# Tiling the surface of a hypersphere with regular simplices

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 $$n \geq 4$$.

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.