If two rotation matrices commute, do their infinitesimal generators commute too? Suppose that $e^A$ and $e^B$ are two rotations in $\mathrm{SO}(n)$. If $e^{A}e^{B} = e^{B}e^{A}$, can we conclude that $e^{A+B}=e^Ae^B$? More importantly, can we say that $AB=BA$?
I'm particularly interested in the cases when $n=2,3,4$ because I was working on a physics problem that this question was brought up. $n=3$ is the most important case for me.
 A: If $A \in \mathrm{SO}(n)$, then $|\lambda|=1$ for each eigenvalue $\lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 \pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
A: Similar questions --- without the requirement that $e^A,e^B\in SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^B\in SO(n,\mathbb R)$ when $n\ge3$. Consider
$$
A=\pmatrix{0&-2\pi&0\\ 2\pi&0&0\\ 0&0&0},\ B=\pmatrix{0&0&0\\ 0&0&-2\pi\\ 0&2\pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=\pmatrix{0&0&4\pi^2\\ 0&0&0\\ -4\pi^2&0&0}\ne0
$$
and the eigenvalues of $A+B$ are $0$ and $\pm\sqrt{8}\pi i$, so that $e^{A+B}$ is similar to $\operatorname{diag}(1,e^{\sqrt{8}\pi i},e^{-\sqrt{8}\pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2\pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $\exp:\overline{\mathbb{M}}_n\to\mathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
