix}, \qquad \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & h_1(z_1) & -h_2(z_1) & h_3(z_1) \\ \pm 1 & h_1(z_2) & h_2(z_2) & h_3(z_2) \\ \pm 1 & h_1(z_3) & h_2(z_3) & h_3(z_3) \\ \end{vmatrix}, \qquad \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & h_1(z_1) & h_2(z_1) & -h_3(z_1) \\ \pm 1 & h_1(z_2) & h_2(z_2) & h_3(z_2) \\ \pm 1 & h_1(z_3) & h_2(z_3) & h_3(z_3) \\ \end{vmatrix} \end{equation*} With $i$ fixed this gives us the four $H$-factors. With the signs fixed the substitutions we get the four $E$-factors.

$E$ In Four Variables

From the Extended Frobenius-Stickelberger theorem \begin{equation} \label{eq:E1} E(u,v,w,x) \space = \space \begin{vmatrix} 1 & \wp(u) & \wp^2(u) & h_1(u)h_2(u)h_3(u) \\ 1 & \wp(v) & \wp^2(v) & h_1(v)h_2(v)h_3(v) \\ 1 & \wp(w) & \wp^2(w) & h_1(w)h_2(w)h_3(w) \\ 1 & \wp(x) & \wp^2(x) & h_1(x)h_2(x)h_3(x) \\ \end{vmatrix} \space = \space - \frac {\sigma(u+v+w+x)\sigma(u-v)\sigma(u-w)\ldots\sigma(w-x)} {\sigma^4(u)\sigma^4(v)\sigma^4(w)\sigma^4(x)} \end{equation}

CROSS CHECK

Taking the product of \eqref{eq:E1} evaluated at $u,\pm v,\pm w,\pm x$ with an even number of minus signs, we have $6 \times 4 / 2 = 12$ occurrances of $\wp$ difference factors which is two Vandermonde determinants worth, leaving $4 \times 4 - 2\times 6 = 4$ powers of $\sigma$ in the denominator \begin{equation} \prod_{\textsf{evens}} E(u,\pm v,\pm w, \pm x) \space = \space \begin{vmatrix} 1 & \wp(u) & \wp^2(u) & \wp^3(u) \\ 1 & \wp(v) & \wp^2(v) & \wp^3(v) \\ 1 & \wp(w) & \wp^2(w) & \wp^3(w) \\ 1 & \wp(x) & \wp^2(x) & \wp^3(x) \\ \end{vmatrix}^2 \cdot \frac {\prod\limits_{\textsf{evens}} \sigma(u \pm v \pm w \pm x)} {\sigma^4(u)\sigma^4(v)\sigma^4(w)\sigma^4(x)} \end{equation} and a similar formula for odds.

$H$ In Four Variables

In preparation for analysing the quartic let \begin{equation} H(u,v,w,x,\delta_1,\delta_2,\delta_3) \space = \space \begin{vmatrix} 1 & e_1 & e_1^2 & \delta_1 \thinspace h_1(u)h_1(v)h_1(w)h_1(x) \\ 1 & e_2 & e_2^2 & \delta_2 \thinspace h_2(u)h_2(v)h_2(w)h_2(x) \\ 1 & e_3 & e_3^2 & \delta_3 \thinspace h_3(u)h_3(v)h_3(w)h_3(x) \\ 0 & 0 & 1 & 1 \\ \end{vmatrix} \end{equation} then \begin{equation} \begin{aligned} H(-u,v,w,x,1,1,1) \space &= \space H(u,v,w,x,-1,-1,-1) \\ H(u + 2\omega_1,v,w,x,1,1,1) \space &= \space H(u,v,w,x,1,-1,-1) \\ H(u + 2\omega_2,v,w,x,1,1,1) \space &= \space H(u,v,w,x,-1,1,-1) \\ H(u + 2\omega_3,v,w,x,1,1,1) \space &= \space H(u,v,w,x,-1,-1,1) \\ \end{aligned} \end{equation} and similar identities for the $v,w,x$ variables. In the arguments $u,v,w,x$ $H$ is an elliptic function of order 4 over the doubled period lattice $2\Omega$. Therefore we can deduce that

where evens means terms with an even number of minus signs, and odds means terms with an odd number.

