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If $f$ and $g$ are two elliptic functions with the same periods, then there exists an algebraic relationship of the form $P(f,g)=0$, where $P$ a polynomial of two variables with constant coefficients.

This is a well known property. As a special case we have $P(f, f^{\prime})=0$, which is satisfied by the Weierstrass elliptic function $\mathfrak{D}$ which is a solution of the differential equation in $\Lambda$

$$(Y^{\prime})^{2}=4(Y)^{3}-g_{2}Y -g_{3}$$

where $\Lambda$ is the lattice generated by the two periods of $\mathfrak{D}$, and $g_{1},g_{2}$ are the invariants of the function $\mathfrak{D}$.

My question is: does there exist an algebraic relationship between two elliptic functions if they don't have the same periods, and if it exists, under which conditions (between periods)?

I have proved the existence of an algebraic relationship between two elliptic functions $f$ and $g$ if the periods of $f$ are $\omega_{1}$ and $\omega_{2}$, and the periods of $g$ are $p\omega_{1}$ and $q\omega_{2}$, where $p,q\in\mathbb{Q}$. This is a generalization of the case where the periods are equal.

The problem is still open for further generalizations. I welcome any suggestions.

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    $\begingroup$ Presumably the answer is that the there is an algebraic relationship if and only if the two elliptic curves are commensurate... $\endgroup$
    – Igor Rivin
    Mar 14, 2012 at 19:55
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    $\begingroup$ Just to confirm Igor's suspicion, the algebraic relation between the functions implies (in fact is equivalent to) the existence of an isogeny between the elliptic curves, which implies that the lattices are commensurate. In the OP's notation, the periods of $g$ are $a_{11}\omega_1+a{12}\omega_2,a_{21}\omega_1+a{22}\omega_2$ where the $a_{ij}$ are rational and $\det (a_{ij}) \ne 0$. $\endgroup$ Mar 14, 2012 at 22:46
  • $\begingroup$ from the remarks of Pr. Felipe Voloch, if there exists a algebraic relationship between tow elliptic curves, then this tow elliptic curves there lattices are commensurate, which give as a relation of equivalence between the existence of an algebraic relationship and "lattices are commensurate". thanks for Pr. Igor Rivin and Igor Rivin for there valuable remarks $\endgroup$ Mar 16, 2012 at 1:16
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    $\begingroup$ If $f$ has periods $\omega_1$ and $\omega_2$ then $f$ is a fortiori $\langle p \omega_1, q \omega_2 \rangle$-periodic, and thus algebraically related with $g$. The same argument shows that more generally if the intersection of the period lattices of $f$ and $g$ is again a lattice then $f,g$ are algebraically dependent. The converse is true too (assuming as always that neither $f$ nor $g$ is constant) but not quite this easy. $\endgroup$ May 9, 2012 at 5:14

1 Answer 1

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There exists an algebraic relation between two elliptic functions if and only if their lattices are commensurable. This means that every period of one function is a rational multiple of periods of another.

Proof. Let $f$ and $g$ be two elliptic functions, let $F(f,g)=0$ be an algebraic relation. Choose a point $a$ such that for each solution $w$ of the equation $F(a,w)=0$ we have $F_w(a,w)\neq 0$. Then there are finitely many holomorphic germs $w=\phi_j(z)$, that describe all solutions of $F(z,w)=0$ near $z=a$.

Let $T$ be a period of $f$. Let $z_0$ be a point such that $f(a_0)=a$. Consider the sequence $z_0, z_0+T, z_0+2T...$ Near these points we must have the relations $g=\phi_j(f)$. Because there are only finitely many $\phi_j$ we will have that the germs of $g$ at $z_0+mT$ and at $z_0+nT$ coinside. That is $(m-n)T$ will be a period of $g$.

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