1

My question really is:

if $e^{2\pi i* g(\theta)}$ is an algebraic function in the variable $e^{2 \pi i \theta}$, what restrictions can we put on g?

My first guess is to say that g is the map that sends everything to zero, or $g(\theta)=n\theta +c$, in which case $e^{2\pi i* g(\theta)}=1$ or $C*(e^{2 \pi i \theta})^{n}$ respectivley.

It seems believable that these would be the only two polynomials, or even rational (or even algebraic?!?) functions that could play the role of g...could a transcendental function do the job of g too?

Any thoughts on where to look?

P.S. I am afraid that I may not be posing the question well...apologies.

flag
What is your domain? Do you want $\theta$ real? – Igor Rivin Dec 15 2010 at 22:34

1 Answer

2

(You should change the tag, this has nothing to do with functional analysis)

By Ax's theorem giving a function field analogue of Schanuel's conjecture, if $g(\theta)$ is an algebraic function of $\theta$ such that $e^{2\pi i \theta}$ and $e^{2 \pi i g(\theta)}$ are algebraically dependent, then $\theta$ and $g(\theta)$ are linearly dependent over $\mathbb{Q}$, so you are almost correct.

http://en.wikipedia.org/wiki/Schanuel%27s_conjecture

link|flag
Thanks...that looks great. – RTodd Dec 16 2010 at 1:17

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.