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In this section we derive an addition formula for the Weiersrass $\sigma$ function.
We also derive several different variants for other sigma products and theta functions.
Sigma Addition Formula
The identity
\begin{equation*}
(e_1-e_2)(e_3-e_4) + (e_1-e_3)(e_4-e_2) + (e_1-e_4)(e_2-e_3) = 0
\end{equation*}
has already appeared in these notes. It implies
\begin{equation*}
[\wp(u)-\wp(v)]\thinspace[\wp(x)-\wp(y)]\space+\space[\wp(u)-\wp(x)]\thinspace[\wp(y)-\wp(v)]\space+\space[\wp(u)-\wp(y)]\thinspace[\wp(v)-\wp(x)] = 0
\end{equation*}
holds for all $u,v,x,y$. Substituting the formula
\begin{equation*}
\wp(u) - \wp(v) = -\frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)}
\end{equation*}
into triplep gives the addition formula for the Weierstrass $\sigma$ function
\begin{equation*}
\sigma(u - v)\sigma(u + v)\sigma(x - y)\sigma(x + y)\space+\space
\sigma(u - x)\sigma(u + x)\sigma(y - v)\sigma(y + v)\space+\space
\sigma(u - y)\sigma(u + y)\sigma(v - x)\sigma(v + x)\space=\space 0
\end{equation*}
Putting
\begin{equation*}
\sigma(x,y)=\sigma(x-y)\sigma(x+y)
\end{equation*}
allows triples to be written as
\begin{equation*}
\sigma(u,v)\sigma(x,y) \space + \space \sigma(u,x)\sigma(y,v) \space + \space \sigma(u,y)\sigma(v,x) \space = \space 0
\end{equation*}
In this form the identity holds in many other variations.
Period Lattices with 4 and 6-fold Rotational Symmetry
For the Weierstrass elliptic function with 3-fold symmetry on a period lattice with 6-fold symmetry we have
\begin{equation*}
\wp'(u) - \wp'(v) = 2\frac {\sigma(u-v)\sigma(u-\zeta^2 v)\sigma(u-\zeta^4 v)} {\sigma^3(u)\sigma^3(v)}
\end{equation*}
where $\zeta = e^{\pi i/3}$ is the sixth root of unity. When we put
\begin{equation*}
\sigma(x,y) = \sigma(x - y)\sigma(x - \zeta^2 y)\sigma(x + \zeta^2 y)
\end{equation*}
this also satisfies addition formula tripleg.
Similarly the elliptic function with 4-fold symmetry
\begin{equation*}
\wp''(u) - \wp''(v) = -6 \frac {\sigma(u-v)\sigma(u-iv)\sigma(u-i^2v)\sigma(u-i^3v)} {\sigma^4(u)\sigma^4(v)}
\end{equation*}
when
\begin{equation*}
\sigma(x,y) = \sigma(x - y)\sigma(x - iy)\sigma(x + y)\sigma(x + iy)
\end{equation*}
also satisfies addition formula tripleg.
And the elliptic function with 6-fold symmetry
\begin{equation*}
\wp''''(u) - \wp''''(v) = -120 \frac {\sigma(u-v)\sigma(u-\zeta v)\sigma(u-\zeta^2 v)\sigma(u-\zeta^3 v)\sigma(u-\zeta^4 v)\sigma(u-\zeta^5 v)} {\sigma^6(u)\sigma^6(v)}
\end{equation*}
with
\begin{equation*}
\sigma(x,y) = \sigma(x - y)\sigma(x - \zeta y)\sigma(x - \zeta^2 y)\sigma(x + y)\sigma(x + \zeta y)\sigma(x + \zeta^2 y)
\end{equation*}
also satisfies addition formula tripleg.
Theta Function Equivalent
Formula tripleg also holds for theta functions.
