Show $\langle \log(R), \log(R^{-1}S) \rangle \geq \langle \log(R), \log(S) - \log(R) \rangle $ for all $R,S \in \mathrm{SO}(3)$

Question

$\DeclareMathOperator\SO{SO}$I have a similar question to one I asked a few days ago. Lately, I've been researching Lie groups equipped with bi-invariant Riemannian metrics. One common object is $\SO(3)$, represented with $3 \times 3$ rotation matrices, equipped with the bi-invariant metric $$\langle \xi, \eta \rangle = \frac{1}{2} tr(\xi^T \eta).$$ Here, $\xi,\eta \in \mathfrak{so}(3)$ are skew-symmetric. During some research, I came across the following inequality I wanted to prove:

$$\langle \log(R), \log(R^{-1}S) \rangle \geq \langle \log(R), \log(S) - \log(R) \rangle$$ for all $R,S \in \SO(3)$ where the above inequality is well-defined. A similar, yet stronger, inequality I noticed was $$\langle \log(R), \log(R^{-1}S)\rangle + \frac{1}{2} \|\log(R^{-1}S)\|^2 \geq \langle \log(R), \log(S)-\log(R) \rangle + \frac{1}{2} \|\log(S) - \log(R)\|^2.$$

I found this quite unintuitive due to the fact that $\|\log(R^{-1}S)\| \leq \|\log(S) - \log(R)\|$. As a result, I would expect the inequality to be flipped. But every pair of rotation matrices I generated, this inequality held.

My attempt at a proof was utilizing the fact that geodesic distance is $\mu$-strongly geodesically convex, and hence is lower bounded by its 2nd order approximation (this was how I originally came across these inequalities).

By strongly convex, I mean the following. Let $(M,g)$ be a complete Riemannian manifold. A differentiable function $f:Q \subset M \to \mathbb{R}$ is called $\mu$-strongly convex on $(M,g)$ if $Q$ is geodesically convex and for all $x,y \in Q$, we have $$f(x) \geq f(y) + \langle \nabla f(y), \log_y(x)\rangle_y + \frac{\mu}{2}\|\log_y(x)\|^2_y.$$

Here, a subset $Q \subset M$ is called geodesically convex if for all $x,y \in Q$, there exists a unique minimizing geodesic connecting $x$ to $y$, and that geodesic is contained in $Q$.

If we fix $y \in M$ and set $f(x):=\frac{1}{2}d^2(x,y)$, one can show that $f$ is $\mu$-strongly geodesically convex on $B \subset M$, where $B$ is a geodesically convex geodesic ball centered at $y$. Here, $\mu$ also has a closed form solution if $M$ has bounded above sectional curvature.

However, I was not able to prove any of these. Any ideas?

Answer 1

Making the substitution $R = e^X$, $R^{-1} S = e^Y$, your first inequality can be rearranged as $$\langle X, \log(e^X e^Y) - X - Y \rangle \leq 0$$ where $X,Y \in \mathfrak{so}(3)$ have eigenvalues of magnitude less than $\pi$ (to be in the range of the logarithm) but are otherwise arbitrary. Since $\log(e^X e^Y) = - \log(e^{-Y} e^{-X})$ this is also equivalent (after replacing $X,Y$ with $-Y,-X$) to $$\langle Y, \log(e^X e^Y) - X - Y \rangle \leq 0.$$

In the range of convergence of the Baker-Campbell-Hausdorff formula $$ \log(e^X e^Y) = X + \int_0^1 \psi( e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} ) Y\ dt$$ with $\psi(x) := \frac{x \log x}{x-1}$, one has $$ \langle Y, \log(e^X e^Y) - X - Y \rangle = \int_0^1 \langle Y, (\psi( e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} )-1) Y \rangle \ dt$$ and the inequality follows from the spectral theorem since $\mathrm{Re}( \psi(e^{i\theta})-1 ) = \frac{\theta/2}{\tan(\theta/2)} - 1\leq 0$ for all $-\pi \leq \theta \leq \pi$ (note that $e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y}$ is unitary on $\mathfrak{so}(3)$).

I don't know however how to extend this argument beyond the range of convergence of the Baker-Campbell-Hausdorff formula.

