# Convex polyhedral decomposition of spheres

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?)

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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 $S^{n+1}$).
Thanks for the answer, but in the $k=4$ case for $S^2$ the triangles contain a great semicircle so that there's a pair of antipodal points in a (and hence each) polyhedron which have a non-unique distance minimizing path between them. (I apologize for the clumsy wording in my original question.) – innerproduct Dec 26 '12 at 13:21
Sure, I corrected the answer. Note that still there is a decomposition of $S^2$ into four convex triangles with angles $2\pi/3$. – Dmitri Dec 26 '12 at 13:26