# Prove that the Log-Euclidean distance is negative-definite

Let $\Bbb{S}_{++}^n$ be the $\frac{n(n+1)}{2}$-dimensional Riemannian manifold of the symmetric positive definite (SPD) $n\times n$ real matrices.

The Log-Euclidean distance between two points of $\Bbb{S}_{++}^n$, i.e. between two SPD matrices $A,B\in\Bbb{S}_{++}^n$, is given by $$d(A,B)=\lVert\log(A)-\log(B)\rVert_{F},$$ where $\|\cdot\|_{F}$ denotes the regular Frobenius norm.

We want to prove that $d\colon\Bbb{S}_{++}^n\times\Bbb{S}_{++}^n\to\Bbb{R}$ is a conditionally negative definite function, as defined below.

For a topological space $\mathcal{X}$, the function $f\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$ is called conditionally negative definite if for any $m\in\Bbb{N}$, $x_1,\ldots,x_m\in\mathcal{X}$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true $$\sum_{i,j=1}^{m}c_ic_jf(x_i,x_j)\leq0.$$

In our case, we want to prove that for any $m\in\Bbb{N}$, $X_1,\ldots,X_m\in\Bbb{S}_{++}^n$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true $$\sum_{i,j=1}^{m}c_ic_j\lVert\log(X_i)-\log(X_j)\rVert_{F}\leq0.$$ Thank you very much in advance!

Edit: In case of the squared Frobenius norm, I think it would be proven more easily, but I still need some help...

• By substituting $Y_i:=\log(X_i)$ you are left with proving the same statement for euclidean spaces, which I guess is true. $\log$ is a bijection from the space of SPD-matricses to the space of symmetric matrices. – user35593 Jul 3 '15 at 17:16
• First, thanks for your comment. Well, I actually tried that, even in the simpler case of $\mathbf{x}_i\in\Bbb{R}^n$: $\|\mathbf{x}_i-\mathbf{x}_j\|_2$. The problem here is the square root. I cannot figure that out; apparently, I cannot use the triangle inequality... If I had the squared norm, i.e. $\|\mathbf{x}_i-\mathbf{x}_j\|_2^2$ thinks become much easier. – nullgeppetto Jul 3 '15 at 17:30
• Why does this have the at.algebraic-topology tag? – Sean Tilson Jul 6 '15 at 10:58
• @SeanTilson, apologies, that's a mistake. I edit it. – nullgeppetto Jul 6 '15 at 10:59
• There is no need for an apology, I was just curious. :) – Sean Tilson Jul 6 '15 at 11:16

First, recall that if $\psi$ is negative definite, then $\exp(-\gamma \psi)$ is positive definite for all $\gamma >0$. Now, from Corollary 2.10 of Harmonic analysis on Semigroups (Berg, Christensen, Ressel) we also know that if $\psi$ is negative definite, then $\psi^\alpha$ is also negative definite for $0 < \alpha < 1$ as long as $\psi(x,x) \ge 0$.
Consider now some map $f: \mathcal{X} \to \mathbb{R}^n$, and set $\psi(x,y) = \|f(x)-f(y)\|^2$. Clearly, $\psi$ is negative definite.
In particular, choosing $\mathcal{X} = \mathbb{S}^n_{++}$, $f \equiv \log$, and $\alpha=1/2$, from the above corollary it follows that $\|\log X - \log Y\|_\text{F}$ is negative definite.
• Thank you very much for your answer! Let me clarify some issues before I know I understand the answer. First, in fact, I need that result in order to prove the positive definiteness of a kernel of the form you mentioned ($\exp(-\gamma d(x,y))$). Now, if I understand your answer, you don't use that fact. The rest of the answer is quite clear to me, I think. So, I prove the negative-definiteness of the squared distance, then I use the corollary you mentioned, am I right? – nullgeppetto Jul 3 '15 at 19:05
• You can just include a proof of the corollary (while still citing that book). The proof follows from the following integral representation: $$\psi^\alpha = \frac{\alpha}{\Gamma(1-\alpha)}\int_0^\infty (1-\exp(-\gamma \psi))\frac{d\gamma}{\gamma^{\alpha+1}},$$ which is valid for $\psi \ge 0$ (the proof follows since $1-\exp(-\gamma \psi)$ is negative definite due to $\exp(-\gamma \psi)$ being pd, and all other terms are nonnegative). – Suvrit Jul 3 '15 at 19:23
• However, I don't want to use the fact that $\exp(-\gamma\psi)$ is positive-definite, since this is exactly I want to prove by the proving the negative-definiteness of $\psi$. Apologies for my ignorance, I just got a bit confused... – nullgeppetto Jul 3 '15 at 19:26