A quote from Wikipedia's article on the Rotation group:

Consider the solid ball in $\mathbb{R}^3$ of radius $\pi$ [...]. Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball. Rotation through angles between $0$ and $-\pi$ correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through $\pi$ and through $-\pi$ are the same. So we identify [...] antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group.

So far, so good. This illustrates $SO(3)\cong \mathbb{RP}^3$.

These identifications illustrate that $SO(3)$ is connected but not simply connected. As to the latter, in the ball with antipodal surface points identified, consider the path running from the "north pole" straight through the center down to the south pole. This is a closed loop, since the north pole and the south pole are identified. This loop cannot be shrunk to a point, since no matter how you deform the loop, the start and end point have to remain antipodal, or else the loop will "break open".

I believe that $SO(3)$ is connected but the "intuitive argument" for $\pi_1(SO(3))\neq 0$ is not clear to me: The starting point at the "north pole" is a rotation of $\pi$ counterclockwise around the $z$ axis. This agrees with the "south pole", a rotation of $\pi$ clockwise around the $z$ axis. So the described path is a full $2\pi$ rotation counterclockwise around the $z$ axis, stating not in the identity position. Why isn't this homotopic to the trivial path? Antipodal points are identified, so what does "start and end point have to remain antipodal, or else the loop will "break open"" mean?

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    $\begingroup$ I changed the title to make it into a question, since "How to demostrate $SO(3)$" didn't make much sense. $\endgroup$ Apr 29 '10 at 13:27
  • $\begingroup$ mathoverflow.net/questions/14776 $\endgroup$ Apr 29 '10 at 13:39
  • $\begingroup$ Voting to close as no longer relevant, as essentially the same question is asked and answered in the link above. $\endgroup$ Apr 29 '10 at 13:44
  • $\begingroup$ I vote not to close, as that question is different (I am the asker). In my case I knew how to compute the fundamental group of SO(3), the problem was more to understand precisely where (in what space) is the loop drawn when the waiter moves the dish. A modelization problem, in some sense. $\endgroup$ Apr 29 '10 at 14:02
  • $\begingroup$ The Dirac belt trick = Feynman plate trick in Andrea's question does demonstrate that SO(3) has a connected double cover --- your quote does this as well. And this cannot happen for simply-connected spaces. $\endgroup$ Apr 29 '10 at 16:46

A loop is homotopically trivial if it can be continuously deformed to the constant loop. This means that at every step of the deformation (every "instant in time") you still have a loop. There are two kinds of loops on the unit ball with antipodal identifications in the boundary: either it's also a loop in the ball (without identifications) or else it starts and ends at antipodal points. The example curve in the question is of the latter kind. It seems intuitive that any continuous deformation of this curve which remains closed has to still connect antipodal points, since you cannot move the ends closer to each other -- which is what you'd have to do in order to get a contractible loop -- while keeping it a closed curve.

  • $\begingroup$ This is a good answer, but it is perhaps worth noting that the first step, as described in the question itself, is identifying SO(3) as being the 3-dimensional real projective space, which is just the 3-dimensional ball with antipodal points on the boundary identified. Then your answer is explaining why real projective space has a homotopically nontrivial loop. $\endgroup$
    – Deane Yang
    Apr 29 '10 at 13:33
  • $\begingroup$ You're right, of course; but my reading of the question was that he OP was OK with the identification of $\mathrm{SO}(3)$ with the antipodal identification of the unit ball. (The further identification with $\mathbb{RP}^3$ is perhaps not relevant.) $\endgroup$ Apr 29 '10 at 13:40
  • $\begingroup$ oops. Yes, you're right. I should have reread the question. $\endgroup$
    – Deane Yang
    Apr 29 '10 at 13:42
  • $\begingroup$ I didn't say that a loop touching the boundary is necessarily nontrivial. Your example is a loop without the antipodal identifications, hence it's trivial. At any rate, I was trying simply to explain why the intuitive argument works. My standard way to prove that SO(3) is not simply connected goes via Clifford algebras or quaternions as in Charlie's answer. $\endgroup$ Apr 29 '10 at 22:38

Don't think about $SO(3)$ to start with, think about the unit quaternions, $S^3\subset \mathbb{R}^4$ where multiplication is given by quaternion multiplication and inverses are given by "complex conjugation". It might help to realize that the dot product in $\mathbb{R}^4$ is given by $q.p=Re(q\overline{p})$.

The unit quaternions act on themselves by conjugation, and this action fixes the identity, inducing an action of the unit quaternions on the tangent space of $S^3$ at $1$ which is canonically isomorphic to $\mathbb{R}^3$. Call the unit quaternions $S^3$, call this action $ad:S^3\times \mathbb{R}^3\rightarrow \mathbb{R}^3$. First prove that this action is as rigid rotations by proving it preserves the dot product on $\mathbb{R}^3$. Next prove that the kernel of the map induced by $ad$, $h:S^3\rightarrow SO(3)$ has kernel $\{\pm 1\}$. Finally, use a dimension count to prove the map is onto. From there you can conclude that $S^3$ is the universal cover of $SO(3)$ with group of deck transformations $\mathbb{Z}_2$.

I usually assign this as homework, though I set the kids up by giving this outline to fill in.

Another fun picture of $SO(3)$ is given by the unit tangent bundle of $S^2$. Notice that this can be described as $$ T_1S^2=\{(\vec{u},\vec{v})\in \mathbb{R}^3\times \mathbb{R}^3|||\vec{u}||=||\vec{v}||=1, \ \vec{u}.\vec{v}=0\} $$

Notice that the matrix with columns $\vec{u},\vec{v},\vec{u}\times \vec{v}$ is in $SO(3)$. This map gives a diffeomorphism between $T_1S^2$ and $SO(3)$. The projection map $p:T_1S^2\rightarrow S^2$ that sends the pair $(\vec{u},\vec{v})$ to $\vec{u}$ is a fibration, and this is the standard fibration that people use to analyze the homotopy groups of $SO(3)$.


You have two different proofs (a.k.a. the cup proof and the belt proof) right here


There are several different demonstrations of the second one (Harrison Brown linked to this video in a comment to an answer I submitted - the cup proof - in the question asking for "proofs w/o words").

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    $\begingroup$ That video convinces me that a 720 degree rotation is contractible. It doesn't convince me that a 360 degree rotation isn't contractible. $\endgroup$
    – Dan Piponi
    Apr 30 '10 at 0:43
  • $\begingroup$ If you want a formal proof, take the SU(2)->SO(3) double cover and be done with it; this is an "informal demonstration". $\endgroup$ Apr 30 '10 at 4:17

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