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So $T^2$ double covers $S^2$, and $S^2$ double covers $\mathbb{R}P^2$. Therefore, $T^2$ quadruple covers $\mathbb{R}P^2$.

I am looking at $\mathbb{R}P^3$ because I am interested in rotations. The Euler angles, for example parametrize $\mathbb{R}P^3$ once you make a restriction on one of the angle's ranges. Originally though, they parametrize $T^3$. Judging from the restriction of the angle, I have come to the conclusion that $T^3$ double covers $\mathbb{R}P^3$. Is this true? And how else would I prove it?

Thanks

Johnver
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    $T^2$ doesn't cover $S^2$ or $\mathbb{R}P^2$. In fact, as $S^2$ is simply connected, the only connected space which covers it is itself. In addition, $T^3$ doesn't cover $\mathbb{R}P^3$ - the only connected covers of $\mathbb{R}P^3$ are itself and $S^3$. – Jason DeVito - on hiatus Jul 28 '16 at 20:18
  • Maybe I'm misusing the language then. Do you understand what I'm saying? What am I looking for? I can put two coordinates on $T^2$. If I use those same coordinates on $S^2$, then they are redundant twice-over at every point in $S^2$. – Johnver Jul 28 '16 at 20:20
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    What do you mean by "use those same coordinates on $S^2$"? How do you use a pair of coordinates to get a point of $S^2$? – Eric Wofsey Jul 28 '16 at 20:20
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    I think I understand. Spherical coordinates give a natural surjective map from $T^2$ to $S^2$, but this map is not a covering as it has some "singularities". For example the preimage of the north pole in $S^2$ has uncoutable cardinality. But, at most points, it is two to one. Is the question about the analgous construction for $T^3\rightarrow S^3$? – Jason DeVito - on hiatus Jul 28 '16 at 20:23
  • Say I use $\phi$ and $\theta$ to parametrize $T^2$. They range from $0$ to $2\pi$ each and are cyclic. Now I want to use the same coordinates on $S^2$. Then I notice that, if $\phi$ is the polar angle, then $(\theta, \phi)$ corresponds to the same point as $(\theta + \pi, 2\pi-\phi)$. Hence what I incorrectly term "a double cover." – Johnver Jul 28 '16 at 20:25
  • Jason, yes, I am trying to draw the analogue. Also I am trying to understand the coordinates that "originate" on $T^3$ on $\mathbb{R}P^3$. – Johnver Jul 28 '16 at 20:27
  • I'm not sure about the connection with Euler angles and T^3, but there is certainly such a thing as spherical coordinates in higher dimensions :https://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates. One of angles goes $0$ to $2\pi$, while all the others go $0$ to $\pi$. So, it seems as though the map $T^3$ to $S^3$ should be generically $4$ to $1$. – Jason DeVito - on hiatus Jul 28 '16 at 20:32
  • Possibly of interest (not a duplicate): Why spherical coordinates is not a covering? – Andrew D. Hwang Jul 30 '16 at 12:13

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There is a map from $T^3$ to $SO(3)$, called "Euler angles" -- basically, you use each angular coordinate to specify "yaw", "pitch", and "roll". This map is, however, singular, much like the map from $T^2$ to $S^2$ discussed in the comments. The singularities are often referred to by the generic name "gimbal lock". You can read about this in books on robotics, or in Wikipedia, or in some computer graphics books.

Note that the euler-angle parametrization, because of its singularities, is not a covering in the sense of covering-spaces, nor does it have many other properties you might hope for, like some sort of equivariance. It's a totally great parameterization near the identity, and gets worse and worse as you get farther from $I$.

What order do pitch, roll, and yaw come in? There are six possible orderings, and I'm pretty sure all have bene used at one time or another. There's no firm consensus.

This particular kind of parameterization of $SO(3)$ turns out to be not very useful for a lot of computer graphics applications, nor for many applications in mathematics. For control of "flying things" in games, it has some value, as it mimics airplane controls (somewhat).

InsideOut
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John Hughes
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  • $SO(3)$ topologically is $\mathbb{R}P^3$. Euler angles start out on $T^3$, but then are seen to provide the same rotation for any given 3-tuple of angles. To fix this, the middle Euler angle, as you say "pitch" is limited to $180^\circ$ of range.

    I guess my question is: how is $\mathbb{R}P^3$ obtained from $T^3$, and does my argument above imply that $T^3$ "double covers" $\mathbb{R}P^3$?

     – Johnver Jul 28 '16 at 20:37
  • Sure. So your map from $T^3$ to "euler angles" involves a divide-by-two in the middle coordinate. That doesn't really change anything. Your argument does NOT show that $T^3$ double-covers $SO(3)$; indeed, the only double-cover of $SO(3)$ corresponds to an index-2 subgroup of $\pi_1(SO(3))$, Since $\pi_1(SO(3)) = \mathbb Z / 2 \mathbb Z$, this subgroup is the trivial one, and the double-cover is $S^3$, which is often described the the unit-quaternion group. Short answer: No. $T^3$ absolutely does not double-cover $SO(3)$ in the sense of covering spaces. – John Hughes Jul 28 '16 at 20:45