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Suppose a function f(u) identically satisfies an equation of the form G{f(u+v),f(u),f(v)}=0 for all u and v and u+v in its domain. Here G(Z,X,Y) is a non vanishing polynomial in the three variables with constant coefficients. Then one says that f admits an ALGEBRAIC ADDITION THEOREM. IF f(u) is cos(u), then

$G(Z,X,Y)=Z^2-2XYZ+X^2+Y^2-1,$

while, if f(u) is the Weierstrass p-function with invariants g_2 and g_3, then

$G(Z,X,Y)=16(X+Y+Z)^2(X-Y)^2 -8(X+Y+Z){4(X^3+Y^3)-g_2(X+Y)-2g_3} +4(X^2+4XY+4Y^2-g_2)^2$

Here is the question: Characterize those polynomials G(Z,X,Y) which express an algebraic addition theorem.

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Just to clarify, do you want the domain and codomain of $f$ to be the complex numbers? Also, what sorts of coefficients of $G$ are we allowed? –  Pace Nielsen Feb 1 '10 at 21:40
    
take the case that f is a meromorphic function and the coefficients of G are complex constants –  Mark B Villarino Feb 1 '10 at 21:41
    
Thanks. I need one more clarification. Let $f$ be the zero function. Would it be correct to state that every nonzero polynomial $G$ with zero constant term expresses an algebraic addition theorem for $f$? –  Pace Nielsen Feb 1 '10 at 21:49
    
yes, although, of course, the question is meant to deal with non trivial meromorphic functions. For example, it is obvious that G should be symmetric in X and Y and homogeneous. But, the degree of homogeneity is related to how many times f takes on a particular value, and that can be complicated...for example if it takes on a particular value n times, the degree of G in Z is n^2. –  Mark B Villarino Feb 1 '10 at 22:05
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It might be slightly nicer to ask for polynomials such that u+v+w=0 implies G(f(u), f(v), f(w)) = 0, since now you have symmetry in all three variables. At least, I'm reasonably certain this version is equivalent. –  Qiaochu Yuan Feb 1 '10 at 22:42

4 Answers 4

The examples listed in David Speyer's answer are all of them. This is equivalent to say that all one dimensional algebraic groups are isomorphic to the additive group, the multiplicative group or an elliptic curve. A proof in the language of "algebraic addition theorems" is given in the old book of H. Hancock, Lectures on the theory of elliptic functions, Ch. XXI.

http://books.google.com/books?id=GDYNAAAAYAAJ&printsec=frontcover&dq=Hancock+elliptic+functions&cd=1#v=onepage&q=&f=false

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Here is a very basic comment no one has made yet: If $f(u)$ is a rational function of $u$, then there will be some nonzero polynomial $G$ such that $G(f(u), f(v), f(u+v))=0$. That's because $\mathbb{C}(u, v, u+v)$ has transcendence degree $2$ over $\mathbb{C}$.

The same argument applies if $f$ is a rational function of $e^u$, or if $f$ is a rational function of $\wp(u)$ and $\wp'(u)$, where $\wp$ is the Weierstrauss $\wp$-function.

Can we show that every example is of one of these forms?

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I took the liberty of \wp-ifying your answer. –  Mariano Suárez-Alvarez Feb 2 '10 at 1:42
    
Thanks for the help! –  David Speyer Feb 2 '10 at 2:01

It is a famous theorem of Weierstrass that the only meromorphic functions admitting an algebraic addition theorem are rational functions, or rational functions of the exponential function, or elliptic functions. What has NOT been answered is: given a polynomial G(Z,X,Y), in the three variables X,Y,Z, is it an addition-theorem polynomial? Which formal characteristics of G characterize it as such a polynomial? As far as I know, this question has never been investigated.

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G=0 will be a rational surface if the group is the additive or multiplicative group, whereas G=0 will be covered by the abelian surface $E \times E$ if the group is the elliptic curve $E$. The surface will also have the symmetry as in Qiaochu Yuan's comment. Beyond that it's not clear what can be said and maybe there is no simple characterization. This is a very old topic so, if there was a simple answer, it would likely be known. Any specific reason for your interest? –  Felipe Voloch Feb 2 '10 at 18:19
    
The specific reason for my interest is that the problem is simply stated, concrete, interesting in itself, and unanswered. Moreover, I am preparing an expository paper on this beautiful classical theory since the presentations in Hancock, and in Forsyth, have definite mistakes and errors, quite apart from being misleading and difuse. Koebe (in his thesis) and Forsyth prove certain properties of G, for example the formula for its degree in Z, but leave the question, there. Since a century has passed, one would hope that fresh insights might lead to further results. This website is ideal. –  Mark B Villarino Feb 2 '10 at 19:07
    
If X' means the derivative with respect to u and Y' that wrsp y, etc., then one condition is: Elimination of Z between G=0 and $X'\frac{\partial G}{\partial Y}=Y'\frac{\partial G}{\partial X}$ leads to only a single equation between X and X' for all values of Y and Y' (see Forsyth, page 357) –  Mark B Villarino Feb 3 '10 at 22:33

If X' means the derivative with respect to u and Y' that wrsp y, etc., then one condition is: Elimination of Z between G=0 and $X'\frac{\partial G}{\partial Y}=Y'\frac{\partial G}{\partial X}$ leads to only a single equation between X and X' for all values of Y and Y' (see Forsyth, page 357) – Mark B Villarino 0 secs ago

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