I'm studying the properties of the manifold of symmetric positive-definite (SPD) matrices and one of the papers that I'm reading mentions that for the case $n = 2$ we have an isometry between its subspace of matrices with unit determinant and $\mathbb{H}_2$, the 2-dimensional hyperbolic space (see [1, p.4-6] and details below).
I'm trying to prove it but it seems to be pretty tedious, so I'm looking for resources, hints, or ideas on how to make it simpler.
The distance between two SPD matrices is $$d_{\textrm{SPD}}(A, B) = \sqrt{ \sum_{i=1}^n \log^2 \lambda_i(A^{-1} B)},$$ where $\lambda_i(A^{-1} B)$ denotes the $i$th eigenvalue of the matrix $A^{-1} B$. The same distance is used when restricting to its subspace of SPD matrices with unit determinant, i.e., $\det(A) = \det(B) = 1$.
The isometric embedding that they propose uses the hyperboloid model of the hyperbolic space which in $\mathbb{R}^3$ consists of points with $x_0^2 - x_1^2 - x_2^2 = 1$ and $x_0 > 0$. The distance function is given by $$d_{\mathbb{H}}(x, y) = \textrm{arccosh}\ b(x, y),\quad\textrm{with}\ \ b(x, y) = x_0 y_0 - x_1 y_1 - x_2 y_2.$$
Their suggested embedding maps a SPD matrix $A = \begin{bmatrix} a&c\\c&b \end{bmatrix}$ with $a > 0$ and $\det(A) = a b - c^2 = 1$ to the point $x$ on the hyperboloid with $$x_0 = \frac{a + b}{2}, \quad x_1 = \frac{a - b}{2},\quad x_2 = c.$$
The claim is that $d_{\textrm{SPD}}(A, B) = d_\mathbb{H}(x, y)$ for all such matrices, using this mapping.
[1]: Chossat, P., & Faugeras, O. (2009). Hyperbolic planforms in relation to visual edges and textures perception. PLoS Computational Biology, 5(12), e1000625.