We all know the spherical excess formula: in a unit sphere, the area of a geodesic triangle is equal to the exceeding from $\pi$ of the sum of the three angles of the triangle.
Is there a similar formula for a geodesic tetrahedron in a 3-sphere? I'm sure there must be, but it seems to be tricky. Thanks in advance.
There are indeed generalization of the angle sums relations to spherical $d$-polytopes.
However, the generalization I know doesn't involve the volume of the polytope for odd $d$.
We have
For the special case of tetrahedron at $d = 3$, independent of whether it is spherical/euclidean/hyperbolic, above formula reduces to
$$\sum_{v \in V} \frac{\Omega_v}{4\pi} - \sum_{e \in E} \frac{\theta_e}{2\pi} +\frac{|F|}{2} - 1 = 0$$ where $V, E, F$ are the set of vertices/edges/facets of tetrahedron, $\Omega_v$ is the solid angle at vertex $v$ and $\theta_e$ is the dihedral angle at edge $e$.
As one can see, this formula doesn't involve the volume for odd number $d = 3$.
I don't really know this stuff, above formula is extracted from the paper Angle sums on polytopes and polytopal complexes by Kristin A. Camenga. Look at $\S 4.2$ there for a more accurate description of the relations and references there for more details.