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 *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.
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?)



Convex polyhedral decomposition of spheres
For any even $k>4$ there is a decomposition of $S^2$ into $k$ congruent triangles with angles $\pi/2,\pi/2, 4\pi/k$. 


For $k=n+2$ in order to get a decomposition of $S^n$ into $k$ congruent simplexes you should just inscribe in $S^n$ the regular simplex and project its hyper-faces to the sphere from its centre.


In general, a sphere of arbitrary dimension $n$ can be decomposed in arbitrary large number of congruent simplexes, for example into $k2^{n-1}$ simplexes with any $k>2$. This can be done by induction. The $n=1$ case is obvious, to go from $S^n$ to $S^{n+1}$ just put $S^n$ into $S^{n+1}$ as the equatorial sphere and consider the suspension of the simplicial decomposition of $S^n$ (a hypersimplex of such decomposition is the convex hull of a union of a hypersimplex in $S^n$ with one of the poles of $S^{n+1}$).


