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Say $P$ ~ $Q (P$ and $Q$ are «projectively equivalent») iff there is a projective transformation $f$ such that $f(P) = Q$. Then ~ is an equivalence relation. I read that the space of inscribed n-gons modulo projective equivalence has dimension n-3. Why is this? Also, are there any related results?

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What is an "inscribed n-gon"? Inscribed in a circle? –  Joseph O'Rourke Aug 5 '10 at 22:43
    
Sorry, I mean inscribed in a conic. –  Adeel Aug 5 '10 at 23:49
    
Conic or circle doesn't make a difference (projectively), but it does matter what you mean by an N-gon. Here it means "an ordered set of N distinct points". Other meanings change the space of polygons, though its dimension is (N-3) under any interpretation. –  T.. Aug 6 '10 at 0:21

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Of the projective transformations fixing a conic, there is a unique one sending any given ordered triple of points to any other given ordered triple of points. So you are free to determine the location of the first 3 vertices of the n-gon, modulo projective equivalence, but any two placements of the remaining (n-3) points are projectively inequivalent.

Thus, the space of n-gons up to projective equivalence can be thought of as the space of (n-3) points on a projective line (or conic, it is isomorphic) punctured at three given points. This has dimension (n-3).

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