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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. – Ryan Budney Apr 24 '11 at 8:17
@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 $1$). – joriki Apr 24 '11 at 19:07
@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. – user8268 Apr 24 '11 at 20:12
There is a Mathoverflow question that is not entirely unrelated here. – yasmar May 1 '11 at 9:50
and also here. – yasmar May 1 '11 at 9:55

Without loss of generality, let the sphere have radius $1$. Let's call the rotation resulting from rolling along a line segment $AB$ "the rotation along $AB$"; this is a rotation by the angle $|AB|$ around an axis in the plane perpendicular to $AB$.

Consider isosceles triangles with long sides $|OP|=|QO|=2\pi n + x$ and short side $|PQ|=y$, where $x,y\in[0,2\pi)$. Rolling along $OP$ and $QO$ is equivalent to rolling only a distance $x$ along these lines. By choosing $n$ sufficiently large, we can make the directions of $OP$ and $QO$ arbitrarily close to each other and to the direction perpendicular to $PQ$ without changing $x$, and the corresponding rotations will be arbitrarily close to inverses of each other.

Then the resulting rotation is arbitrarily close to the conjugate of a rotation along $PQ$ by a rotation around $PQ$, both with arbitrary angles. The direction of $PQ$ can also be freely chosen by rotating the entire triangle around the origin. Thus, we can get arbitrarily close to any rotation: Choose the direction of $PQ$ perpendicular to the desired rotation axis, choose $y$ as the desired rotation angle, and choose $x$ such that the conjugation rotates the desired rotation axis into the plane.

This reduces the global non-linear problem to a local linear one: It only remains to be shown that the linear transformation locally mapping the four generators of the translations of $P$ and $Q$ in the plane to the three generators of the rotations has full rank; then we can correct for the arbitrarily small parallax error. For sufficiently large $n$, the three translation generators corresponding to the parameters $x$, $y$ and the direction of $PQ$ map to three linearly independent rotation generators unless the rotation is a rotation in the plane; in that case, changing the direction of $PQ$ doesn't change the rotation. But user8268 has already shown in the comments that we can obtain these rotations. Thus it follows that all rotations can be obtained and the map is surjective.

[Edit in response to user8268's comment:] Of course user8268 had only proved that some rotations in the plane are in the image, not all. Fortunately, we have one more translation generator left, and it turns out it maps to an independent rotation generator. For constellations approximating rotations in the plane, the translation generators corresponding to the parameters $x$ and $y$ map to rotation generators that tilt the rotation axis around $PQ$ and change the rotation angle, respectively. The translation generator corresponding to changing the direction of $PQ$ maps to zero. These three can be described as symmetric translation perpendicular to $PQ$ and antisymmetric and symmetric translation in the direction of $PQ$, respectively. That leaves antisymmetric translation perpendicular to $PQ$, i.e. moving $P$ away from the origin and $Q$ towards the origin (or vice versa). This multiplies the rotation along $PQ$ being conjugated by two additional rotations around $PQ$, both in the same direction. By symmetry, the rotation axis of this product cannot have a component perpendicular to the plane. Thus, its rotation axis lies in the plane, and the conjugation rotates it into the plane through $PQ$ perpendicular to the plane. Thus, this translation generator maps to a rotation generator that tilts the rotation axis in that plane, and this is linearly independent from the other two. It follows that in this case, too, the transformation has full rank and we can correct for the arbitrary small parallax error; so the map is indeed surjective.

[Further edit in response the comment:] I didn't spell the full-rank argument out in rigorous detail, and I won't do so now, but I think it's OK and I'll try to justify it. I think it shows more than just that the image is open.

Given a rotation $f_0$ for which we want to find a preimage, we can define a sequence of points $P_n,Q_n$ as described above such that their images approach $f_0$. Then we can look at the images of points in neighbourhoods of these points. Denoting the map in question by $f$, we can write (in suitable, locally non-degenerate coordinates for the codomain):

$$ \begin{eqnarray} f(P_n+\Delta P,Q_n+\Delta Q) &=& f(P_n,Q_n)+f_{1n}(\Delta P,\Delta Q)+f_{2n}(\Delta P,\Delta Q) \\ &=& f_0+e_n+f_{1n}(\Delta P,\Delta Q)+f_{2n}(\Delta P,\Delta Q) \;, \end{eqnarray} $$

where $f_{1n}$ is the linear approximation to $f$ at $(P_n,Q_n)$, $f_{2n}$ is the error term for that approximation and $e_n$ is the error $f(P_n,Q_n)-f_0$.

The linear approximations $f_{1n}$ depend on $n$ but converge to the linear map we get by assuming $OP\parallel QO$ and $OP\perp PQ$. (I implicitly used that in reasoning about their rank.) Likewise, their at most quadratic error terms $f_{2n}$ depend on $n$ but converge to their $OP\parallel QO$ versions. Thus, for sufficiently large $n$, a neighbourhood of $(P_n,Q_n)$ of constant size independent of $n$ is surjectively covered. This is where the situation differs from one in which a set is dense and open but not complete: In that case, the open neighbourhoods necessarily shrink as the missing points are approached. In the present case, by contrast, we can choose $n$ large enough that $-e_n$ lies within the image of $f_{1n}+f_{2n}$ in a neighbourhood of constant size, and at that point the map takes on the value $f_0$.

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there is one point that I don't understand: why can we obtain rotation around the $z$-axis by arbitrary angle? – user8268 Apr 25 '11 at 13:32
@user8268: You're right; I misremembered your comment when I wrote that :-) I'll try to fix that... – joriki Apr 25 '11 at 13:34
@joriki: one more question: if I understand correctly, you prove first that the image is dense (your approximation argument) and then that it's open (your full rank argument). But a dense and open subset is not necessarily the whole $SO(3)$. Is there something I'm missing? (nice arguments btw) – user8268 Apr 25 '11 at 19:04
@user8268: I tried to flesh out that argument -- it's still not completely rigorous, but I hope it's clearer now that I was trying to show more than just that the image is open? – joriki Apr 25 '11 at 22:09
@joriki: sorry, my last comment was probably not clear enough. I understand that you showed: you get a subset $S\subset SO(3)$ which is open and $S\supset SO(3)-SO(2)$ (where $SO(2)$ is the subgroup of rotations around the $z$-axis). Moreover, $S\cap SO(2)$ is dense in $SO(2)$. However, I don't see why $SO(2)\subset S$. – user8268 Apr 26 '11 at 18:41

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