Define the theta function
\begin{equation*}
\theta(z) = \sum_{n=-\infty}^{\infty}(-1)^n z^n q^{\tfrac 1 2 n(n-1)}
\end{equation*}
where $|q| \lt 1$. Define
\begin{equation*}
\theta(x,y) = \frac 1 x \thinspace \theta(xy) \thinspace \theta(x/y)
\end{equation*}
then we have the following identity, which can also be proven using triple
\begin{equation*}
\theta(u,v)\theta(x,y) \space + \space \theta(u,x)\theta(y,v) \space + \space \theta(u,y)\theta(v,x) \space = \space 0
\end{equation*}
This simple algebraic proof of tripletheta makes use of identity triple.
Substituting theta into theta2 gives
\begin{equation*}
\theta(x,y) = \sum_{n_1,n_2 \in \Z} (-1)^{n_1+n_2} \space x^{n_1+n_2-1} \space y^{n_1-n_2} \space q^{\tfrac 1 2 n_1(n_1-1) + \tfrac 1 2 n_2(n_2-1)}
\end{equation*}
Making the change of variables $m_1=n_1+n_2-1,\space m_2=n_1-n_2$ gives
\begin{equation*}
\theta(x,y) = \sum_{\substack{m_1\not\equiv m_2 \mod 2 \\ m_1,m_2 \in \Z}} (-1)^{m_1+1} \space x^{m_1} \space y^{m_2} \space q^{\tfrac 1 4 (m_1^2 + m_2^2 - 1)}
= \sum_{m_1,m_2 \in \Z} -\tfrac 1 2\left[(-1)^{m_1} - (-1)^{m_2}\right] \space x^{m_1} \space y^{m_2} \space q^{\tfrac 1 4 (m_1^2 + m_2^2 - 1)}
\end{equation*}
and so
\begin{equation*}
\theta(u,v)\theta(x,y) = \sum_{m_1,m_2,m_3,m_4 \in \Z} \tfrac 1 4\left[(-1)^{m_1} - (-1)^{m_2}\right]\left[(-1)^{m_3} - (-1)^{m_4}\right]
\space u^{m_1} \space v^{m_2} \space x^{m_3} \space y^{m_4} \space q^{\tfrac 1 4 (m_1^2 + m_2^2 + m_3^2 + m_4^2 - 2)}
\end{equation*}
When this formula is substituted into tripletheta the terms in square brackets annihilate because of triple yielding the desired identity.
Theta Product Equivalent
Define the theta product
\begin{equation*}
\phi(x,y) = (1-xy)(\frac 1 x - \frac 1 y)\prod_{n=1}^{\infty}(1 - xy q^n)(1 - \frac 1 {xy} q^n)(1 - \frac x y q^n)(1 - \frac y x q^n)
\end{equation*}
then
\begin{equation*}
\phi(u,v)\phi(x,y) \space + \space \phi(u,x)\phi(y,v) \space + \space \phi(u,y)\phi(v,x) \space = \space 0
\end{equation*}
Substituting Jacobi's triple product identity
\begin{equation*}
\theta(z) = \sum_{n=-\infty}^{\infty}(-1)^n z^n q^{\tfrac 1 2 n(n-1)} = (1-z)\prod_{n=1}^{\infty}(1- z q^n)(1-\frac 1 z q^n)(1-q^n)
\end{equation*}
into theta2 we have
\begin{equation*}
\theta(x,y) = \phi(x,y) \prod_{n=1}^{\infty}(1 - q^n)^2
\end{equation*}
and the result then follows using tripletheta
Putting $x,y,u,v$ equal to various powers of $q$ in triplephi gives several well known theta product formulae as special cases.
EXAMPLE 1
Putting $q=q^2$ then $x=1,\space y=-1,\space u=q,\space v=-q$ gives
\begin{equation*}
16q\prod_{n=1}^{\infty}\left(1 + q^{2n}\right)^8 + \prod_{n=1}^{\infty}\left(1 - q^{2n-1}\right)^8 - \prod_{n=1}^{\infty}\left(1 + q^{2n-1}\right)^8= 0
\end{equation*}
EXAMPLE 2
Putting $a=x+1/x, \space b=y+1/y$ in thetaprod gives
\begin{equation*}
\phi(a,b) = (a-b)\prod_{n=1}^{\infty}\left[1 - ab q^n + (a^2+b^2-2) q^{2n} - ab q^{3n} + q^{4n}\right]
\end{equation*}
which also satisfies addition formula triplephi.
