Elliptic Functions Of Order 2

In this section we derive formulae for order 2 elliptic functions. All of the results are deduced from standard properties of general elliptic functions.

In this section $f$ and $g$ are order 2 elliptic functions with respect to some fixed lattice $\Omega$. For $f$ the two poles are placed symmetrically about the origin, while for $g$ they are in an arbitrary position.

Let $f(z)$ be an elliptic function of order 2 with two simple poles at $\pm\rho$ with residues $\pm\nu$ and $\rho \ne 0$ and $\nu \ne 0$. Then $f(z)$ and $f(-z)$ have Laurent power series expansions at $z = \rho$ with the same initial term

\begin{aligned} f(z) &= {\nu \over z - \rho} + \bigO(1) \\[1em] f(-z) &= {-\nu \over -z + \rho} + \bigO(1) \end{aligned}

from which it follows that the elliptic function $f(z) - f(-z)$ has no pole at $z = \rho$. By a similar argument it has no pole at $z = -\rho$ and therefore it has no poles at all and is a constant. By evaluating at $z = 0$ it follows that this constant is zero and that

which is to say that $f$ is an even function if the midpoint of it's two poles lies at the origin.

By considering the power series expansions of $f^n(z)$ at $z = \rho$ it can be seen that the set of functions $\{1, f, f^2, f^3, f^4\}$ is a basis for the vector space $\mathcal{V}$ of all even elliptic functions with poles of order at most 4 at $\{\rho,-\rho\}$ and no poles elsewhere.

By differentiating power series power it follows that $f'$ has poles at $\pm\rho$ of order 2. And by differentiating identity even it follows that $f'$ is an odd function.

Therefore $f'(z)^2 \in \mathcal{V}$ and can be written as a linear combination of its basis elements

#fold f'(z)^2 \sp = \sp a f^4(z) + b f^3(z) +$ c f^2(z) + d f(z) + e

This is an ordinary first order differential equation for $f$. From the power series expansion at $z = \rho$ it is easily determined that $\nu = \pm1 / \sqrt{a}$.

By considering $f(z_1)-f(z_2)$ as an order 2 elliptic function in $z_1$ with poles at $\pm\rho$ and zeroes at $\pm z_2$ we can determine that the ratio of the two sides of the equation below has no poles or zeroes and is therefore a constant.

#fold f(z_1) - f(z_2) \sp = \sp -\nu \cdot \frac {\sigma(2\rho)\sigma(z_1 - z_2)\sigma(z_1 + z_2)} {\sigma(z_1 - \rho)\sigma(z_1 + \rho)\sigma(z_2 - \rho)\sigma(z_2 + \rho)}

Further that constant can be obtained by considering the power series expansion of both sides at $z_1 = \rho$.

Dividing sigma_product_f by $z_1 - z_2$ and then taking the limit as $z_2 \rightarrow z_1$ yields this sigma product for $f'(z)$

#fold f'(z) \sp = \sp -\nu \cdot \frac {\sigma(2\rho)\sigma(2z)} {\sigma^2(z - \rho)\sigma^2(z + \rho)}

#fold \sigma(2z) = 2 \cdot \frac {\sigma(z)\sigma(z + \omega_1)\sigma(z + \omega_2)\sigma(z + \omega_3)} {\sigma(\omega_1)\sigma(\omega_2)\sigma(\omega_3)}

where $\omega_i$ are three distinct points on the half-period grid chosen so that $\omega_1 + \omega_2 + \omega_3 = 0$, it is possible to verify a second form of the differential equation for $f$

#fold f'(z)^2 \sp = \sp a \cdot\prod_{i=1}^4 \left[f(z) - f(\omega_i)\right]

From this it follows that if $e_i$ are the four roots of the polynomial $ax^4 + bx^3 + cx^2 + dx + e = 0$ then $e_i = f(\omega_i)$.

Substituting formula sigma_product_f into the cross ratio formula, all the poles and constants cancel to leave

#fold \crossratio{f(z_1),f(z_2),f(z_3),f(z_4)} \sp = \sp \frac {\sigma(z_1 - z_2)\sigma(z_1 + z_2)\sigma(z_3 - z_4)\sigma(z_3 + z_4)} {\sigma(z_1 - z_3)\sigma(z_1 + z_3)\sigma(z_2 - z_4)\sigma(z_2 + z_4)}

This coinicides nicely with the fact that both the cross-ratio and differential equation are invariant under Möbius transformations.

Similar to the basis $\{1,f,f^2,f^3,f^4\}$ for even functions constructed above, $\{1,f,f^2,f'\}$ is a basis for the vector space of all (not necessarily even) elliptic functions with poles of order at most 2 at ${\rho,-\rho}$. We can extend this basis indefinitely by adding two elements $f^n$ and $f^{n-2}f'$ for $n = 2,3 \ldots$ to the initial basis $\{1,f\}$.

