# Isomorphism between the hyperbolic space and the manifold of SPD matrices with constant determinant

I'm studying properties of the manifold of symmetric positive-definite (SPD) matrices and I've learnt about the following connection to the hyperbolic space [1, Section 2.2], $$\mathcal{P}(n) = \mathcal{SP}(n) \times \mathbb{R}^+,\qquad \mathcal{SP}(n) \cong \mathbb{H}^p,$$ where $$\mathcal{P}(n)$$ is the space of SPD matrices, $$\mathcal{SP}(n) = \{ A \in \mathcal{P}(n) : \lvert A \rvert = 1 \}$$, and $$p = n (n + 1) / 2 - 1$$. In words,

$$\mathcal{P}(n)$$ is a foliated manifold whose codimension-one leaves are isomorphic to the hyperbolic space $$\mathbb{H}^p$$.

I wanted to see a proof of this isomorphism (and what it means more precisely) and I found  and its follow up  which focus on $$2 \times 2$$ SPD matrices. They mention certain connections between the standard metrics in these spaces, e.g. [3, p4], \begin{align}\tag{1}\label{eq:1} d_{\mathcal{P}}(X_1,X_2) = \sqrt{\frac{1}{2} \big(\log \lvert X_1 \rvert - \log \lvert X_2 \rvert \big)^2 + d_{\mathbb{D}}^2(y_1,y_2)}, \end{align} where $$d_{\mathcal{P}}$$ is the canonical metric on $$\mathcal{P}(n)$$ (see [2, eq. (6)]), $$d_{\mathbb{D}}(y_1,y_2)$$ is the distance in the Poincaré disk between two points that "correspond" to the SPD matrices $$X_1, X_2$$ (more precisely, their scaled versions $$\tilde{X}_1,\tilde{X}_2 \in \mathcal{SP}(n)$$; see [2, p4-6]).

That being said, I couldn't find a proof for the general case of $$n \times n$$ SPD matrices. Is anyone aware of other relevant resources? Or is it really obvious? Does a generalized form of \eqref{eq:1} still hold?

Thank you.

: Moakher, M. (2005). A differential geometric approach to the geometric mean of symmetric positive-definite matrices. SIAM Journal on Matrix Analysis and Applications, 26(3), 735-747.

: Chossat, P., & Faugeras, O. (2009). Hyperbolic planforms in relation to visual edges and textures perception. PLoS Computational Biology, 5(12), e1000625.

: Faye, G., Chossat, P., & Faugeras, O. (2011). Analysis of a hyperbolic geometric model for visual texture perception. The Journal of Mathematical Neuroscience, 1(1), 4.

• It is just false for $n\ge 3$; what you get is a different symmetric space. – Misha May 28 '19 at 13:43
• @Misha You're saying $\mathcal{SP}(n) \cong \mathbb{H}^{n (n + 1) / 2 - 1}$ is false for $n \ge 3$? Could you please elaborate? – Călin May 28 '19 at 13:52

This 'isomorphism' does not hold for $$n>2$$, in the sense that the 'natural' $$\mathrm{SL}(n,\mathbb{R})$$-invariant metric on what you are calling $$\mathcal{SP}(n)$$ and the constant sectional curvature hyperbolic metric on $$\mathbb{H}^p$$ for $$p = n(n{+}1)/2-1$$ are only isometric when $$n=1$$ (the trivial case) and $$n=2$$.
The reason is that $$\mathcal{SP}(n) = \mathrm{SL}(n,\mathbb{R})/\mathrm{SO}(n)$$ as symmetric spaces while $$\mathbb{H}^p = \mathrm{SO}(p,1)/\mathrm{SO}(p)$$ as symmetric spaces. In fact, for $$n>2$$, the symmetric space $$\mathrm{SL}(n,\mathbb{R})/\mathrm{SO}(n)$$ does not have constant sectional curvature, while hyperbolic space $$\mathrm{SO}(p,1)/\mathrm{SO}(p)$$ has constant sectional curvature for all $$p$$.
When $$n=2$$, it's one of the 'accidental' isomorphisms that $$\mathrm{SO}(2,1)$$ is double-covered by $$\mathrm{SL}(2,\mathbb{R})$$. For $$p>2$$, the group $$\mathrm{SO}(p,1)$$ is not evenly-covered by any $$\mathrm{SL}(n,\mathbb{R})$$ for any $$n$$.
• @CălinCruceru: What's true is that both spaces are diffeomorphic to $\mathbb{R}^p$ as smooth manifolds, but that is a very weak statement (and essentially obvious to boot). When you asked about 'isomorphism', I assumed that you were asking about 'isometry'. Because they are both spaces of non-positive sectional curvature (and simply-connected) there is a 'canonical' diffeomorphism between them (once base points have been chosen in each) got by comparing their exponential maps at their respective base points, but, when $n>2$, that map is very far from being an isometry. – Robert Bryant May 28 '19 at 16:58
• The reason I used "isomorphism" is because that's what the author of the first reference uses and I wasn't sure myself what it means. After I found the 2nd and 3rd references, it was clear that there is an isometry for the case $n=2$, but I was still wondering what he means for $n > 2$. – Călin May 29 '19 at 9:11