Connection to Pfaffians
There is a natural way to extend triple to $n$ variables although it doesn't really give any further identities.
Define the $n\times n$ skew-symmetric matrix $E$ by
\begin{equation*}
E_{ij} = e_i - e_j
\end{equation*}
then $E$ has rank 2 (because it is the difference of two rank 1 matrices) and therefore $\det(E)=0$ for $n > 2$, and by Cayley's theorem on
Pfaffians $\pfaffian(E)=0$.
For $n=4$ this is equation triple.
If we put $e_i=\wp(x_i)$ then apply wpsigma and multiply out the denominator we get
\begin{equation*}
\pfaffian\left(\sigma(x_i,x_j)\right) = 0
\end{equation*}
which is an addition formula in $n$ variables. For $n = 4$ this is equation tripleg.
Frobenius-Stickelberger Addition Formula
In their 1877 paper , Frobenius and Stickelberger start with the ingenious zeta-sigma determinant formula:
\begin{equation*}
\begin{vmatrix}
0 & 1 & 1 & \ldots & 1 \\
1 & \zeta(u_1+v_1) & \zeta(u_1+v_2) & \ldots & \zeta(u_1+v_n) \\
1 & \zeta(u_2+v_1) & \zeta(u_2+v_2) & \ldots & \zeta(u_2+v_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \zeta(u_n+v_1) & \zeta(u_n+v_2) & \ldots & \zeta(u_n+v_n) \\
\end{vmatrix}
\space = \space - \frac {\sigma\left(\sum\limits_{i=1}^n (u_i + v_i)\right) \space \prod\limits_{1 \le i \lt j \le n} \space \sigma(u_i - u_j) \sigma(v_i - v_j)}
{\prod\limits_{i=1}^n \prod\limits_{j=1}^n \sigma(u_i + v_j)}
\end{equation*}
The derivation of this formula goes something like this.
The LHS is an elliptic function in the $u_i$ because $\zeta(u_i + v_j + 2\omega_k) = \zeta(u_i + v_j) + 2\eta_k$ and the $2\eta_k$'s can be removed by an elementary row operation.
It is also an elliptic function in the $v_i$ for a similar reason.
It has $n-1$ obvious zeroes at $u_1 = u_2, \ldots u_n$ and $n$ obvious poles at $u_1 = -v_1, \ldots -v_n$ and hence a non-obvious zero at $u_1 = -(u_2 + \ldots u_n + v_1 + \ldots v_n)$
because for any elliptic function the sum of zeroes is congruent to the sum of poles modulo the period lattice.
The constant multiplier on the RHS needs to be determined.
For $n=1$ it is easily seen to be $-1$.
The following iterative process shows it is the same for every $n$ by reducing the formula to the $n-1$ case.
Put $v_n=0$ then multiply the last column on the LHS by $u_n$ and let $u_n \rightarrow 0$ so that all elements in the last column become 0 except the last which will be 1.
Then expanding the determinant by the last column reduces it to the $n - 1$ version.
On the RHS the $u_n + v_n$ term will disappear from the first $\sigma$ term in the numerator.
In the product, terms containing $u_n$ or $v_n$ reduce to $\prod_{i=1}^{n-1} \sigma(u_i)\sigma(v_i)$.
In the denominator there will be a similar set of terms which will cancel them out.
Also there will be one additional term in the denominator $\sigma(u_n)/u_n$ which will go to 1.
Then the whole formula has been reduced to the $n - 1$ version.
From this they obtain their classic addition formula
\begin{equation*}
\begin{vmatrix}
1 & \wp(u_1) & \wp'(u_1) & \ldots & \wp^{(n-2)}(u_1) \\
1 & \wp(u_2) & \wp'(u_2) & \ldots & \wp^{(n-2)}(u_2) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \wp(u_n) & \wp'(u_n) & \ldots & \wp^{(n-2)}(u_n) \\
\end{vmatrix}
\space = \space
(-1)^{n-1} \space \prod\limits_{i=1}^{n-1} i! \space \frac {\sigma\left(\sum\limits_{i=1}^n u_i\right) \space
\prod\limits_{1 \le i \lt j \le n} \sigma(u_j-u_i)} {\prod\limits_{i=1}^n \sigma(u_i)^n}
\end{equation*}
essentially by taking the limit of fsz as $v_i \rightarrow 0$.
