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Cubic

The symmetric addition formula on 3 variables can be extended to $3n$ variables.
To get to 6 add the monomials $f^2,fg,g^2$ to $1,f,g$. Then we have $3 \times 2 = 6$ poles and a basis with 6 elements - all good.
To get to 9 add $f^3,f^2g,fg^2$. Note that $g^3$ is redundant because it is just a linear combo of the rest, due to the cubic relation. So again all good.
To get to $3n$ add $f^{i+1},f^{i}g,f^{i-1}g^2$ for $i=1\ldots n-1$. So the number of poles is $0,3,3,6,6,6,9,9,9,12,12,12\ldots$
Can we do a 4 and 5 variable version?
I don't think so. Obviously we could just put $z_5=0$ in the 6 variable formula. But in general $f(0),g(0)$ will not be a rational function of the coeffs of $F$ and there is no other point that would in general work either. And there isn't any way to construct a 5-element basis from monomials, or by any other method - so NO.
However it's a different story for the $y^2=ax^3\ldots$ cubic. Because there $f(0),g(0)$ is "rational". And you can reduce.
You can see that another way as well. Just use the basis $1,f,g,f^2,fg,f^3,f^2g,\dots$. That has $0,2,3,4,5,6,7\ldots$ poles, and the same number of basis elements. So you can directly apply EFS, there is no need to reduce higher order formulae.

Quartic

For the $x^2y^2$ curve add $1,f,g,fg,f^2,g^2,f^2g,fg^2,f^3,g^3,f^3g,fg^3\ldots$ to get to $0,2,4,4,6,8,8,8,10,12,12,12\ldots$. Here $f^2g^2$ is redundation because of the quartic relation. So repeating this pattern we can do $4n$.
For the $y^2=x^4\dots$ curve add $1,f,f^2,g,f^3,fg,f^4,f^2g\dots$ to get $0,2,4,4,6,6,8,8\dots$ poles. So we can do $2n$ for $n \ge 2$.

General Nonsense

We now look at a general irreducible genus one curve $F(x,y) = 0$. \begin{equation} F(x,y) \space = \space \sum_{i=0}^n C_i x^{m_i} y^{n_i} \end{equation} Let $M_i(x,y) = x^{m_i} y^{n_i} \space \text{for}\space i = 1 \ldots n$ be the monomial terms that make up $F(x,y)$.
\begin{equation} F(x,y) \space = \space \sum_{i=1}^{n} C_i M_i(x,y) \end{equation} For convenience write $M_i(z) \space = \space M_i(f(z),g(z)) \space = \space f(z)^{m_i} g(z)^{n_i}$.
If we were naive we might attempt to get a formula for the $C_i$ by using the method of undetermined coefficients.
That is by evaluating at $n-1$ points and then solving the linear equations to obtain \begin{equation} \label{eq:efs6} \begin{vmatrix} M_1(z_1) & M_2(z_1) & \cdots & M_n(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_{n-1}) & M_2(z_{n-1}) & \cdots & M_n(z_{n-1}) \\ M_1(z) & M_2(z) & \cdots & M_n(z) \\ \end{vmatrix} \space = \space 0 \end{equation} where the $C_i$ are given by the cofactors of the last row, that is \begin{equation} \label{eq:efs7} C_1 \space = \space (-1)^{n} \begin{vmatrix} M_2(z_2) & \cdots & M_n(z_1) \\ \vdots & \ddots & \vdots \\ M_2(z_{n-1}) & \cdots & M_n(z_{n-1}) \\ \end{vmatrix} \qquad\qquad \cdots \qquad\qquad C_n \space = \space (-1)^{n-1} \begin{vmatrix} M_1(z_1) & \cdots & M_{n-1}(z_1) \\ \vdots & \ddots & \vdots \\ M_1(z_{n-1}) & \cdots & M_{n-1}(z_{n-1}) \\ \end{vmatrix} \end{equation} Now for the magic.
Each $C_i$ cannot be a different function of the $z_i$ because that would imply $f,g$ satisfy multiple different equations.
Each $C_i$ must therefore be a constant multiple of some function $W(z_1,\ldots z_{n-1})$.
By applying the Extended Frobenius Stickelberger formula to the formula for $C_i$ we can discover that it is \begin{equation} C_i = K_i \frac {\sigma(z_1 + z_2 + \ldots z_{n-1} - s) \prod\limits_{1 \leq i \lt j \leq n} \sigma(z_i - z_j)} {\prod\limits_{i=1}^{n} \prod\limits_{j=1}^{m} \sigma(z_i - \rho_j)} \end{equation} where $s = \sum \rho_j$ and $\rho_j$ are the locations of the $m$ poles.
So the level 1 symmetric addition formula for the curve in $n$ variables is simply the curve itself \begin{equation} \label{eq:efs8} \begin{vmatrix} M_1(z_1) & M_2(z_1) & \cdots & M_n(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_{n-1}) & M_2(z_{n-1}) & \cdots & M_n(z_{n-1}) \\ M_1(z_n) & M_2(z_n) & \cdots & M_n(z_n) \\ \end{vmatrix} \space = \space 0 \end{equation} The level 2 symmetric addition formula is \begin{equation} f(s - z_1 - z_2 \space \ldots \space - z_{n-1}) \space = \space \frac 1 {f(z_1)f(z_2) \ldots f(z_{n-1})} \prod\limits_{i=1}^{M} \frac {\begin{vmatrix} M_1(z_1) & M_2(z_1) & \cdots & M_n(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_{n-1}) & M_2(z_{n-1}) & \cdots & M_n(z_{n-1}) \\ M_1(\alpha_i) & M_2(\alpha_i) & \cdots & M_n(\alpha_i) \\ \end{vmatrix}} {\begin{vmatrix} M_1(z_1) & M_2(z_1) & \cdots & M_n(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_{n-1}) & M_2(z_{n-1}) & \cdots & M_n(z_{n-1}) \\ M_1(\beta_i) & M_2(\beta_i) & \cdots & M_n(\beta_i) \\ \end{vmatrix}} \end{equation} where $g(\alpha_i)$ are the $M = \max\{m_j\}$ conjugate roots of $F(0,y)=0$ and $g(\beta_i)$ are the $M$ conjugate roots of $F(\infty,y)=0$. And since the formula is symmetric in the roots it is a rational function of $z_1 \dots z_{n-1}$ and $C_1 \ldots C_n$ ! WRONG # terms-1 (n+1)(m+1)-1 < # poles 2mn so it doesn't work in general only when n = m = 2

