Michael Kleber is correct. When I scanned some of this thread, at first I thought that maybe Poncelet's porism would generalize (as Victor Miller suggested:)
Is what you want related to Poncelet's Porism? http://mathworld.wolfram.com/PonceletsPorism.html
There's an argument that would have imply that if there were exactly two solutions to arcs of length L inscribed in an ellipse, for every L, the same phenomenon as in Poncelet's porism would hold: the map of the ellipse to itself gotten by stepping off with steps of length L would be diffeomorphically conjugate to a rotation because the algebraic curve consisting of secants of length L would be an elliptic curve. But feeding the algebra to Mathematica showed this is not true: for each point on the ellipse, there's a degree 4 equation for the other endpoints of the secant, and the curve parametrizing secants of length L has high genus. (approximately genus 16, but I'm not clear enough on algebraic geometry to do it exactly without more work than I'm up for). The complexification of an ellipse is topologically an annulus, and given a length, L, and a starting point, x, you get an image of a 4-way-branching tree mapped to the annulus. As you vary L and x, the branches sometimes intersect and form loops. Sometimes these are on the real ellipse, sometimes just on the complexification. There are typically 32 critical points for the process, where two of the 4 branches coincide. When L is just a little longer than the minor axis of the ellipse so that there are "less obvious" secants of length L, there are 4 real critical points where the secant is "jammed" --- one end is perpendicular to the ellipse, the other makes an angle, and the perpendicular end can be perturbed in either direction while moving the non-jammed end slightly forward. There are lots of other non-obvious complex critical points Anyway, for each n, there is an exponential-in-n number of complex values of L with complex inscribed polygons. Even for the case of real polygons where each secant is chosen to advance the obvious way in the smallest possible step, the polygon can be any of the (n-1) n-pointed stars (where retracing is allowed, e.g. going around a triangle 3 times would count as a 9-pointed star). If Kleber's non-obvious secants are allowed, I think the number of n-gons is exponential-in-n even in the real case. That's because there's always an open set of {(starting point p, length L)} where there are four secants, when p is near an end of the minor axis and L is just longer than the minor axis. For any L < major axis, there's a rotation number associated with the process of advancing counterclockwise to the nearest point at distance L. Choose L so that the rotation number is irrational, and L is just greater than the minor axis. Every orbit for L' near L enters the open set with 4 solutions at regular intervals. By making branching choices at those times, then proceeding "obviously" for a while, and eventually adjusting L slightly to make the polygon close, I think you get exponentially many solutions. Bill Thurston On Nov 20, 2010, at 9:21 AM, Michael Kleber wrote:
On Sat, Nov 20, 2010 at 12:44 AM, David Wilson <davidwwilson@comcast.net>wrote:
I read your notes and did some imagining, and here are my unjustified beliefs:
For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse.
I disagree -- I think there are "less obvious" inscribed equilateral n-gons in ellipsis with high eccentricity, in which one or two edges take a short-cut, circumventing a large piece of the ellipse. (Given a point P on the ellipse and a distance d, there may be *four* points on the ellipse at distance d from P, not just two.)
This objection carries through to all the following statements also, I think...