They also obtain the Kiepert formula
\begin{equation*}
\begin{vmatrix}
\wp'(u) & \wp''(u) & \ldots & \wp^{(n-1)}(u) \\
\wp''(u) & \wp'''(u) & \ldots & \wp^{(n)}(u) \\
\vdots & \vdots & \ddots & \vdots \\
\wp^{(n-1)}(u) & \wp^{(n)}(u) & \ldots & \wp^{(2n-3)}(u) \\
\end{vmatrix}
\space = \space
(-1)^{n-1} \space \left(\prod\limits_{i=1}^{n-1} i!\right)^2 \space \frac {\sigma(n u)} {\sigma(u)^{n^2}}
\end{equation*}
by taking the limit of fsn as $u_i \rightarrow u$.
COROLLARY
Substituting fsn with $n=2$ and fslimit with $n=2$
\begin{equation*}
\wp(u)-\wp(v) \space = \space -\frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)} \qquad\qquad\qquad
\wp'(u) \space = \space - \frac {\sigma(2u)} {\sigma^4(u)}
\end{equation*}
into fsn with $n=3$
\begin{equation*}
2 \frac {\sigma(u+v+w)\sigma(v-u)\sigma(w-u)\sigma(w-v)} {\sigma^3(u)\sigma^3(v)\sigma^3(w)} \space = \space
\left[\wp(w)-\wp(v)\right]\wp'(u) \space + \space
\left[\wp(u)-\wp(w)\right]\wp'(v) \space + \space
\left[\wp(v)-\wp(u)\right]\wp'(w)
\end{equation*}
gives a three variable addition formula for the Weierstrass sigma function
\begin{equation*}
\sigma(u+v+w) \space = \space \frac 1 2 \left[ \frac {\sigma(v)\sigma(w)\sigma(v+w)} {\sigma(u)\sigma(u-v)\sigma(u-w)} \sigma(2u) \space + \space
\space \frac {\sigma(u)\sigma(w)\sigma(u+w)} {\sigma(v)\sigma(v-u)\sigma(v-w)} \sigma(2v) \space + \space
\space \frac {\sigma(u)\sigma(v)\sigma(u+v)} {\sigma(w)\sigma(w-u)\sigma(w-v)} \sigma(2w) \right]
\end{equation*}
Substituting into fsn with $n=4$
\begin{equation*}
-12 \frac {\sigma(u+v+w+x)\sigma(v-u)\sigma(w-u)\sigma(x-u)\sigma(w-v)\sigma(x-v)\sigma(x-w)} {\sigma^4(u)\sigma^4(v)\sigma^4(w)\sigma^4(x)} \space = \space
6\left[\wp(w)-\wp(v)\right]\left[\wp(x)-\wp(v)\right]\left[\wp(x)-\wp(w)\right]\wp'(u) \space + \space \ldots
\end{equation*}
gives a four variable addition formula for the Weierstrass sigma function
\begin{equation*}
\sigma(u+v+w+x) \space = \space \frac 1 2 \left[ \frac {\sigma(v+x)\sigma(w+x)\sigma(v+w)} {\sigma(u-x)\sigma(u-v)\sigma(u-w)} \sigma(2u) \space + \space
\space \frac {\sigma(u+x)\sigma(w+x)\sigma(u+w)} {\sigma(v-x)\sigma(v-u)\sigma(v-w)} \sigma(2v) \space + \space
\space \frac {\sigma(u+x)\sigma(v+x)\sigma(u+v)} {\sigma(w-x)\sigma(w-u)\sigma(w-v)} \sigma(2w) \space + \space
\space \frac {\sigma(u+v)\sigma(u+w)\sigma(v+w)} {\sigma(x-u)\sigma(x-v)\sigma(x-w)} \sigma(2x) \right]
\end{equation*}
Using the $\zeta$ addition formula we can also write fsz in terms of $\wp$ as follows
\begin{equation*} (\tfrac 1 2)^{n-1}
\begin{vmatrix}
