Jacobian Elliptic Functions

Near the start of the nineteenth century Jacobi had the extremely fruitful idea of constructing quasi-periodic theta functions, and from their ratio, the elliptic functions which now bear his name. In the middle of the century Riemann used the same idea, extended to $n$ dimensions, to solve the Jacobi inversion problem for general Abelian integrals. Finally at the end of the century Poincaré generalised the idea in a slightly different way to construct what he called theta-fuchsian functions^[1] and used them to solve the uniformisation problem for algebraic curves in two variables.

The subject of Jacobian elliptic functions is treated exhaustively in several online works, see for example Wikipedia and NIST whose notation I am following here.

Like all elliptic functions, the Jacobian elliptic functions are completely defined by their periods and principal parts at the poles (and one constant term). With respect to the period lattice $\lattice{4K,4iK'}$ all three Jacobian elliptic functions are order 4 elliptic functions with four simple poles at $iK', -iK', 2K + iK', -(2K + iK')$. For convenience let's label these poles $\rho_1,\rho_2,\rho_3,\rho_4$ and tabulate the principal parts.

The principal parts at the other poles can be obtained from the equivalence classes of the poles and the fact that the residues of elliptic functions always sum to zero. For example $\sn$ has the same principal part at $\rho_2$ as $\rho_1$ and the opposite sign on the residues at $\rho_3$ and $\rho_4$.

There are three different ways of halving a period lattice $\lattice{\omega_1,\omega_2}$ - the three lattices $\lattice{\omega_1, \sfrac 1 2 \omega_2}$, $\lattice{\omega_1, \sfrac 1 2 (\omega_1+\omega_2)}$ and $\lattice{\sfrac 1 2 \omega_1,\omega_2}$. If we put $\omega_1=4K$ and $\omega_2=4iK'$ and look at the table above, we can see that each of the three Jacobian elliptic functions $\sn,\cn,\dn$ is an order 2 elliptic function, with respect to one of these lattices. This is the main reason why it is natural to have three Jacobian elliptic functions.

Table 1
Function	Fundamental Periods	Principal Part At $\rho_1$	Equivalence Classes Of Poles Modulo The Fundamental Periods
$\sn u$	$4K,2iK'$	$\displaystyle \frac 1 k (u-\rho_1)^{-1} \space + \space \bigO(u - \rho_1)$	$\{\rho_1,\rho_2\}\quad\{\rho_3,\rho_4\}$	two single poles
$\cn u$	$4K,2K+2iK'$	$\displaystyle -\frac i k (u-\rho_1)^{-1} \space + \space \bigO(u - \rho_1)$	$\{\rho_1,\rho_4\}\quad\{\rho_2,\rho_3\}$	two single poles
$\dn u$	$2K,4iK'$	$\displaystyle -i \thinspace (u-\rho_1)^{-1} \space + \space \bigO(u - \rho_1)$	$\{\rho_1,\rho_3\}\quad\{\rho_2,\rho_4\}$	two single poles
$\sn^2 u$	$2K,2iK'$	$\displaystyle \frac 1 {k^2} (u-\rho_1)^{-2} \space + \space \bigO(1)$	$\{\rho_1,\rho_2,\rho_3,\rho_4\}$	one double pole

