[math-fun] Ideal polygons in the hyperbolic plane
In the Poincaré model, the hyperbolic plane H is viewed as an open unit disk D in the (ordinary) plane, with the geodesics being any arc of a circle that's perpendicular to the circular boundary bd(D). For any geodesic polygon in H, its area is A = [sum of its exterior angles] - 2 pi. One can also have ideal (geodesic) polygons, all of whose vertices lie on "the circle at infinity", i.e., bd(D). Since each exterior angle of such a polygon must be pi, its area is a function only of the number of sides N: A = (N-2) pi. Also, all the sides of an ideal polygon are of infinite length. QUESTION: When are two such ideal hyperbolic N-gons isometric (i.e., congruent in H) ? Note that the space of all such N-gons is N-dimensional (depending on the vertices on bd(D), whereas the group of isometries of H is the same as the group of conformal (or anti-conformal) automorphisms of D -- a 3-dimensional group. Hence there are N-3 dimensions of congruence classes of such ideal N-gons.
From this it's easy to see that all ideal triangles in H are congruent. Given the 4 vertices on bd(D) of each of two ideal squares in H, what's a concise algorithm to determine whether the two squares are congruent?
--Dan
Two ideal hyperbolic quadrilaterals are congruent iff they have equal angles between their diagonals. I don't see how to extend this idea to (5+N)-gons, though. I'm also very interested in Dan's general question. - Scott
In the Poincaré model, the hyperbolic plane H is viewed as an open unit disk D in the (ordinary) plane, with the geodesics being any arc of a circle that's perpendicular to the circular boundary bd(D).
For any geodesic polygon in H, its area is A = [sum of its exterior angles] - 2 pi.
One can also have ideal (geodesic) polygons, all of whose vertices lie on "the circle at infinity", i.e., bd(D). Since each exterior angle of such a polygon must be pi, its area is a function only of the number of sides N: A = (N-2) pi. Also, all the sides of an ideal polygon are of infinite length.
QUESTION: When are two such ideal hyperbolic N-gons isometric (i.e., congruent in H) ?
Note that the space of all such N-gons is N-dimensional (depending on the vertices on bd(D), whereas the group of isometries of H is the same as the group of conformal (or anti-conformal) automorphisms of D -- a 3-dimensional group. Hence there are N-3 dimensions of congruence classes of such ideal N-gons.
From this it's easy to see that all ideal triangles in H are congruent. Given the 4 vertices on bd(D) of each of two ideal squares in H, what's a concise algorithm to determine whether the two squares are congruent?
--Dan
.. the hyperbolic plane ... ha[s] ideal (geodesic) polygons, all of whose vertices lie on "the circle at infinity"... When are two such ideal hyperbolic N-gons isometric (i.e., congruent...)? --Dan
Two ideal hyperbolic quadrilaterals are congruent iff they have equal angles between their diagonals. - Scott
I think Scott's observation is it. Ideal triangles are always congruent. If N>3: Given an N-sided ideal hyperbolic polygon, take any four consecutive vertices A B C D, and consider the diagonals AC and BD. Call the point of intersection E, and find the angle BEC. Associate that angle with the polygon-side BC. There are as many such angles as sides: N of them. Clearly, if two such N-gons are congruent, the associated angles are the same. It is less obvious that if two such N-gons are incongruent, some of the associated angles are different. But if it's true for 4-gons, one should be able to construct an inductive proof on the number of sides. Contrapositively, if the sequences of associated angles are the same [different], the polygons are congruent [incongruent]. ----- A vertex of an ideal N-gon can be represented as its direction (in some co-ordinate system) from the origin. Let A, B, C, D be the directions of four consecutive vertices. Then the angle X (associated with BC) is cos [(D-B)/2] cos [(C-A)/2] - cos [(D-C+B-A)/2] cos X = ----------------------------------------------- sin [(D-B)/2] sin [(C-A)/2] (Can someone double-check?) ----- When are the sequences of associated angles the same? Here are two such sequences (not necessarily in radians :-) 2 6 4 8 2 9 3 8 4 6 2 3 9 2 They might took different, but they're really the same. (In the second, start with the first '2', and go backwards.) Is there an O(n) algorithm for cyclically comparing two sequences of numbers? -- Don Reble djr@nk.ca
participants (3)
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asimovd＠aol.com -
Don Reble -
Scott Huddleston