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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.

stressed out
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  • For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{\theta S(v)}$ so $A=\theta S(v)$ – Widawensen Jan 08 '19 at 11:54
  • @Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$? – stressed out Jan 08 '19 at 11:57
  • If general form of both matrices is $A=\theta_1 S(v)$ and $B=\theta_2 S(v)$ they must commute. – Widawensen Jan 08 '19 at 11:59
  • @Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis? – stressed out Jan 08 '19 at 12:01
  • From this I do not know but it can be concluded from quaternion representation of rotation . See https://math.stackexchange.com/questions/1863176/the-relation-between-axes-of-3d-rotations – Widawensen Jan 08 '19 at 12:02
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    Sort of obvious, but if $a\in so(3)$ and $||a||=\pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$. – Charlie Frohman Jan 08 '19 at 12:21

2 Answers2

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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.

user1551
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  • Wait... Aren't all elements in $\mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $\exp(\cdot): \mathrm{Skew}(n) \to \mathrm{SO}(n)$ is not surjective in general? – stressed out Jan 08 '19 at 15:31
  • @stressedout The mapping is surjective. Is there anything wrong with that? – user1551 Jan 08 '19 at 15:49
  • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric? – stressed out Jan 08 '19 at 16:20
  • @stressedout I don't understand your point. The $A$ in my answer is skew-symmetric. – user1551 Jan 08 '19 at 16:31
  • Paragraph 2: "The answer is still no even if you require that $A,B$ are skew-symmetric"... If $e^A,e^B \in \mathrm{SO}(n)$, then $A,B$ are automatically skew-symmetric. Aren't they? So, why the word "even"? Why paragraph 2? – stressed out Jan 08 '19 at 16:33
  • @stressedout Surely, when talking about infinitesimal generators of members of the special orthogonal group, we mean skew-symmetric matrices, but in general, if $e^A\in SO(n,\mathbb R)$ for some real square matrix $A$, $A$ is not necessarily skew-symmetric. That's why the second paragraph is phrased that way. – user1551 Jan 08 '19 at 16:41
  • Oh, so what you're saying is that $\exp(\cdot): \mathrm{GL_n}(\mathbb{R}) \to \mathrm{SO}(n)$ is well-defined and it fails to be injective. Have I understood you correctly this time? I thought that $\exp(\cdot)$ was well-defined only on $\mathrm{Skew}(n)$. – stressed out Jan 08 '19 at 16:50
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    @stressedout Close but not exactly. I mean $\exp(\cdot):M_n(\mathbb R)\to GL_n(\mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^X\in SO_n(\mathbb R)$, such as $X=\pmatrix{0&-2\pi\ 2k^2\pi&0}$ where $k$ is any integer strictly greater than $1$. – user1551 Jan 08 '19 at 16:59
  • Thanks. But shouldn't we show that $\exp(\cdot): \mathrm{M_n}(\mathbb{R}) \to \mathrm{SO}(n)$ fails to be injective? Why $\mathrm{GL_n}(\mathbb{R})$ instead of $\mathrm{SO}(n)$? Maybe all the points in the co-domain that have more than one pre-image in $M_n(\mathbb{R})$ are in $\mathrm{GL_n}(\mathbb{R})-\mathrm{SO}(n)(\mathbb{R})$? Where can I learn about his stuff properly? – stressed out Jan 08 '19 at 17:03
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    @stressedout We don't write $\exp:M_n(\mathbb R)\to SO_n(\mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $\exp\pmatrix{0&1\ 0&0}=\pmatrix{1&1\ 0&1}\notin SO(2)$). As for learning, sorry, I haven't any suggestions. – user1551 Jan 08 '19 at 17:09
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    OK. Thank you for answering all my questions patiently. I appreciate it lot. :) – stressed out Jan 08 '19 at 17:17
  • @user1551 We have3d rotations by angle $pi$ with perpedincular axes. They commute ( for x and z axis matrices are diagonal), but their skew-symmmetric matrices for axes ( take them for x and z axis) don't commute with argumentation similar to presented by you. What this mean for the problem? Congruencies for $\pi$ should also be included ( not only $2\pi$) ? Spectra A and B should have been $\pi i $ - congruence free ? ( for perpenpicular axes) – Widawensen Jan 14 '19 at 15:15
  • @Widawensen I see. So, you mean $K_1=\pmatrix{0&0&0\ 0&0&-\pi\ 0&\pi&0}$ does not commute with $K_2=\pmatrix{0&-\pi&0\ \pi&0&0\ 0&0&0}$, but $e^{K_1}=\operatorname{diag}(1,-1,-1)$ commutes with $e^{K_2}=\operatorname{diag}(-1,-1,1)$. This does not contradict what was mentioned in the answer, because the spectra of $K_1$ or $K_2$ are not $2\pi i$-congruence free. You see, the eigenvalues of $K_1$ are $(\lambda_1,\lambda_2,\lambda_3)=(0,\pi i,-\pi i)$ and $\lambda_2-\lambda_3$ is an integer multiple of $2\pi i$. – user1551 Jan 14 '19 at 15:41
  • @user1551 ok. Situation now is clear.Thank you for explanation. – Widawensen Jan 14 '19 at 15:42
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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

Fred
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