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Is there a rotation representation that can also represent "turns", instead of collapsing coincident rotations into the same representation?

In 2D, a simple angle satisfies this, as it can have additional multiples of $2\pi$. For example, rotating by a turn and a half would be $3\pi$.

Is there something similar for 3D rotations? Does the concept even make sense there? Quaternions don't work for this since they only have two representations of any given rotation. Rotation vectors ($\theta\hat{e}$) seem to work, though they are very hard to work with.

EDIT: My objective with this is to extend quaternion spherical interpolation to rotations of more than 180° in terms of beginning and end "orientation with turns" objects, so you could, for example, interpolate over an entire revolution using the same machinery as you would with normal small rotation interpolation.

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There is an awful lot about rotations in computer graphics, and correct use of quaternions, at – Will Jagy Mar 22 '13 at 1:25
Evidently robotics as well. – Will Jagy Mar 22 '13 at 1:28
up vote 5 down vote accepted

Since you haven't described what you plan to do with this representation, I'm not sure what method would work well.

One of the problems with representing "turns" in more than two dimensions is that you don't have much in the way of discrete invariants. This is because the fundamental group of $SO(n)$ only has two elements when $n \gt 2$. Indeed, a double rotation in any direction can be continuously deformed to the identity rotation. This means that two turns are indistinguishable from zero turns, if you consider paths that can be deformed to each other to be equivalent.

If you want to keep track of how your rotation came about, here are two suggestions:

  1. You can use based paths in the group of rotations, i.e., continuous maps from a real interval into rotation matrices.
  2. If your paths are smooth, you can take the derivative, and use paths in velocity space, i.e., the Lie algebra of the rotation group.

Both methods give you points in an infinite dimensional path space.

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Please leave an explanation when downvoting. – S. Carnahan Mar 21 '13 at 7:54
Regarding Scott's suggestion 2: Just be careful to note that, if you have a differentiable curve of $n$-by-$n$ rotation matrices $A(t)$ with $A(0)=I_n$ and you set $a(t) = A(t)^{-1}A'(t)$, then $a(t)$ will, indeed, be a curve in the vector space of $n$-by-$n$ skew-symmetric matrices, but you won't (usually) have $A(t) = \mathrm{exp}\bigl(a(t)\bigr)$ when $n>2$, so this doesn't really define any sort of 'lift' (even multi-valued) for $\mathrm{SO}(n)$ when $n>2$. You'll usually need the whole $a(t)$ for $0\le t\le T$ to figure out $A(T)$. – Robert Bryant Mar 21 '13 at 12:41
I updated the question with what I'm trying to achieve. – Forrest Mar 21 '13 at 23:05

The resolution in 2D that you suggested may also be viewed as going from the circle to its universal covering space: $\mathbb{R}\to S^1$.

SO the the same trick should work: take the universal cover of SO(3).

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In posing the question, Forrest pointed out that there are only two quaternions representing a given rotation, i.e. he already thought about the universal covering space of $SO(3)$ and noticed that it doesn't do what he wants. – Ben McKay Mar 21 '13 at 18:04

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