0 & 1 & 1 & \ldots & 1 \\
1 & \frac {\wp'(u_1) - \wp'(v_1)} {\wp(u_1) - \wp(v_1)} & \frac {\wp'(u_1) - \wp'(v_2)} {\wp(u_1) - \wp(v_2)} & \ldots & \frac {\wp'(u_1) - \wp'(v_n)} {\wp(u_1) - \wp(v_n)} \\
1 & \frac {\wp'(u_2) - \wp'(v_1)} {\wp(u_2) - \wp(v_1)} & \frac {\wp'(u_2) - \wp'(v_2)} {\wp(u_2) - \wp(v_2)} & \ldots & \frac {\wp'(u_2) - \wp'(v_n)} {\wp(u_2) - \wp(v_n)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \frac {\wp'(u_n) - \wp'(v_1)} {\wp(u_n) - \wp(v_1)} & \frac {\wp'(u_n) - \wp'(v_2)} {\wp(u_n) - \wp(v_2)} & \ldots & \frac {\wp'(u_n) - \wp'(v_n)} {\wp(u_n) - \wp(v_n)} \\
\end{vmatrix}
\space = \space - \frac {\sigma\left(\sum\limits_{i=1}^n (u_i + v_i)\right) \space \prod\limits_{1 \le i \lt j \le n} \space \sigma(u_i - u_j) \sigma(v_i - v_j)}
{\prod\limits_{i=1}^n \prod\limits_{j=1}^n \sigma(u_i + v_j)}
\end{equation*}
Extended Frobenius-Stickelberger Formula
The classic Frobenius-Stickelberger formula fsn can be extended to general elliptic functions.
The set of elliptic functions with poles of order at most $n_j \gt 0$ at $a_j$ for $j=1\dots k$ form a vector space of dimension $n = \sum n_j$.
If $f_1 \dots f_n$ is a basis for that vector space then
\begin{equation*}
\begin{vmatrix}
f_1(u_1) & f_2(u_1) & \ldots & f_n(u_1) \\
f_1(u_2) & f_2(u_2) & \ldots & f_n(u_2) \\
\vdots & \vdots & \ddots & \vdots \\
f_1(u_n) & f_2(u_n) & \ldots & f_n(u_n) \\
\end{vmatrix}
\space = \space K \space \frac {\sigma\left(\sum\limits_{i=1}^n u_i - \sum\limits_{j=1}^k n_j a_j\right) \space
\prod\limits_{1 \le i \lt j \le n} \sigma(u_j - u_i)} {\prod\limits_{i=1}^n \prod\limits_{j=1}^k \sigma(u_i - a_j)^{n_j}}
\end{equation*}
This formula follows from the fact that the LHS of fsx is an order $n$ elliptic function of $u_1$ with $n-1$ obvious zeroes
at $u_1 = u_2, \ldots, u_n$ and obvious poles at $u_1 = a_j$ of order $n_j$ (for almost all values of $u_i$),
and a non-obvious zero at $u_1 = (n_1 a_1 + \dots + n_k a_k)-(u_2 + \ldots + u_n)$.
This last zero follows from the fact that for any elliptic function the sum of zeroes equals the sum of poles (modulo the period lattice).
where $K$ is a constant independent of the $u_i$.
An explicit formula for $K$ can be obtained as follows.
Let $R$ be the matrix of the coefficients of the negative powers in the Laurent expansions of $f_i$ at $a_j$.
Each column of $R$ corresponds to one of the $f_i$.
Each row of $R$ corresponds to the coefficients of $(z-a_j)^l$ in the Laurent expansions of $f_i(z)$ at one particular point $a_j$ and for one particular $l \lt 0$.
