Let $S^{n-1} = \{x \in \mathbb{R}^n : x\_1^2 + \cdots + x\_n^2 = 1\}$. Consider a regular spherical simplex, obtained e.g. by taking a hyperspherical cap, picking $n$ equally-spaced points $P = \{p\_1, \dots, p\_n\}$ on the boundary (which is isomorphic to $S^{n-2}$), then taking the boundaries of the simplex to be the great (hyper)circles through each $P \setminus \{p\_i\}$.


For $n \geq 4$, we know that $\mathbb{R}^{n-1}$ cannot be tiled by congruent regular simplices. Despite this, I see two ways to tile $S^{n-1}$ with congruent regular spherical simplices: (1) the orthants are regular spherical simplices of side length $\pi/2$ (there are $2^n$ of them); (2) inscribing a regular simplex in $S^{n-1}$, then taking the natural partition of the surface tiles $S^{n-1}$ with $n+1$ spherical simplices.


Are there other ways to tile $S^{n-1}$ with congruent regular spherical simplices? Feel free to assume $n \geq 4$.



Tiling the surface of a hypersphere with regular simplices
The regular simplex construction is related to 3-3-...-3 Coxeter groups, and the orthant construction is related to 3-3-...-4 Coxeter groups. Along the lines, there are constructions related to the 3-5 and 3-3-5 Coxeter groups:


For the 3-5 Coxeter group, project a regular icosahedron to $S^2$ with the same centre. 


For the 3-3-5 Coxeter group, project a 600-cell to $S^3$ with the same centre. 


The images of the faces (hyperfaces resp.) are regular spherical simplices.


