Bounds on the angles between four unit-length vectors in three dimensional Euclidean space

Question

Let's consider four unit-length vectors $\mathbf{s}_i$, $i=1,2,3,4$, in three-dimensional Euclidean space. Let $\theta_{ij}$ be the angle between $\mathbf{s}_i$ and $\mathbf{s}_j$. Given the set of six angles, $\theta_{12},\theta_{13},\theta_{14},\theta_{23},\theta_{24},\theta_{34}$, there may or may not exist $\mathbf{s}_1,\mathbf{s}_2,\mathbf{s}_3,\mathbf{s}_4$ for which the angles between them are given by the said set. To put it slightly differently, the following equations about $\mathbf{s}_i$ may or may not have solutions for given $\theta_{ij}$:

$\mathbf{s}_i\cdot\mathbf{s}_j = \cos\theta_{ij}$, $1\le i<j\le 4$.

My questions are the following:

(1) Is there an efficient way of determining whether the above equations have a solution?

(2) Are there non-trivial bounds on $\theta_{ij}$ for arbitrary $\mathbf{s}_i$?

Answer 1

If $A$ denotes the $3 \times 4$ matrix having the (unknown) vectors' coordinates as columns, the coefficients of $G=A^T A$ are all the dot products $s_i.s_j$, i.e., coefficients $\theta_{ij}$ (with $\theta_{ii}=1$ on the diagonal). This "Gram matrix" https://en.wikipedia.org/wiki/Gramian_matrix is known to have a null determinant. This provides a (necessary) compatibility condition.

There is an auxiliary condition that should also be fulfilled due to the fact that this $G=A^TA$ is semi-definite positive with rank 3 (I assume there that you want that your four vectors generate $\mathbb{R^3}$). This condition is that the determinant of the $3 \times 3$ principal block obtained by taking the first $3$ lines and $3$ columns of $G$ is $>0$ (usually with a small value).

The last step, being able to construct a matrix $A$ knowing matrix $G$ looks to me rather simple: build any spherical triangle $M_1M_2M_3$ with angles $\theta_{1,2}$, $\theta_{2,3}$ and $\theta_{3,1}$, and then check that there is a solution $M_4=(x,y,z)$ such that $\vec{OM_4}.\vec{OM_1}=cos \theta_{14}$, and 2 more equations of the same type, giving a system of 3 equations with 3 unknowns for $x,y,z$ with the supplementary condition that $x^2+y^2+z^2=1$.