Second Attempt

From formulae like \eqref{eq:fwp2} we can, in principle, derive a birational mapping from the Weierstrass curve to the curve representing the algebraic relation of two elliptic functions $f$ and $g$. For example let $f$ and $g$ be two order four elliptic functions with simple poles on the half period grid, then \begin{equation} f(z)\space = \space f_0 \space + \space \frac {Q_1\left(\wp(z)\right)} {\wp'(z)} \qquad\qquad g(z)\space = \space g_0 \space + \space \frac {Q_2\left(\wp(z)\right)} {\wp'(z)} \end{equation} where $Q_1,Q_2$ are quadratic functions. This implies \begin{equation} \wp(z) \space = \space Q\left(f(z)/g(z)\right) \qquad\qquad \wp'(z) \space = \space ? \end{equation} May be better to start with \begin{gather} f(z) \space = \space \sn(z) \space + \space a \cn(z) \implies (f - s)^2 = a^2 - a^2 s^2 \implies (a^2+1)s^2 - 2sf + f^2 - a^2 = 0 \implies s = f + a\sqrt{a^2 + 1 - f^2} \\ g(z) \space = \space \sn(z) \space + \space b \dn(z) \implies (g - s)^2 = b^2 - b^2 k^2 s^2 \implies (b^2k^2+1)s^2 - 2sg + g^2 - b^2 = 0 \implies s = g + b\sqrt{b^2 k^2 + 1 - k^2 g^2} \\ f - g = \sqrt{A + Bf^2} + \sqrt{C + Dg^2} \implies (1-B)f^2 - 2fg + (1-D)g^2 - (A+C) = 2 \sqrt{A + Bf^2} \sqrt{C + Dg^2} \end{gather}

For example let $f$ and $g$ share $m$ simple poles, then the number of monomials of total degree $n$ is $\tfrac 1 2 (n+1)(n+2)$. The dimension the vector space of elliptic functions with the same poles up to order $n$ is $mn$. Setting the number monomials to be one greater than the dimension of this vector space gives \begin{equation} mn + 1 = \tfrac 1 2 (n+1)(n+2) \qquad\implies\qquad n = 2m - 3 \end{equation} So $f$ and $g$ satisfy an algebraic equation of the form \begin{equation} \sum_{0 \le i+j \le 2m - 3} f(z)^i g(z)^j \space = \space 0 \end{equation} Now let $r = mn = m(2m-3)$ and let $M_i(z)$ be the first $r$ monomials and construct an alternant determinant from them. Then we have the following Frobenius-Stickelburger type formula \begin{equation} \begin{vmatrix} M_1(z_1) & M_2(z_1) & \ldots & M_r(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_r) & M_2(z_r) & \ldots & M_r(z_r) \\ \end{vmatrix} \enspace = \enspace const. \frac {\sigma\left(\sum_1^r z_i - n\sum_1^m \rho_j \right)\prod\limits_{i \le j} \sigma(z_j - z_i)} {\prod\limits_{i,j} \sigma(z_i - \rho_j)^n} \end{equation} Misleading deduction - in general the algebraic relation between $f$ and $g$ has total degree $m$ ??. Apart from the case $m \le 3$ the constant on the right hand side is zero and the determinant vanishes because columns are trivially linearly dependent.
From this we have a formula analogous to the classic symmetric addition for $\wp$ \begin{equation} \sum_1^r z_i \equiv n\sum_1^m \rho_j \mod \Omega \qquad\qquad \implies \qquad\qquad \begin{vmatrix} M_1(z_1) & M_2(z_1) & \ldots & M_r(z_1) \\ \vdots & \vdots & \ddots & \vdots \\ M_1(z_r) & M_2(z_r) & \ldots & M_r(z_r) \\ \end{vmatrix} \enspace = \enspace 0 \end{equation}

