Closed-form for the volume of a spherical tetrahedron with dihedral angle arcsec(4)

Prove that the volume f(5,1) of the spherical tetrahedron T_1 in the 3-sphere S^3, arising in the n = 5 standard normal case of the Approval Voting analysis (Example 4.2) where all face-pair dihedral angles are arcsec(4), equals arccos(61/64) multiplied by π/5. Equivalently, establish that f(5,1) = (π/5)·arccos(61/64), which would imply P(W_5 = 1) = P(W_5 = 4) = arccos(61/64)/(2π) via equation (4.10).

Background

In Example 4.2, the authors analyze the case where the utilities U_1,…,U_n are i.i.d. standard normal variables. Writing Y_i = U_i − U (with U the sample mean), they show that P(W_n = k) can be expressed as the ratio of the volume of a spherical polyhedron T_k in the unit sphere S^{n−2} to the sphere’s surface area. Specifically, for n = 5 and k = 1, T_1 is a spherical tetrahedron whose bounding hyperplanes yield equal face-pair dihedral angles arcsec(4).

Using Murakami’s formula, a high-precision numerical evaluation of the volume f(5,1) was obtained, and it was empirically observed (and numerically corroborated to hundreds of decimal places) that this volume appears to equal a simple closed form. However, no proof is known. Establishing this identity would give an exact closed-form expression for the probabilities P(W_5 = 1) and P(W_5 = 4) in the normal case by the geometric representation formula (4.10).