The order of the rows is taken as, first the coefficients of $(z-a_1)^{-n_1} \ldots (z-a_1)^{-1}$ at $a_1$, then of $(z-a_2)^{-n_2} \ldots (z-a_2)^{-1}$ at $a_2$ etc.
Under this ordering we have
\begin{equation*}
K \space = \space
\begin{vmatrix}
R_{11} & R_{12} & \ldots & R_{1n} \\
R_{21} & R_{22} & \ldots & R_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
R_{n-1,1} & R_{n-1,2} & \ldots & R_{n-1,n} \\
f_1(u) & f_2(u) & \ldots & f_n(u) \\
\end{vmatrix}
\quad \prod\limits_{1 \le i \lt j \le k} \sigma(a_i - a_j)^{n_i n_j}
\end{equation*}
Let $D$ be the determinant on the LHS of fsx. The following process can be used to convert $D$ to $R$ one row at a time:
- Multiply the first row of $D$, and the RHS, by $(u_1 - a_1)^{n_1}$ and then let $u_1 \rightarrow a_1$.
- On the LHS this will convert the first row of $D$ to the first row of $R$.
- On the RHS it will replace all $u_1$ by $a_1$ except the terms $\sigma(u_1-a_1)^{n_1}$ in the denominator which will go to 1.
- Subtract the first row, multiplied by $(u_2 - a_1)^{-n_1}$, from the second. This will remove any powers of $(u_2-a_1)^{-n_1}$ from the Laurent expansion of $f_i(u_2)$ at $a_1$.
- Multiply the second row of $D$ and the RHS by $(u_2 - a_1)^{n_1 - 1}$ and then let $u_2 \rightarrow a_1$.
- On the LHS this will convert the second row of $D$ to the second row of $R$.
- On the RHS it will cancel out the term $\sigma(u_2 - a_1)$, which appeared in the numerator due to the previous row processing, and $n_1$ similar terms in the denominator.
- Repeat the process another $n_1 - 2$ times with $u_i \rightarrow a_1$ converting the first $n_1$ rows of $D$ to $R$.
- Repeat the process $n_2$ times with $u_i \rightarrow a_2$ converting the next $n_2$ rows of $D$ to $R$.
- Repeat for the other $a_i$ up to the second to last row of $D$, converting $n - 1$ rows of $D$ to $R$.
- After processing the second to last row, the numerator on the RHS will contain $\sigma(u_n - a_k) \prod_{j=1}^{n - 1} \sigma(u_n - A(u_j))$ where
the function $A$ maps each $u_i$ to its corresponding $a_j$.
The denominator will contain $\prod_{j=1}^k\sigma(u_n - a_j)^{n_j}$.
These terms will cancel each other out, leaving no $u_n$ terms on the RHS!
- Replace the $u_n$ in the last row of $D$ by $u$ and we are done, except we need to tally up all the terms on the RHS like $\sigma(a_j-a_i)$.
- From the numerator there will be $\prod_{i \lt j}\sigma(a_j - a_i)^{n_i n_j}$.
From the denominator there will be a similar term and another with the indices reversed ie. $\prod_{i \lt j}\sigma(a_i - a_j)^{n_i n_j}$.
The first two products cancel leaving just the term with reversed indices in the denominator.
As a sanity check we see that if we swap $a_1$ and $a_2$ then, to restore to the original formula, we need to make $n_i n_j$ row swaps in the determinant and $n_i n_j$ sign changes in the
product.
As expected these sign changes will cancel each other out.
Similar if we swap $a_{k-1}$ and $a_k$ but this time we also need to manipulate the row of residues for $a_{k-1}$ to turn them into the residues for $a_k$.
As a further check we see that if we if we make a linear change of basis in fsx both sides of the equation will be multiplied by the same determinant.
Note $\rank(R) = n - 1$ because the sum of the $k$ rows corresponding to the residues of $f_i$ is a row of zeroes.
Due to this the last row of residues of $f_i$ at $a_k$, in the formula for $K$, ends up being replaced by the basis functions evaluated at a completely arbitrary point $u$.
This may seem surprising but all it is really saying is that a certain linear combination of the basis functions $f_i(u)$ is constant.
