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:

- A
*polyhedron*is defined as an area enclosed by a piecewise-geodesic simple closed curve. *Decomposition*is meant in the usual sense of polyhedral decomposition.- (
*Reworded*) We impose a strict form of convexity which, in this case, means*no individual polyhedron may contain 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. *Congruence*just means the existence of an isometry.- 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?)