EDIT: a similar analysis also works for the second inequality, again in the range of convergence of (two applications) of BCH. Writing $Z = \log(e^X e^Y) - X - Y$, the inequality can be written as $$ \langle X, Y \rangle + \frac{1}{2} \|Y\|^2 \geq \langle X, Y+Z \rangle + \frac{1}{2} \| Y+Z \|^2$$ which rearranges as $$ \langle X,Z \rangle + \langle Y,Z \rangle + \frac{1}{2} \langle Z,Z \rangle \leq 0. \quad (1)$$ As before we have $$ Z = \int_0^1 (\psi( e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} )-1) Y\ dt$$ in the range of the BCH formula, and by replacing $X,Y$ with $-Y,-X$ as before we also can check that $$ Z = \int_0^1 (\psi( e^{-\mathrm{ad}_Y} e^{-s \mathrm{ad}_X} )-1) X\ ds$$ (in the range of a second BCH formula) and so we can write the left-hand side of (1) as $$ \int_0^1 \int_0^1 \langle X, (\psi( e^{-\mathrm{ad}_Y} e^{-s \mathrm{ad}_X} )-1) X \rangle + \langle Y, (\psi( e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} )-1) Y \rangle + \frac{1}{2} \langle (\psi( e^{-\mathrm{ad}_Y} e^{-s \mathrm{ad}_X} )-1) X, (\psi( e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} )-1) Y \rangle\ ds dt.$$ Estimating the last term using the arithmetic mean-geometric mean inequality $|\langle u,v\rangle| \leq \|u\| \|v\| \leq \frac{1}{2} \|u\|^2 + \frac{1}{2} \|v\|^2$ followed by the spectral theorem, we see that we will be done if we can establish the inequality $$ \frac{1}{4} |\psi(e^{i\theta})-1|^2 \leq \mathrm{Re}(1 - \psi(e^{i\theta})) \quad (2)$$ for $-\pi \leq \theta \leq \pi$, which can be checked numerically:

Answer 2

This is more an answer to the corresponding question for the quaternions: Can one show that, for unit quaternions $p,q\in S^3\subset\mathbb{H}\simeq\mathbb{R}^4$, one has the inequality $$ \langle\log p,\log p + \log q - \log (pq)\rangle \ge 0 $$ when it makes sense, i.e., when $p$, $q$ and $pq$ are not equal to $-1\in S^3$?

The answer is 'yes', and I suspect that this can be used to prove the originally requested inequality on $\mathrm{SO}(3)$ (as rewritten by Terry Tao): $$\langle\log R,\log R + \log S - \log (RS)\rangle \ge 0$$ when $\mathrm{tr}(R)$, $\mathrm{tr}(S)$, and $\mathrm{tr}(RS)$ are all greater than $-1$. (In other words, one doesn't have to assume anything about the range of convergence of the BCH formula.)

The above claim is based on the fact that $\log$ can be defined on $S^3\setminus\{-1\}$ by the formula $$ \log q = \frac{\cos^{-1}\bigl(\mathrm{Re}(q)\bigr)}{\sqrt{1-\mathrm{Re}(q)^2}}\,\mathrm{Im}(q). $$ (Note that the function $f(x) = \frac{\cos^{-1}(x)}{\sqrt{1-x^2}}$, nominally defined only on $(-1,1)$, extends smoothly and analytically to $(-1,\infty)$, and I assume that this extension is made).

Since the inequality clearly holds when either $p$ or $q$ is equal to 1, one can assume that $p = e^{au}=\cos a + \sin a \,u$, $q=e^{bv}=\cos b + \sin b\,v$ where $u$ and $v$ are unit imaginary quaternions satisfying $u\cdot v = c = \cos\theta$, and $0<a,b<\pi$ while $0\le \theta\le \pi$, i.e., $|c|\le 1$. The condition that $pq\not=-1$, i.e $q\not=-\bar p$ translates to the inequality $\cos a\cos b- (\sin a\sin b)\,c \ge -1$ being strict. Then the desired inequality above translates into the inequality $F(a,b,c)\ge 0$ for $0<a,b<\pi$, $|c|\le 1$ and $\cos a\cos b- (\sin a\sin b)\,c > -1$, where $$ F(a,b,c) = a(a{+}bc) - {\textstyle\frac{a\cos^{-1}\bigl(\cos a\cos b{-}(\sin a\sin b)\,c\bigr)}{\sqrt{1-\bigl(\cos a\cos b{-}(\sin a\sin b)\,c\bigr)^2}}}\bigl(\sin a\cos b + (\cos a\sin b)c\bigr). $$

With some work and elementary calculus, the inequality $F(a,b,c)\ge 0$ can be established via the following steps:

Set $$ G(a,b,c) = \frac{\sqrt{1-\bigl(\cos a\cos b{-}(\sin a\sin b)\,c\bigr)^2}}{a} F(a,b,c), $$ so that $G$ is continuous in the desired ranges of $a$, $b$, and $c$. Then the following four properties of $G$ are easily established:

1. $G(a,b,-1) = 0$ when $0<a,b<\pi$ and $G(a,b,\epsilon-1)>0$ for $\epsilon>0$ sufficiently small.

2. $G(a,b,1) = 0$ when $0<a,b<\pi$ and $a+b\le \pi$

3. $G(a,b,1) = 2\pi|\sin(a{+}b)|>0$ when $0<a,b<\pi$ and $\pi < a{+}b < 2\pi$

4. When $0<a,b<\pi$, $\frac{\partial G}{\partial c}(a,b,c)=0$ has at most one root $c$ in the range $-1<c<1$.

From these facts, it follows that $G(a,b,c)\ge0$ for $0<a,b<\pi$ and $|c|\le 1$, and the same conclusion then follows for $F(a,b,c)$.