There is a generic way of constructing the Jacobi elliptic functions from an arbitrary order 2 elliptic function. Every such function $f$ has four double points $e_1,e_2,e_3,e_4$. If a Möbius transformation is applied to $f$ to move one of those points to zero and another to infinity the resulting function is a perfect square, and its square root $g$ is another order 2 elliptic function (with respect to a sublattice). \begin{equation} g_{ij}(z) \space = \space\sqrt{ \frac {f(z) - e_i} {f(z) - e_j} } \end{equation} There are twelve different ways to do this and the resulting functions are, more or less, the twelve Jacobian elliptic functions in Glaisher's $\operatorname{pq}$ notation. The Glaisher identity \begin{equation} \operatorname{pq}(z) \space = \space \frac {\operatorname{pr}(z)} {\operatorname{qr}(z)} \space = \space \frac 1 {\operatorname{qp}(z)} \end{equation} is simply the identity \begin{equation} g_{ij}(z) \space = \space \frac {g_{ik}(z)} {g_{jk}(z)} \space = \space \frac 1 {g_{ji}(z)} \end{equation} To get the definitions to line up exactly requires a bit of fiddling around. More precisely, let $f$ be an order 2 elliptic function with half periods $\omega_1,\omega_2$ and with its $z$-argument scaled and translated so that it satisfies the differential equation \begin{equation} {f'}^2 \space = \space \frac {4 (f-e_1)(f-e_2)(f-e_3)(f-e_4)} {(e_1-e_2)(e_3-e_4)}\qquad\text{with}\quad f(0) = e_4,\space f(\omega_1) = e_1,\space f(\omega_2) = e_2,\space f(\omega_1+\omega_2) = e_3 \end{equation} Define \begin{equation} \label{eq:generic} \operatorname{s}(z) = \sqrt{\frac {f(z) - e_4} {f(z) - e_2} \cdot \frac {e_2 - e_1} {e_4 - e_1}},\qquad \operatorname{c}(z) = \sqrt{\frac {f(z) - e_1} {f(z) - e_2} \cdot \frac {e_2 - e_4} {e_1 - e_4}},\qquad \operatorname{d}(z) = \sqrt{\frac {f(z) - e_3} {f(z) - e_2} \cdot \frac {e_2 - e_4} {e_3 - e_4}},\qquad k = \sqrt{\frac {e_3 - e_2} {e_3 - e_4} \cdot \frac {e_4 - e_1} {e_2 - e_1}} \end{equation} then it is easily verified that \begin{equation} \operatorname{s}(0) \space = \space 0,\qquad {\operatorname{s}(z)}^2 \space + \space {\operatorname{c}(z)}^2 \space = \space 1,\qquad k^2 {\operatorname{s}(z)}^2 \space + \space {\operatorname{d}(z)}^2 \space = \space 1,\qquad {\operatorname{s}'(z)}^2 \space = \space {\operatorname{c}(z)}^2 {\operatorname{d}(z)}^2 \end{equation} and therefore the three functions $\operatorname{s},\operatorname{c},\operatorname{d}$ are exactly, provided you choose the correct square roots, the three Jacobian elliptic functions $\sn,\cn,\dn$ with modulus $k$. And the half periods $\omega_1,\omega_2$ are the Jacobian quarter periods $K,iK'$.

Using the basic identities $\sn^2 u + \cn^2 u = 1$ and $k^2 \sn^2 u + \dn^2 u = 1$ it can easily be verified that the Jacobian elliptic functions are uniformising elliptic functions for the following curves:

Table 2
Curve	$x(u)$	$y(u)$	$g_2$	$g_3$	$\Delta$
$y^2 = (1 - x^2)(1 - k^2x^2)$	$\sn u$	$\cn u\dn u$	$\frac {1} {12} (k^4 + 14k^2 + 1)$	$\frac {1} {216} (k^2 + 1)(k^4 - 34k^2 + 1)$	$\frac {1} {16} k^2(k^2 - 1)^4$
$y^2 = (1 - x^2)(1 - k^2 + k^2x^2)$	$\cn u$	$-\sn u\dn u$	$\frac {1} {12} (16k^4 - 16k^2 + 1)$	$\frac {1} {216} (2k^2 - 1)(32k^4 - 32k^2 - 1)$	$\frac {1} {16} k^2(k^2 - 1)$
$y^2 = (1 - x^2)(k^2 - 1 + x^2)$	$\dn u$	$-k^2\sn u\cn u$	$\frac {1} {12} (k^4 - 16k^2 + 16)$	$\frac {1} {216} (k^2 - 2)(k^4 + 32k^2 - 32)$	$-\frac {1} {16} k^8(k^2 - 1)$
$y^2 = x(1 - x)(1 - k^2x)$	$\sn^2 u$	$\sn u\cn u\dn u$	$\frac {1} {12} (k^4 - k^2 + 1)$	$\frac {1} {432} (k^2 - 2)(2k^2 - 1)(k^2 + 1)$	$\frac {1} {256} k^4(k^2 - 1)^2$

