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Suppose one starts with a sphere $S$ resting on a ($2$-dimensional) plane $H$ at the origin. A "move" consists of the following: Let $P$ and $Q$ be two points in $H$. Roll the sphere $S$ along a straight line on the plane from the origin to $P$, then roll the sphere along the plane from $P$ to $Q$, and then from $Q$ back to the origin. This gives a map from $\mathbf{R}^4 = H^2$ to $\mathrm{SO}(3,\mathbf{R})$. Is this map surjective? I imagine one could answer this question by a "coordinate bash", but I was hoping for a slicker argument that might shed light on the nature of this map $\mathbf{R}^4 \rightarrow \mathrm{SO}(3,\mathbf{R})$.

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    Nice question. My initial guess would be the answer is no. I'm not seeing a way to generate (non-trivial) rotations in the plane $H$ via this process. It would be nice to have a solution that avoids big computations.2011-04-24
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    @joriki Thanks for pointing out my misunderstanding of the question and the enlightening remark.2011-04-24
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    @Ryan: I've done some numerical experiments, and they suggest that the map is surjective. Randomly choosing points seems to fill the space roughly uniformly. In particular, there seems to be nothing special about rotations in the plane. For example, if $H$ is the $x,y$-plane, choosing $P=(5.7784334335523795,9.040940896629237)$ and $Q=(10.951206469558093,1.6865258234864382)$ approximately yields a rotation through an angle of $1.346444309742066$ around the $z$-axis (assuming a radius of $1$).2011-04-24
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    So the Lie algebra $so(3)$ of infinitesimal rotations is 3-dimensional, and we've chosen a 2-dimensional subspace of it corresponding to rolling along the plane. The question then amounts to asking, for $p, q, r$ in this subspace such that $p + q + r = 0$, does $\exp(p)\exp(q)\exp(r)$ cover $SO(3)$? Unfortunately, this is where my understanding of Lie theory ends.2011-04-24
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    @joriki, yeah, I think I'm starting to change my mind now too.2011-04-24
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    @Rahul: yes, that's the abstract formulation of the problem. I think there's a relatively intuitive way to approach this problem as a comparison problem between Euclidean and spherical geometry, where you are comparing the geometry of spherical and planar triangles.2011-04-24
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    @Ryan: Could you elaborate on that? I had that same idea but couldn't make it work.2011-04-24
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    @Ryan+joriki: If the lengths of the sides of the triange (origin-P-Q) are $2n_1\pi$, $(2n_2+1)\pi$, $(2n_3+1)\pi$ ($n_i$'s are integers) then we get a rotation around the $z$-axis.2011-04-24
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    @joriki: if I find a slick argument I'll certainly post an answer.2011-04-25
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    There is a Mathoverflow question that is not entirely unrelated [here](http://mathoverflow.net/questions/62156).2011-05-01
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    and also [here](http://mathoverflow.net/questions/63574/rolling-a-random-walk-on-a-sphere).2011-05-01
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    A sphere can be rolled in a straight line using any axis of rotation perpendicular to that line (except the axis also perpendicular to the plane). (For example every billiards player has seen a ball with "spin" traveling in a straight line.) I assume you mean to exclude this, i.e. the axis of rotation should be parallel to the plane?2014-08-23

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