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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 [2] and its follow up [3] 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.


[1]: 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.

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

[3]: 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.

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    $\begingroup$ It is just false for $n\ge 3$; what you get is a different symmetric space. $\endgroup$ – Misha May 28 '19 at 13:43
  • $\begingroup$ @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? $\endgroup$ – Călin May 28 '19 at 13:52
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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 $n$.

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  • $\begingroup$ I see. Thanks a lot, Robert. It's funny that the author states the property in a pretty generic way in several places, though. $\endgroup$ – Călin May 28 '19 at 14:20
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    $\begingroup$ @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. $\endgroup$ – Robert Bryant May 28 '19 at 16:58
  • $\begingroup$ 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 $n > 2$. $\endgroup$ – Călin May 29 '19 at 9:11

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