As can be seen from the table each of these curves has a different $j$-invariant, reflecting the fact that the period ratio's of the uniformising functions are different. If we let $\tau = iK' / K$ then, from the first and last line of both Table 1 and Table 2, we have \begin{gather*} j(\tfrac 1 2 \tau) \space = \space 16 \frac {(k^4 + 14k^2 + 1)^3} {k^2(k^2 - 1)^4} \space = \space 16 \frac {(\lambda^2 + 14\lambda + 1)^3} {\lambda (1 - \lambda)^4}\\\\ j(\tau) \space = \space 256 \frac {(k^4 - k^2 + 1)^3} {k^4(k^2 - 1)^2} \space = \space 256 \frac {(\lambda^2 - \lambda + 1)^3} {\lambda^2 (1 -\lambda)^2} \end{gather*} where $\lambda = k^2$ and we have a rational parameterisation of the level 2 modular equation $X(j(\tau),j(2\tau)) = 0$.

From Table 1 we can see that the three Jacobian elliptic functions, along with the constant function, form a basis for the vector space of all elliptic functions with periods $4K, 4iK'$ and a pole of order at most one at $\rho_1,\rho_2,\rho_3,\rho_4$. Therefore we can apply the Extended Frobenius-Stickelberger formula to obtain

\begin{equation} \label{eq:add4x} \begin{vmatrix} \sn(u_1) & \cn(u_1) & \dn(u_1) & 1 \\ \sn(u_2) & \cn(u_2) & \dn(u_2) & 1 \\ \sn(u_3) & \cn(u_3) & \dn(u_3) & 1 \\ \sn(u_4) & \cn(u_4) & \dn(u_4) & 1 \\ \end{vmatrix} \space = \space \frac 4 {k^2} \space \frac {\sigma(u_1+u_2+u_3+u_4) \space \prod\limits_{i \lt j} \sigma(u_j - u_i) \prod\limits_{i \lt j} \sigma(\rho_j - \rho_i)} {\prod\limits_{i,j} \sigma(u_i - \rho_j)} \end{equation}

In the first $\sigma$ term we use the fact $\rho_1+\rho_2+\rho_3+\rho_4=0$. The only real calculation required is the leading constant $C$. From the power series expansions at the poles we get \begin{equation*} C \space = \space \frac 1 k \cdot \frac {-i} k \cdot {-i} \cdot \begin{vmatrix} 1 & 1 & 1 & 0 \\ 1 & -1 & -1 & 0 \\ -1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{vmatrix} \space = \frac 4 {k^2} \end{equation*}

Putting $u_1+u_2+u_3+u_4 = 0$ in \eqref{eq:add4x} gives the level 1 addition formula which may be written as \begin{equation} \label{eq:add4} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \\ \end{vmatrix} \space = \space 0 \end{equation} In this form the addition formula has a geometric interpretation in 3 dimensions, similar to that for cubic curves in 2 dimensions. Consider three points $(x_1,y_1,z_1),\space(x_2,y_2,z_2),\space(x_3,y_3,z_3)$ on the 3-D space curve given implicitly by \begin{equation} \label{eq:curve} x^2 + y^2 = 1,\quad k^2x^2 + z^2 = 1 \end{equation} If you run a plane through those three points, it will intersect the curve at a fourth point. That fourth point is given by solving \eqref{eq:add4} and \eqref{eq:curve} for $(x_4,y_4,z_4)$. It's not obvious but the coordinates of this point will be rational functions of the coordinates of the other three points and $k^2$. These are the level 2 addition formulae which we will now derive.

\begin{equation} \label{eq:addx4} x_4 \space = \space \frac 1 {x_1 x_2 x_3} \frac {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 0 & 1 & 1 & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 0 & -1 & 1 & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 0 & 1 & -1 & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 0 & -1 & -1 & 1 \\ \end{vmatrix}} {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & -ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & -ik & 0 \\ \end{vmatrix}} \space = \space \frac {E_2^4 + E_3^4 + E_4^4 \space - \space 2E_2^2E_3^2 - 2E_2^2E_4^2 - 2E_3^2E_4^2} {x_1x_2x_3\Big(E_1^4 + E_2^4 + k^4 E_3^4 \space + \space 2 E_1^2E_2^2 + 2 k^2E_1^2E_3^2 - 2k^2 E_2^2E_3^2\Big)} \end{equation}

This follows from the formula \begin{equation} \prod_\text{all signs} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & \pm y_4 & \pm z_4 & 1 \\ \end{vmatrix} \space = \space (x_1-x_4)(x_2-x_4)(x_3-x_4)\Big(N(x_1,\ldots z_3) \space - \space D(x_1,\ldots z_3) x_4\Big) + Z(x_1, \ldots z_4) \end{equation} where $Z$ is a polynomial which vanishes on \eqref{eq:add4} and \eqref{eq:curve}, and $N$ and $D$ are polynomials which do not depend on $x_4,y_4,z_4$.

