An element of $SO(n)$ represents an rotation so that it must have identity with $2\pi$-like additional rotation. On the other hand, the elements of lie algebra $so(n)$ construct an noncompact vector space. So I think $so(n)$ does not have specific periodic structure naturally.

I want to use $SO(n)$ with exponential coordinates, but cannot see the all points on $\log(SO(n))$ as independent ones because of the structures above. Could I have any suggestions or topics to study for handling the situations correctly?

edit1: I created a simple example below.

Consider $SO(3)$, we can use Rodrigues' formula for mapping lie algebra $\Omega \in so(3)$ to a rotational matrix ($\in SO(3)$) as follows, $$R = I + \frac{\sin \theta}{\theta}\Omega + \frac{1-\cos \theta}{\theta^2}\Omega^2,$$ $$\Omega = \begin{pmatrix}0 & \omega_1 & \omega_2 \\ -\omega_1 & 0 & \omega_3 \\ -\omega_2 & -\omega_3 & 0\end{pmatrix}, \quad \theta = \sqrt{\omega_1^2 + \omega_2^2 + \omega_3^2}.$$ Then consider the case where $\theta = 2\pi k, \ k=1,2,\cdots$. We have the same rotations (identity matrices) for various $\Omega$s satisfying $$\omega_1^2 + \omega_2^2 + \omega_3^2 = (2\pi k)^2, \quad k=1,2,\cdots.$$ Because the lie algebra is a vector space, I think this represents multiple "sphere shells". So, when we handle the $SO(3)$ through exponential coordinates, we cannot see the all points in $so(3)$ as different rotations, but have to take the identity for each sphere shells into account.

I've got a speculation from the creation of the example that we could have consistency by considering only $|\Omega| < 2\pi$ points. Is that correct? Does the set $\{R = \exp(\Omega) \ | \ \Omega \in so(3), |\Omega| < 2\pi\}$ cover $SO(3)$?

edit2: I've had the linear algebraic viewpoint of my issue.

Any real skew symmetric matrix $\Omega$ can be decomposed as follows, $$\Omega = U\Lambda U^T$$ where $U$ is a unitary matrix and $\Lambda$ is a diagonal matrix valued as $$\Lambda = \text{diag}(i\theta_1, -i\theta_1, \cdots, i\theta_{n/2}, -i\theta_{n/2}).$$ (for $SO(2n+1)$, the remaining dimension of diagonal matrix is padded with a zero) We can easily compute exponential mapping of these decomposition as, $$R = U \exp(\Lambda) U^T $$ where $$\exp(\Lambda) = \text{diag}(e^{i\theta_1}, e^{-i\theta_1}, \cdots, e^{i\theta_{n/2}}, e^{-i\theta_{n/2}}).$$ From this result, following skew symmetric matrices lead the same rotational matrix by exponential mapping, $$U\Lambda({k_i})U^T \\ \Lambda({k_i}) = \text{diag}(i(\theta_1 + 2\pi k_1), -i(\theta_1 + 2\pi k_1), \cdots, i(\theta_{n/2} + 2\pi k_{n/2}), -i(\theta_{n/2} + 2\pi k_{n/2})) \\ k_1, \cdots, k_{n/2} \in \mathbb{Z}. $$ These matrices are not the same in $so(n)$. Now we have the infinite number of points on lie algebra that correspond to the same point on lie group through exponential mapping. I speculate that, with all possible unitary matrix $U$ and $\theta_1 \cdots \theta_{n/2} \in [-\pi, \pi)$, we could cover $SO(n)$ (at least, without at most countable points) bijectively.

I still wonder wheter such results are able to lie-theoretically interpreted. But most of my questions have been resolved. Thank you all for reply.

I'm glad if I have another suggestions or comments.

  • $\begingroup$ Can you be more specific about what you want to do? $\endgroup$ – Qiaochu Yuan Nov 28 '14 at 9:26
  • 1
    $\begingroup$ I'm not entirely sure what you want to do, exactly, but note the following: (1) Since $SO(n)$ is compact, the geodesic distance of any point from $I$ is bounded above. In general (for $n > 1$) there is no open subset of $\mathfrak{so}(n)$ that maps homeomorphically via $\exp$ or any other map onto $SO(n)$, so there is no hope of "consistency" in your sense. (2) The group $Spin(n)$ has the same Lie algebra as $so(n)$ but its geodesics are "longer", i.e., the behavior you call its "periodic structure" is different. $\endgroup$ – Travis Nov 29 '14 at 12:17
  • $\begingroup$ Thank you for your reply. I've got your advice. I think I have to study more about differential geometric view of Lie groups. Now I've edited another view of my question and got some answer. If you would give me additional advice, I was happy. $\endgroup$ – kohta Dec 1 '14 at 13:21

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