I understand that an $SO(3)$ elements corresponds to a $3 \times 3$ rotation matrix, and it could be mapped to $\omega \in \mathfrak{so}(3)$ (lie algebra), quaternion and Euler angles.

But how to plot an $SO(3)$ element to unit sphere (2-sphere or $S_2$)? Since many paper visualised $SO(3)$ on a sphere, e.g.:

  1. Figure.1 in Evan, S. Gawlik, Embedding-Based Interpolation on the Special Orthogonal Group

  2. Figure 2. in T. Shingel, Interpolation in Special Orthogonal Groups.

Current, I apply a rotation to three axes $\{1,0,0\}$, $\{0,1,0\}$, and $\{0,0,1\}$ respectively and subsequently an $SO(3)$ element corresponds to three points on the sphere instead of one.

  • $\begingroup$ You could model it as the set of ordered pairs of orthogonal points on the sphere I think $\endgroup$ – Akiva Weinberger Aug 6 '17 at 8:02
  • $\begingroup$ Do you mean the three orthogonal axes appearing in many 3D software? $\endgroup$ – whitegreen Aug 6 '17 at 8:08
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    $\begingroup$ Both of your referenced figures clearly explain in their captions: what is being plotted is just the action of the rotations on a fixed vector; i.e. $R(t)v_0$ where $R(t) \in SO(3)$ and $v_0 \in S^2$. $\endgroup$ – Anthony Carapetis Aug 6 '17 at 8:08
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    $\begingroup$ I don't need to answer your question, since Anthony Carapetis has already done that. I just wanted to call your a attention to a confusion that you have shown us. You seem to think that there is a natural map from $SO(3)$ into $\mathfrak{so}(3)$. Actually, it's the other way around: there is a natural map (the exponential map) from $\mathfrak{so}(3)$ onto $SO(3)$. $\endgroup$ – José Carlos Santos Aug 6 '17 at 8:14
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    $\begingroup$ It's possible to identify $SO(3)$ with points on the unit ball, with antipodal boundary points identified. The line going through the origin and a given point is the axis, and the (signed) distance of the point from the centre gives the angle of rotation about that axis. As has been mentioned, since $SO(3)$ has dimension 3, and the unit sphere has dimension 2, there won't be a nice identification between these. $\endgroup$ – Joppy Aug 6 '17 at 14:57

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