While \eqref{eq:addx4} is a neat and concise formula, it is also in a certain sense, a very bad formula. The numerator and denominator share many common factors modulo the relations \eqref{eq:add4} and \eqref{eq:curve}. Eliminating them systematically requires a very large algebraic computation more or less equivalent to computing the level 4 addition formula.

To put it another way \eqref{eq:addx4} should reduce to the standard addition formula for $\sn u$ namely \begin{equation} \label{eq:addsn} x_4 \space = \space \frac {x_1 y_2 z_2 \space - \space x_2 y_1 z_1} {k^2 x_1^2 x_2^2 \space - \space 1} \end{equation} under the limiting substitution $(x_3,y_3,z_3) \rightarrow (0,1,1)$. To get to \eqref{eq:addsn} requires "eliminating the surds" $y = \sqrt{1 - x^2}$ and $z = \sqrt{1 - k^2x^2}$ from the denominator, followed by normalising the numerator modulo \eqref{eq:add4} and \eqref{eq:curve}, and finally elimination of many common factors.

\begin{equation*} y_4 \space = \space \frac 1 {y_1 y_2 y_3} \frac {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & 0 & k' & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ -1 & 0 & k' & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & 0 & -k' & 1 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ -1 & 0 & -k' & 1 \\ \end{vmatrix}} {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & -ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & -ik & 0 \\ \end{vmatrix}} \space = \space \frac {E_1^4 + {k'}^4 E_3^4 + E_4^4 \space - \space 2{k'}^2 E_1^2E_3^2 - 2 E_1^2E_4^2 - 2{k'}^2 E_3^2E_4^2} {y_1y_2y_3\Big(E_1^4 + E_2^4 + k^4 E_3^4 \space + \space 2 E_1^2E_2^2 + 2 k^2E_1^2E_3^2 - 2k^2 E_2^2E_3^2\Big)} \end{equation*} where $k^2 + {k'}^2 = 1$ \begin{equation*} z_4 \space = \space \frac 1 {z_1 z_2 z_3} \frac {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & k' & 0 & k \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ -1 & k' & 0 & k \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -k' & 0 & k \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ -1 & -k' & 0 & k \\ \end{vmatrix}} {\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & i & -ik & 0 \\ \end{vmatrix} \space \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ 1 & -i & -ik & 0 \\ \end{vmatrix}} \space = \space \frac {E_1^4 + {k'}^4 E_2^4 + k^4 E_4^4 \space - \space 2{k'}^2 E_1^2E_2^2 - 2k^2 E_1^2E_4^2 - 2k^2{k'}^2E_2^2E_4^2} {z_1z_2z_3\Big(E_1^4 + E_2^4 + k^4 E_3^4 \space + \space 2 E_1^2E_2^2 + 2 k^2E_1^2E_3^2 - 2k^2 E_2^2E_3^2\Big)} \end{equation*}

While \eqref{eq:addx4} is a neat and concise formula, it is also in a certain sense, a very bad formula. The numerator and denominator share many common factors modulo the relations \eqref{eq:curve} and eliminating them requires a massive algebraic computation more or less equivalent to computing the level 4 addition formula. \begin{equation*} \prod_\text{all signs} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & \pm y_2 & \pm z_2 & 1 \\ x_3 & \pm y_3 & \pm z_3 & 1 \\ x_4 & \pm y_4 & \pm z_4 & 1 \\ \end{vmatrix} \space = \space k^{16} \thinspace {k'}^{16} \thinspace \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} \quad\ldots \end{equation*}

If you look at this table of the twelve Jacobian elliptic functions you will see that $\sn, \ns, \cd, \dc$ all have the same fundamental periods.