We may take $u$ to be one of the poles $a_j$ in which case the last row will consist of the coefficients of the $(z-a_j)^0$ terms in the Laurent expansion of $f_i$ at $a_j$.
Nomenclature
I have called fsx the Extended Frobenius-Stickelberger formula because I need a name for it and don't want to confuse it with the original
Frobenius-Stickelberger formula fsn.
It is more or less equivalent to formula fsz and is a generalisation of formula fsn.
Note there are already Generalised Frobenius-Stickelberger formulae in the literature but these refer to the generalisation of fsn to hyperelliptic $\sigma$ functions.
For example see .
Formula fsx is also closely related to the Vandermonde determinant formula.
COROLLARY
For the basis $2n$-dimensional basis $1,\wp, \dots \wp^n, \wp', \ldots \wp'\wp^{n-2}$ the extended Frobenius-Stickelberger gives, after applying the
sub-determinant expansion formula to the left $n+1$ columns
\begin{equation*}
\sum\limits_{P\in\mathfrak{P}} \left( (-1)^{\order(P)} \prod_{\substack{i \gt j \\ i,j\in P}} \Big(\wp(u_i) - \wp(u_j)\Big)
\prod_{\substack{k \gt l \\ k,l\in P'}} \Big(\wp(u_k) - \wp(u_l)\Big) \prod\limits_{m \in P'} \wp'(2u_m) \right)
\space = \space 2^{n-1} \frac {\sigma\left(u_1 + u_2 + \ldots u_{2n}\right) \prod\limits_{i \gt j} \sigma(u_i - u_j)} {\prod\limits_{i=1}^{2n} \sigma(u_i)^{2n}}
\end{equation*}
where $P$ ranges over all subsets $\mathfrak{P}$ of size $n+1$ choosen from the set of integers $1 \ldots 2n$.
$P'$ is the complement of $P$ and $\order(P)$ is the number of transpositions required to shuffle the elements of $P$ to the left.
From this the addition formula for the $\sigma$ function can be deduced
\begin{equation*}
\sigma\left(u_1 + u_2 + \ldots u_{2n}\right) \space = \space \frac 1 {2^{n-1}}
\sum\limits_{P\in\mathfrak{P}} \left( \frac {\prod\limits_{\substack{i \lt j \\ i,j\in P}} \sigma(u_i + u_j) \prod\limits_{\substack{k \lt l \\ k,l\in P'}} \sigma(u_k + u_l)}
{\prod\limits_{i\in P \space j\in P'} \sigma(u_i - u_j)} \prod\limits_{m \in P'} \sigma(2u_m) \right)
\end{equation*}
using
\begin{equation*}
\prod\limits_{i \gt j} \sigma(u_i - u_j) \space = \space (-1)^{\order(P)}
\prod\limits_{\substack{i \lt j \\ i,j\in P}} \sigma(u_i - u_j) \prod\limits_{\substack{k \lt l \\ k,l\in P'}} \sigma(u_k - u_l)
\prod\limits_{i\in P \space j\in P'} \sigma(u_i - u_j)
\end{equation*}
Let $u_{2n-1}, u_{2n} \rightarrow 0$.
When they are both in $P$ the summand is zero.
When they are both in $P'$ the summand is zero.
When one is in $P$ and the other is in $P'$ the summands combine to a formula of the form (after judiciously setting a few $u_{2n-1}$ and $u_{2n}$ to zero)
\begin{equation*}
RHS \space = \space \frac 1 2 \sigma(u_1+\ldots u_{2n-2}) \frac {\sigma(2u_{2n-1}) - \sigma(2u_{2n})} {\sigma(u_{2n-1} - u_{2n})} \longrightarrow \sigma(u_1+\ldots u_{2n-2})
\end{equation*}
verifying the value of the leading coefficient.
