riemannian geometry – Identify an ordered-eigenvalue simplex with tetrahedral dihedral angle \$cos ^{-1}left(frac{1}{3}right)\$ in volume, area formulas

Let us order the four eigenvalues ($$lambda_1 geq lambda_2 geq lambda_3 geq lambda_4$$, $$Sigma_{i-1}^4 lambda_i=1$$) of a Hermitian, trace-one, positive-definite $$4 times 4$$ matrix (a “two-qubit density matrix”, in quantum parlance). Those density matrices for which the inequality
$$lambda_1-lambda_3 < 2 sqrt{lambda_2 lambda_4}$$ holds are termed “absolutely separable” (that is, can not be entangled by any global unitary transformation).
In terms of the Hilbert-Schmidt probability distribution $$9081072000 Pi_{j on the simplex of ordered eigenvalues (HilbertSchmidt), the volume/probability of the absolutely separable states has been found to be (Evaluate Explicit)
$$begin{equation} frac{29902415923}{497664}-frac{50274109}{512 sqrt{2}}-frac{3072529845 pi }{32768 sqrt{2}}+frac{1024176615 cos ^{-1}left(frac{1}{3}right)}{4096 sqrt{2}} approx 0.00365826 end{equation}$$
while the area/probability of the boundary absolutely separable states (the locus of $$lambda_1-lambda_3 =2 sqrt{lambda_2 lambda_4}$$ ) has been further shown to be (Confirm)
$$begin{equation} -frac{837276448115}{663552}+frac{86031670725 pi }{32768 sqrt{2}}-frac{86031670725 cos ^{-1}left(frac{1}{3}right)}{16384 sqrt{2}} approx 0.15339, end{equation}$$
where, notably, $$cos ^{-1}left(frac{1}{3}right)$$ is the dihedral angle of the regular tetrahedron.

Might these formulas provide any possible insight into the underlying geometry of the two-qubit absolutely separable states?