Another example let $f$ have $m_1$ simple poles and $g$ have $m_2$ simple poles distinct from $f$'s. Then the number of monomials of degree $n_1$ in $f$ and $n_2$ in $g$ is $(n_1 + 1)(n_2 + 1)$. The dimension the vector space of elliptic functions with the same $f$ poles up to order $n_1$ and $g$ poles up to order $n_2$ is $m_1n_1 + m_2n_2$. Setting the number monomials to be one greater than the dimension of this vector space gives \begin{equation} m_1n_1 + m_2n_2 + 1 = (n_1 + 1)(n_2 + 1) \qquad\implies\qquad n_1 = 2m_2 - 2, \enspace n_2 = 2m_1 - 2 \end{equation} So $f$ and $g$ satisfy an algebraic equation of the form \begin{equation} \sum_{\substack{0 \le i \le 2m_2 - 2 \\ 0 \le j \le 2m_1 - 2}} f(z)^i g(z)^j \space = \space 0 \end{equation}

Algebraic Relation Between Two Elliptic Functions With Same Poles

Let $f$ be an elliptic function with $n$ simple poles. The dimension of the space of all such functions is $n$ poles + $n - 1$ residues plus $1$ additive constant plus $2$ periods which totals $2n + 2$. Let $g$ be another such function with the same poles and periods then the dimension of the space of all such ordered pairs is $3n + 2$. Therefore the dimension of the space of all such curve parameterisations is $3n$, because applying a scale and translation to $z$ gives the same curve.

Let $C$ be a curve of total degree $n$. The dimension of the space of such curves is $\tfrac 1 2 (n+1)(n+2) - 1$ since one coefficient is redundant due to arbitrary scaling. The genus of the curve is in general $\tfrac 1 2 (n-1)(n-2)$. Therefore there are effectively $\tfrac 1 2 n^2 + \tfrac 3 2 n$ parameters minus $\tfrac 1 2 n^2 - \tfrac 3 2 n$ constraints to force the curve to be genus $1$. Therefore the dimension of the space of all such curves with genus $1$ is $3n$.

From this we infer that the algebraic relation between two elliptic functions with $n$ shared poles is a curve of total degree $n$ constrained so that it has genus $1$. And conversely that any curve of genus one and total degree $n$ can be parameterised by a pair of elliptic functions with the same $n$ poles.

Algebraic Relation Between Two Elliptic Functions With Different Poles

Let $f$ be an elliptic function with $n_1$ simple poles and let $g$ be an elliptic function with the same periods but with $n_2$ different poles. The dimension of the space of all such order pairs $2n_1 + 2n_2 + 2$. The dimension of the space of all such curve parameterisations is $2n_1 + 2n_2$.

Let $C$ be a curve of maximum degree $n_2$ in $x$ and $n_1$ in $y$. The number of such monomials is $(n_1 + 1)(n_2 + 1)$ and therefore dimension of the space of such curves is this number minus one for the arbitrary scaling factor. The genus of the curve is in general $(n_1 - 1)(n_2 - 1)$. Therefore there are effectively $n_1n_2 + n_1 + n_2$ parameters minus $n_1n_2 - n_1 - n_2$ constraints to force the curve to be genus $1$. Therefore the dimension of the space of all such curves with genus $1$ is $2n_1 + 2n_2$.

From this we infer that the algebraic relation between two elliptic functions one with $n_1$ poles and the other with $n_2$ different poles is a curve of maximum degree $n_2$ in $x$ and $n_1$ in $y$ constrained so that it has genus $1$. And conversely that any such curve can be parameterised by a pair of elliptic functions, the first with $n_1$ poles and the second with $n_2$ different poles.

Monomial Basis Sequence For Two Elliptic Functions With Same Poles

The space of elliptic functions with up to $n$ poles of order $m$ is $mn$, because one degree of freedom is lost because the residues must sum to zero, and one degree of freedom is gained because of an arbitrary additive constant. Now the number of monomials in $f$ and $g$ of total degree at most $n$ is $\tfrac 1 2 (n+1)(n+2)$ with one linear relation between them. So the dimension of the vector space of all monomials with total degree at most $m$ is $\tfrac 1 2 n(n+3) + (m - n)n = mn - \frac 1 2 n(n-3)$. Thus with just two functions the sequence of monomials is never a basis sequence, except for the case $n=3$.

So to construct a monomial basis sequence with eilliptic functions of order $4$ we require three elliptic functions say $f,g,h$ giving a basis sequence \begin{equation} 1,f,g,h; \quad f^2,fg,g^2,fh; \quad f^3,f^2g,fg^2,g^3; \quad f^4,f^3g,f^2g^2,fg^3; \end{equation}

References

[1] Euler, Leonhard (1761) Observationes de comparatione arcuum curvarum irrectificibilium Euler Archive - All Works. 252.