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 $S^1$?

  • $\begingroup$ 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. $\endgroup$ – Richard Rast Sep 27 '11 at 22:22
  • $\begingroup$ 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. $\endgroup$ – Richard Rast Sep 27 '11 at 22:25
  • $\begingroup$ @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). $\endgroup$ – George Lowther Sep 27 '11 at 22:59
  • $\begingroup$ @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. $\endgroup$ – Richard Rast Sep 28 '11 at 4:06
  • $\begingroup$ @George: post deleted. I read the question too quickly :) $\endgroup$ – Piero D'Ancona Sep 28 '11 at 12:32

@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 $[0,1]$.


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