CROSS CHECK

Appying a simple translation and negation of the $u$-arguments in \eqref{eq:Isigma} gives the other sign variants \begin{equation} \label{eq:I2} H(u,v,w,x,1,-1,-1) \enspace = \enspace 2\kappa \cdot \frac {\prod\limits_{\textsf{evens}} \sigma(\frac u 2 \pm \frac v 2 \pm \frac w 2 \pm \frac x 2 + \omega_1)} {\sigma(u + 2\omega_1)\sigma(v)\sigma(w)\sigma(x)} \qquad\qquad\qquad H(u,v,w,x,-1,1,1) \enspace = \enspace -2\kappa \cdot \frac {\prod\limits_{\textsf{odds}} \sigma(\frac u 2 \pm \frac v 2 \pm \frac w 2 \pm \frac x 2 + \omega_1)} {\sigma(u + 2\omega_1)\sigma(v)\sigma(w)\sigma(x)} \end{equation} By taking the product of \eqref{eq:Isigma} and \eqref{eq:I2} over the $0,\omega_1,\omega_2,\omega_3$ variants we get

Finally taking the product of the two formula in \eqref{eq:I3} gives \begin{equation} \label{eq:I4} \prod_{\textsf{all signs}} H(u,v,w,x,\pm 1,\pm 1, \pm 1) \enspace = \enspace \kappa^4 \cdot \frac {\prod\limits_{\textsf{all signs}} \sigma(u \pm v \pm w \pm x)} {\sigma^8(u)\sigma^8(v)\sigma^8(w)\sigma^8(x)} \end{equation}

Four Variable Summary

Using the convention that root extraction takes the root whose meromorphic expansion have residue $+1$ at zero, we have

\begin{equation*} \begin{aligned} \begin{vmatrix} 1 & e_1 & e_1^2 & \hphantom{+}h_1(u)\thinspace h_1(v)\thinspace h_1(w)\thinspace h_1(x) \\ 1 & e_2 & e_2^2 & \hphantom{+}h_2(u)\thinspace h_2(v)\thinspace h_2(w)\thinspace h_2(x) \\ 1 & e_3 & e_3^2 & \hphantom{+}h_3(u)\thinspace h_3(v)\thinspace h_3(w)\thinspace h_3(x) \\ 0 & 0 & 1 & 1 \\ \end{vmatrix} \enspace &= \enspace 2 \kappa \cdot \frac {\sigma(\frac u 2 + \frac v 2 + \frac w 2 + \frac x 2) \thinspace \sigma(\frac u 2 + \frac v 2 - \frac w 2 - \frac x 2) \thinspace \sigma(\frac u 2 - \frac v 2 + \frac w 2 - \frac x 2) \thinspace \sigma(\frac u 2 - \frac v 2 - \frac w 2 + \frac x 2)} {\sigma(u) \thinspace \sigma(v) \thinspace \sigma(w) \thinspace \sigma(x)} \\\\\\ \begin{vmatrix} 1 & e_1 & e_1^2 & -h_1(u)\thinspace h_1(v)\thinspace h_1(w)\thinspace h_1(x) \\ 1 & e_2 & e_2^2 & -h_2(u)\thinspace h_2(v)\thinspace h_2(w)\thinspace h_2(x) \\ 1 & e_3 & e_3^2 & -h_3(u)\thinspace h_3(v)\thinspace h_3(w)\thinspace h_3(x) \\ 0 & 0 & 1 & 1 \\ \end{vmatrix} \enspace &= \enspace 2 \kappa \cdot \frac {\sigma(-\frac u 2 + \frac v 2 + \frac w 2 + \frac x 2) \thinspace \sigma(\frac u 2 - \frac v 2 + \frac w 2 + \frac x 2) \thinspace \sigma(\frac u 2 + \frac v 2 - \frac w 2 + \frac x 2) \thinspace \sigma(\frac u 2 + \frac v 2 + \frac w 2 - \frac x 2) } {\sigma(u) \thinspace \sigma(v) \thinspace \sigma(w) \thinspace \sigma(x)} \end{aligned} \end{equation*}

References

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