Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here a question that has me stumped. Maybe someone familiar with algebraic or differential curves can help. Suppose that $\gamma:[0,1] \rightarrow \mathbb{C}$ is an analytic function. Is it true that the range of $\gamma$ is either homeomorphic to a line segment or contains a subset homeomorphic to $S^1$?

share|improve this question
    
I believe so; let S={x | there exists y < x such that f(y)=f(x)}. Then assuming S has a first element y, or is empty, gamma[0,y] is S^1, or gamma[0,1] is an interval. If we assume gamma has an analytic continuation to some open neighborhood in C, then S is finite or gamma is a constant function, but I can't remember if real analytic is enough to guarantee that. I think it is, but I won't add this as an answer since I'm not sure. –  Richard Rast Sep 27 '11 at 22:22
    
Wait, I meant gamma[x,y] in the above, where x is the point where f(x)=f(y). But I can't edit comments. –  Richard Rast Sep 27 '11 at 22:25
    
@Richard: How does that argument apply to $\gamma(t)=\sin\pi t$? Then $S$ is not finite and doesn't have a first element (but the image of $\gamma$ is a line segment). –  George Lowther Sep 27 '11 at 22:59
    
@George Good call, hm. In my mind that was bookkeeping; in fact it was just wrong. Was thinking about zero sets, and it doesn't quite generalize. –  Richard Rast Sep 28 '11 at 4:06
    
@George: post deleted. I read the question too quickly :) –  Piero D'Ancona Sep 28 '11 at 12:32

1 Answer 1

@Richard: Real analytic is the same as complex analytic since locally the power series expansions converge on disks. Yes, $\gamma$ is analytic (or has analytic extension) to a neighborhood of $[0,1]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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