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Suppose we have three rotations $r_1$, $r_2$ and $r_3$ in $SO(3)$. Each $r_i$ is a rotation about axis $a_i$ by an irrational multiple of $\pi$.

When considered as unit vectors $a_1$, $a_2$, $a_3$, these axes are not collinear (so they span $\mathbb R^3$).

Does $\{r_1,r_2,r_3\}$ densely generate all of $SO(3)$?

Clearly each $r_i$ densely generates all rotations about axis $a_i$. So my question boils down to: by composing rotations about three non-collinear axes, can one create any rotation in $SO(3)$?

Note that these three axes are not necessarily orthogonal.

Thanks for your help.

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This might be relevant:… – joriki Dec 16 '11 at 23:55
up vote 4 down vote accepted

Already two such rotations are enough. You have rotations through all angles about an axis $a$, so all you need to show is that you can bring any other axis arbitrarily close to $a$; then you can use conjugation to get rotations through all angles about that other axis.

That this is possible is perhaps easiest to see if you represent the axes by their intersections with the unit sphere. Assume there is some open set on the unit sphere that you can't rotate into $a$. You can sweep that set around one of the axes to get a whole annulus that you can't rotate into $a$. You can sweep that annulus around the other axis to get a bigger annulus. By repeating this, alternating between the axes, you can eventually cover the whole sphere, contradicting the assumption that it isn't possible to rotate the open set into $a$.

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I agree that two rotations suffice. Now, what if one rotation is by a rational multiple of pi, and the other is by an irrational multiple of pi? Do these two rotations still densely generate $SO(3)$? – Adam Bouland Jan 26 '12 at 20:09

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