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

$$\DeclareMathOperator\SO{SO}$$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?

• I’m not certain, but the convexity properties of geodesic distance has been extensively studied in optimal transport. In particular, the following paper by Guillen and Kitagawa might be relevant: arxiv.org/pdf/1212.4865.pdf. If this is indeed what you need, your question would reduce to showing this property in constant positive curvature. I don’t know the original reference offhand, but that fact is shown in “Nearly round spheres look convex.” May 12 at 2:00
• I haven't seen the inequality you posted, but it is reminiscent of inequalities I mentioned. And if you can prove your inequality under the additional assumption that geodesic distance is strongly convex, those references might establish the convexity properties you need. Sorry for the inconclusive remarks; if I am able to verify the inequality I'll post an answer. May 12 at 17:27
• @GabeK Ah, I see! I'm currently reading through "nearly round spheres look convex", which I think is great since $SO(3)$ is a subset of the 3-sphere (really working with the unit quaternions might be easier). However, the paper is quite beyond my current understanding of Riemannian geometry. I would greatly appreciate a point to any theorem in that paper that could help me prove this inequality! May 12 at 17:27
• If you can clarify the exact form of "geodesic distance is $\mu$-strongly geodesically convex" that you need to establish, I can try to see if this follows from that paper. The relevant part would be Subsection 5.2, although I agree that paper is very challenging to read. (I have tried and failed many times to really understand it.) May 12 at 17:36
• @GabeK I just now included a definition for geodesic strong convexity in my post. I wrote the version when $f$ is differentiable. There is a non-differentiable version of strong geodesic convexity, but since squared geodesic distance is smooth (on a small enough domain), then we can just use that much nicer version. Thanks for the relevant part. Maybe I can check out citations of that paper to find something simpler! May 12 at 17:44

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:

• Very nice argument! May 13 at 0:14

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 $$0 while $$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, $$|c|\le 1$$ 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 $$0 and $$G(a,b,\epsilon-1)>0$$ for $$\epsilon>0$$ sufficiently small.

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

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

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

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