For example for $n=6$
\begin{equation*}
\sigma(u_1+u_2+u_3+u_4+u_5+u_6) \space = \space \frac 1 4 \frac {\sigma(u_1+u_2)\sigma(u_3+u_4)\sigma(u_3+u_5)\sigma(u_3+u_6)\sigma(u_4+u_5)\sigma(u_4+u_6)\sigma(u_5+u_6)}
{\sigma(u_1-u_3)\sigma(u_2-u_3)\sigma(u_1-u_4)\sigma(u_2-u_4)\sigma(u_1-u_5)\sigma(u_2-u_5)\sigma(u_1-u_6)\sigma(u_2-u_6)} \sigma(2u_1)\sigma(2u_2) \quad + \quad \textsf{14 more terms}
\end{equation*}
Now put $u_1=u_3=0$ to get
\begin{equation*}
\frac 1 4 \frac {\sigma(u_2)\sigma(u_4)\sigma(u_5)\sigma(u_6)\sigma(u_4+u_5)\sigma(u_4+u_6)\sigma(u_5+u_6)}
{\sigma(u_2)\sigma(-u_4)\sigma(u_2-u_4)\sigma(-u_5)\sigma(u_2-u_5)\sigma(-u_6)\sigma(u_2-u_6)} \frac {\sigma(2u_1)\sigma(2u_2)} {\sigma(u_1-u_3)} \ldots
\end{equation*}
EXAMPLE
The classic Frobenius-Stickelberger formula fsn can be obtained from fsx by letting $f_i = 1, \wp, \wp', \dots ,\wp^{[n-2]}$.
Then $f_i$ is a basis of the vector space of elliptic functions with a single pole of order at most $n$ at $0$.
Applying formula fsxK with $u=0$ gives the determinant of an anti-diagonal
matrix
\begin{equation*}
K = \det\left(\antidiagonal((-1)^{n}(n-1)!\thinspace, \ldots -4!\thinspace, 3!\thinspace, -2!\thinspace, 1!\thinspace, 1)\right) =
(-1)^{\tfrac 1 2 n(n-1)} \space (-1)^{\tfrac 1 2 (n-1)(n-2)} \space 1! \ldots (n-1)! = (-1)^{n-1} \prod_{i=1}^{n-1} i!
\end{equation*}
and the rest of the formula follows upon putting $a_i=0$.
Extended Kiepert Formula
Letting $u_i \rightarrow u$ in fsx yields an extension of the Kiepert formula:
\begin{equation*}
\begin{vmatrix}
f_1(u) & f_2(u) & \ldots & f_n(u) \\
f_1'(u) & f_2'(u) & \ldots & f_n'(u) \\
\vdots & \vdots & \ddots & \vdots \\
f_1^{(n-1)}(u) & f_2^{(n-1)}(u) & \ldots & f_n^{(n-1)}(u) \\
\end{vmatrix}
\space = \space K \space \prod\limits_{i=1}^{n-1} {i!} \space \frac {\sigma\left(n u - \sum\limits_{j=1}^k n_j a_j\right)} {\left(\prod\limits_{j=1}^k \sigma(u - a_j)^{n_j}\right)^n}
\end{equation*}
In more detail in formula fsx
- Replace the $f_i(u_2)$'s in the second row by their taylor expansions at $u_1$ ie. $f_i(u_2) = f_i(u_1) + (u_2-u_1)f_i'(u_1) + \frac 1 {2!} (u_2-u_1)^2f_i''(u_1)+\ldots$
- Remove the leading $f_i(u_1)$ terms by subtracting the first row from the second.
- Divide by $(u_2 - u_1)$ and let $u_2 \rightarrow u_1$. This reduces the second row to $f_i'(u_1)$ and cancels out the $\sigma(u_2-u_1)$ term on the right hand side.
- Repeat the process for the third row but this time remove the first two terms by subtracting suitable multiples of the first two rows from the third.
- Divide by $(u_3-u_1)^2$ and let $u_3 \rightarrow u_1$. This reduces the third row to $\frac 1 {2!} f_i''(u_1)$ and cancels out the two $\sigma(u_3-u_1)$ terms that are now on the right
hand side.
- Repeat a similar process for the rest of the rows and finally factor out the factorial constants and replace $u_1$ by $u$.
where $K$ is the constant given in fsxK.
References