#list 1, f, f^2, f', f^3, ff', \ldots

The fact that all elements are monomials is not crucial to the following arguments but does have the advantage of generating considerably simpler formulae.

Using the first 4 elements of monomial_basis we can apply the extended Frobenius-Stickelberger formulae to obtain

#fold,rowgap=1em \begin{vmatrix} 1 & f(z_1) & f^2(z_1) & f'(z_1) \\ 1 & f(z_2) & f^2(z_2) & f'(z_2) \\ 1 & f(z_3) & f^2(z_3) & f'(z_3) \\ 1 & f(z_4) & f^2(z_4) & f'(z_4) \\ \end{vmatrix}\sp = \sp {\sigma^4(2\rho) \over a^2} \cdot \frac {\sigma(z_1 + z_2 + z_3 + z_4) \prod\limits_{i > j} \sigma(z_i - z_j)} {\prod\limits_{i}\sigma^2(z_i - \rho)\sigma^2(z_i + \rho)}

From symmetric_sigma we obtain the symmetric addition formula for $f$, if the $z_i$ are distinct

The following identity is a sigma product analog of an algebraic identity which arises when solving the quartic curve / parabola intersection equation. It can be directly verified by expanding the LHS as a sigma product using sigma_product_f and the RHS as a sigma product using symmetric_sigma.

#fold \big[f(z) - f(z_1 + z_2 + z_3)\big] \prod_{i=1}^3 \big[f(z) - f(z_i)\big] \sp = \sp $ K(z_1,z_2,z_3) \cdot \prod_{\textsf{signs}} \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) & f(z)^2 & \pm f'(z) \\ \end{vmatrix}

where $K$ is a function of $z_1,z_2,z_3$ but not of $z$. Letting $z \rightarrow \rho$ in poly_res_f gives

#fold K(z_1,z_2,z_3) \sp = \sp $ \prod_{\textsf{signs}} \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 & 0 & 0 & \pm \sqrt{a} \\ \end{vmatrix}^{-1}

Taking the product of poly_res_f over $z \in \{0, \omega_1,\omega_2,\omega_3\}$ and using de_2 gives

#fold {1 \over a^4} \cdot f'(z_1 + z_2 + z_3)^2 \prod_{i=1}^3 f'(z_i)^2 \sp = \sp $ K^4(z_1,z_2,z_3) \cdot \prod_{i=1}^4 \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 & e_i & e_i^2 & 0 \\ \end{vmatrix}^2

Putting $z$ equal to a zero of $f$ in equation poly_res_f and using poly_res_f_K yields the following three variable addition formula for $f(z)$

#flow,rowgap=1em f(z_1 + z_2 + z_3) \sp = \sp {1 \over f(z_1)f(z_2)f(z_3)} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & f(z_1) & f^2(z_1) & f'(z_1) \\ 1 & f(z_2) & f^2(z_2) & f'(z_2) \\ 1 & f(z_3) & f^2(z_3) & f'(z_3) \\ 1 & 0 & 0 & \pm\sqrt{e} \\ \end{vmatrix} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & f(z_1) & f^2(z_1) & f'(z_1) \\ 1 & f(z_2) & f^2(z_2) & f'(z_2) \\ 1 & f(z_3) & f^2(z_3) & f'(z_3) \\ 0 & 0 & 1 & \pm\sqrt{a} \\ \end{vmatrix}^{-1}

Similarly taking the square root of poly_res_df gives a three variable addition formula for $f'(z)$ with the correct sign being determined by guess work

#flow,rowgap=1em f'(z_1+z_2+z_3) \sp = \sp - {a^2 \over f'(z_1)f'(z_2)f'(z_3)} \cdot $ \prod_{i=1}^4 \begin{vmatrix} 1 & f(z_1) & f^2(z_1) & f'(z_1) \\ 1 & f(z_2) & f^2(z_2) & f'(z_2) \\ 1 & f(z_3) & f^2(z_3) & f'(z_3) \\ 1 & e_i & e_i^2 & 0 \\ \end{vmatrix} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & f(z_1) & f^2(z_1) & f'(z_1) \\ 1 & f(z_2) & f^2(z_2) & f'(z_2) \\ 1 & f(z_3) & f^2(z_3) & f'(z_3) \\ 0 & 0 & 1 & \pm\sqrt{a} \\ \end{vmatrix}^{-2}

where $e_1,e_2,e_3,e_4$ are the roots of the polynomial equation $ax^4 + bx^3 + cx^2 + dx + e = 0$.

