Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a decomposition of $S^2$ into $k$ (geodesically) convex polyhedra that are congruent to each other? What about $S^n$ for $n>1$?

Remarks:

  1. A polyhedron is defined as an area enclosed by a piecewise-geodesic simple closed curve.
  2. Decomposition is meant in the usual sense of polyhedral decomposition.
  3. (Reworded) We impose a strict form of convexity which, in this case, means that no individual polyhedron contains a pair of antipodal points. (So that, in particular, $k>2$.) The idea is that for any two points in a polyhedron, the distance minimizing geodesic is unique and lies completely within the polyhedron.
  4. Congruence just means the existence of an isometry.
  5. Can this be done with $k=4$ in the case of $S^2$? (Or $k=n+2$ for $S^n$?)

I strongly suspect that the answer is no. This prompts the question what's the best one can do (e.g., are there any interesting polyhedral decompositions of $S^2$ that satisfy all but one of these requirements?)

share|improve this question
    
Please note that this question has already been (mostly) answered on MathOverflow. –  innerproduct Dec 26 '12 at 13:39
    
    
If you've already received an answer elsewhere, you should post it as a Community Wiki answer here and accept it so the question can be marked as answered. By the way, this is why simultaneous cross-posting is discouraged. –  Rahul Jan 2 '13 at 16:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.