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Consider this polygon as the setting for a dynamical billiard:

The polygon in question

When it's drawn in the plane, the polygon intersects itself; it is non-simple. However, I don't want to embed the polygon in the plane! I want to make it the boundary of a flat Riemannian surface homeomorphic to a disk (just like the area enclosed by an ordinary simple polygon). The billiard map on this surface will "see" only one "flap" at a time:

Two views of the billiard on the polygon

The vertex in the middle of the digram has an interior angle $< -180^\circ$, or equivalently an exterior angle $> 360^\circ$, hence the title question.

  • What would a geometer call this surface and the polygon that bounds it? Is it even a "polygon" if we're not embedding it in the plane?
  • Would it be clearer to call it a Riemann surface? I only care about geodesics and reflection angles. (Assume that my complex-analysis-fu is weak.)
  • What should the unusual vertex be called: a "cusp", a "branch point", an angle $> 360^\circ$, or...?
  • What would make the diagrams clearer?

I just want to refer to this billiard table in an offhand example, so I don't want to spend too much time describing it. On the other hand, I don't want it to be confused with a self-intersecting polygon in the plane, and ideally, I don't want to sound like a crazy person. :)

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up vote 3 down vote accepted
  1. Do not call it a Riemann surface, since this notion is too weak for your purposes.

  2. You can call it "a flat surface with conical singularities" or just a "flat surface".

  3. You can call the point with total angle $>2\pi$ a "singular point" or "cone point".

  4. Surfaces like this do appear routinely in the theory of billiards, see e.g. "Rational billiards and flat structures" by Masur and Tabachnikov, In Handbook Dynamical Systems, Elsevier,

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Thanks!! One question: If I call the point a "conical singularity" or a "cone point", doesn't that suggest to the reader that I'm gluing its adjacent edges together? (Which I'm not.) – Chris Culter Aug 27 '13 at 1:23
@ChrisCulter: No, it does not. It suggests that local geometry of the surface at this point is one of a cone. (This is a very standard terminology, just google "conical singularity + flat surface".) – studiosus Aug 27 '13 at 7:31
Hmm, the articles I'm finding speak of conical singularities in the interior of a surface, not on the boundary. (I'm not unfolding the polygon, and even if I did, the angle isn't rational, so the unfolded surface still doesn't look like a cone in a neighborhood of that vertex.) Also, cone angles may be less than or greater than $2\pi$. That said, we can use "cone point" more broadly to mean "a point in a flat surface with a neighborhood that doesn't embed in the plane". In that sense, the vertex in question is indeed the unique cone point on the surface. Is that what you mean? – Chris Culter Aug 27 '13 at 15:42
@ChrisCulter: Yes, I agree, this is not a completely standard terminology. You can add "with cone angle $>2\pi$" to make everything completely clear (although, a bit long). The novelty of your case is that you have a flat surface with boundary (and people usually consider surfaces without boundary). If you double your surface along the boundary, you will obtain a cone point with cone angle $>4\pi$ at the singular point in question. I do not know of a special name for such conical points on surfaces without boundary, otherwise, I would suggest you use it. – studiosus Aug 27 '13 at 15:55
Here is another possible name: "Overtwisted boundary point". The terminology "overtwisted" comes from contact geometry and I do not think there will be any confusion since the context is very different. – studiosus Aug 28 '13 at 6:40

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