These formulae express $f(z_1+z_2+z_3)$ and $f'(z_1+z_2+z_3)$ as a rational functions of $f(z_1),f'(z_1),$ $f(z_1),f'(z_2),$ $f(z_3),f'(z_3)$ and $a,b,c,d,e$. The latter five being rational because the algebraic expressions involving them are symmetric.

All the previous formulae remain true in some form for arbitrary order 2 elliptic functions. Let $g$ be such a function and $\rho_1,\rho_2$ be its poles with residues $\pm\nu$. Then $g$ may be written very simply in terms of an even function $f$ by translating it's argument to the midpoint $T$ of the two poles as follows

g(z) \sp = \sp f(z + T)

where $T = -\tfrac 1 2 (\rho_1 + \rho_2)$ and $\rho = \tfrac 1 2 (\rho_1 - \rho_2)$. Note that $g(z)$ also satisfies differential equation de.

Now it is just a matter of going through the previous equations and replacing $\rho$ by $\tfrac 1 2 (\rho_1 - \rho_2)$ and $z_i$ by $z_i + T$ followed by replacing $f(z_i + T)$ by $g(z_i)$.

#fold g(z_1) = g(z_2) \iff $ z_1 + z_2 \equiv \rho_1 + \rho_2 \mod \Omega

#fold g(z_1) - g(z_2) \sp = \sp -\nu \cdot \frac {\sigma(\rho_1 - \rho_2) \sigma(z_1 + z_2 - \rho_1 - \rho_2)\sigma(z_1 - z_2)} {\sigma(z_1 - \rho_1)\sigma(z_1 + \rho_1)\sigma(z_2 - \rho_2)\sigma(z_2 + \rho_2)}

The symmetric addition formula symmetric_sigma becomes, if the $z_i$ are distinct

#flow,rowgap=1em \begin{vmatrix} 1 & g(z_1) & g^2(z_1) & g'(z_1) \\ 1 & g(z_2) & g^2(z_2) & g'(z_2) \\ 1 & g(z_3) & g^2(z_3) & g'(z_3) \\ 1 & g(z_4) & g^2(z_4) & g'(z_4) \\ \end{vmatrix}\sp = \sp 0 $ \iff $ z_1 + z_2 + z_3 + z_4 \equiv 2\rho_1 + 2\rho_2 $ \mod \Omega

Finally the three variable addition formula for even order 2 elliptic functions can be converted into a two variable addition formulae for arbitrary order 2 elliptic functions.

For equation addition_f the process outlined above does not quite work out because on the LHS we are left with $f(z_1 + z_2 + z_3 + 3T)$ instead of the desired $f(z_1 + z_2 + z_3 + T)$. This can be remedied by changing the mapping to

(z_1, z_2, z_3) \mapsto (u + T, v + T, -T)

and noting using f_g and even that $f(z_3) = f(-T) = f(T) = g(0)$ and $f'(z_3) = f'(-T) = -f'(T) = -g'(0)$ introducing a minus sign on all the $f'(z_3)$ term's and yielding

#flow,rowgap=1em g(u + v) \sp = \sp $ {1 \over g(u)g(v)g(0)} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & g(u) & g^2(u) & g'(u) \\ 1 & g(v) & g^2(v) & g'(v) \\ 1 & g(0) & g^2(0) & -g'(0) \\ 1 & 0 & 0 & \pm\sqrt{e} \\ \end{vmatrix} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & g(u) & g^2(u) & g'(u) \\ 1 & g(v) & g^2(v) & g'(v) \\ 1 & g(0) & g^2(0) & -g'(0) \\ 0 & 0 & 1 & \pm\sqrt{a} \\ \end{vmatrix}^{-2}

#flow,rowgap=1em g'(u + v) \sp = \sp $ {a^2 \over g'(u)g'(v)g'(0)} \cdot $ \prod_{i=1}^4 \begin{vmatrix} 1 & g(u) & g^2(u) & g'(u) \\ 1 & g(v) & g^2(v) & g'(v) \\ 1 & g(0) & g^2(0) & -g'(0) \\ 1 & e_i & e_i^2 & 0 \\ \end{vmatrix} \cdot $ \prod_{\textsf{signs}} \begin{vmatrix} 1 & g(u) & g^2(u) & g'(u) \\ 1 & g(v) & g^2(v) & g'(v) \\ 1 & g(0) & g^2(0) & -g'(0) \\ 0 & 0 & 1 & \pm\sqrt{a} \\ \end{vmatrix}^{-4}

These formulae express $g(u+v)$ and $g'(u+v)$ as a rational functions of $g(u),g'(u),$ $g(v),g'(v),$ $g(0),g'(0)$ and $a,b,c,d,e$.

M. Popoviciu Draisma 2014 Invariants of binary forms. http://dx.doi.org/10.5451/unibas-006268669 Doctoral Dissertation: Philosophisch-Naturwissenschaftlichen Fakultät der Universität Basel