$\DeclareMathOperator\SO{SO}$I asked this initially in math stack exchange, but thought to ask it here since it is more advanced and related to my research topic. I study optimization on Lie groups from a Riemannian geometry viewpoint. Naturally, one of the Lie groups I use the most is $\SO(3) \subset \mathbb{R}^{3 \times 3}$ equipped with the bi-invariant metric $\langle \xi, \eta\rangle = \frac{1}{2} \operatorname{tr}(\xi^T\eta )$, where $\xi,\eta \in \mathfrak{so}(3)$ are $3 \times 3$ skew-symmetric matrices. The way I parameterize $\SO(3)$ and $\mathfrak{so}(3)$ is not important ofc. I could just as easily use unit quaternions.


Lately in my studies, I've been interested in *commutation error*, and ways to bound it.


I was especially curious if there existed $K > 0$ such that $$\left\|\log(R^{-1}S)\right\| \leq K \left\|\log(S) - \log(R)\right\|$$ for all $R,S \in \SO(3)$ such that $\log(S), \log(R), \log(R^{-1}S)$ exists. Here, $\log(\cdot)$ is the Lie logarithm. Under my parameterization, it coincides with the matrix logarithm.


After generating random rotation matrices $R,S$ in a script, to my surprise I found $K=1$ *every single time*. I have yet to randomly generate or analytically derive a counterexample. Once again, I am only focusing on pairs $(R,S)$ such that each term above exists.


If true, this is a very powerful inequality! I'm surprised to see very little discussion about online. Surely, someone must've noticed this. Any idea how I could prove this?



Is it true $\left\|\log(RS)\right\|≤\left\|\log(R)+\log(S)\right\|$ for all $R,S \in \mathrm{SO}(3)$, where $\|\cdot\|$ is the Frobenius norm?
Now that I have written out the completely elementary proof above for the quaternions and for $\mathrm{SO}(3)$, I feel that I should point out that the statement $|\log(ab)|\le |\log(a) + \log(b)|$, suitably interpreted, is true for any compact connected Lie group $G$ endowed with a biïnvariant Riemannian metric. The proof is not completely elementary; but it only relies on facts that are covered in any first course on Riemannian geometry.


Let $G$ be a compact, connected Lie group with Lie algebra ${\frak{g}}=T\_eG$, and let $<,>:\frak{g}\times\frak{g}\to\mathbb{R}$ be an $\mathrm{Ad}(G)$-invariant positive definite inner product on $\frak{g}$. Let $|v| = <v,v>^{1/2}$ for $v\in\frak{g}$, as usual. There is a unique biïinvariant Riemannian metric $g$ on $G$ that equals $g\_0=<,>$ on ${\frak{g}} = T\_eG$. Let $d:G\times G\to \mathbb{R}$ be the associated distance function of $g$. It satisfies $d(ac,bc)=d(ca,cb)=d(a,b)$ for all $a,b,c\in G$.


Because the sectional curvature of $g$ is non-negative, the Lie group exponential map (which is equal to the exponential map of $g$ at $e\in G)$, i.e., $\exp:{\frak{g}}\to G$, satisfies $\exp^\*(g)\le g\_0$ (as smooth quadratic forms on $\frak{g}$). (For a proof, see, for example, Helgason's *Differential Geometry, Lie Groups, and Symmetric Spaces*.)


Now, Let $U\subset G$ be the connected, dense open set of points $a\in G$ for which there is a unique $g$-length-minimizing geodesic from $e$ to $a$. (The complement of $U$ is a closed algebraic set in $G$ called the *cut locus* of $G$.) Then for each $a$ in $U$, there is a unique element $\log a\in\frak{g}$, such that the curve $\gamma(t) = exp\bigl(t\log(a)\bigr)$ for $0\le t\le 1$ is that unique $g$-length-minimizing geodesic. Then $\log:U\to\frak{g}$ is a smooth inverse to $\exp$ in the sense that $\exp(\log a) = a$.


By construction, $|\log(a)| = d(a,e)$, so, in particular, the function $a\mapsto|\log(a)|$ extends continuously to all of $G$, even though $\log$ does not.


Now, for $a,b\in U$, we have the following estimate:
$$
|\log(ab^{-1})| = d(e,ab^{-1}) = d(b,a) = d(a,b)
$$
and the righthand number is less than or equal to the length of the curve
$\gamma(t) = \exp\bigl((1{-}t)\log(a)+t\log(b)\bigr)$ for $0\le t\le 1$. However, because $\exp^\*(g)\le g\_0$, the $g$-length of $\gamma$ is less than or equal to the $g\_0$-length of the line segment $\alpha(t) = (1{-}t)\log(a)+t\log(b)$ in $\frak{g}$, which is $|\log(a)-\log(b)|$. (Note that $\alpha(t)$ does not have to lie in the image of $\log$ for all $0<t<1$ in order for this statement to hold.)


Thus, $|\log(ab^{-1})| \le |\log(a)-\log(b)|$ for all $a,b\in U$, i.e., for all $a$ and $b$ for which $\log$ is defined. Replacing $b$ by $b^{-1}$ and using the fact that $\log(b^{-1}) = -\log b$ yields
$$
|\log(ab)| \le |\log(a)+\log(b)|
$$
for all $a,b$ in the domain of $\log:U\to \frak{g}$. (It doesn't matter whether $ab$ is in the domain of $\log$.)


