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I hope this is serious enough. It is a well-known fact that $\pi_1(SO(3)) = \mathbb{Z}/(2)$, so $SO(3)$ admits precisely one non trivial covering, which is 2-sheeted.

Another well known fact is that you can hold a dish on your hand and perform two turns (one over the elbow, one below) in the same direction and come back in the original position.

These facts are known to be related, and I more or less can guess why. Some configuration of the system (hand + dish) must draw a path in $Spin(3)$ whose projection in $SO(3)$ is the closed non trivial loop pointed at the identity.

The problem is that I cannot make this precise, since it is not clear to me which is the variety which parametrizes the position of the elbow and the hand. Is there a clean way to see how $Spin(3)$ comes into play?

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Spin(3) comes into the play only as the covering space of SO(3), I think. You do everything in SO(3). Draw a curve through your body from a stationary point, like your foot, up the leg and torso and out the arm, ending at the dish. Each point along the curve traces out a curve in SO(3), thus defining a homotopy. After you have completed the trick and ended back in the original position, you now have a homotopy from the double rotation of the dish with a constant curve at the identity of SO(3). You can't stop at the halfway point, lock the dish and hand in place, now at the original position, and untwist your arm: This reflects the fact that the single loop in SO(3) is not null homotopic.

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  • $\begingroup$ Is there a rigorous proof that you can't untwist your arm? This trick describes a specific loop $\gamma$ in $SO(3)$ and shows (rigorously) that $\gamma^2$ is nullhomotopic. I would love to see a rigorous proof that $\gamma$ is not nullhomotopic. Just knowing that $\pi_1 (SO(3)) = Z/2$ certainly doesn't do it. Does $\gamma$ live inside a copy of $RP^2$ that can serve as the 2-skeleton in a CW decomposition of $SO(3)$? Can one somehow explicitly lift this loop to the double cover (viewed as $S^3$, or maybe $SU(2)$) and see that the lift is no longer a loop? Is there some other argument? $\endgroup$
    – Dan Ramras
    Commented Jan 26, 2011 at 23:31
  • $\begingroup$ @Dan: Indeed, the loop lifts to a non-loop in the double cover. Perhaps the easiest way to see this is to represent the double cover as the unit quaternions and start with the curve given by $\hat\gamma(t)=\cos t+i\sin t$ with $t\in[0,\pi]$. A unit quaternion $q$ acts as a rotation on the imaginary quaternions via $x\mapsto qx\bar q$; this action is the covering map onto $SO(3)$. You can now compute the resulting path in $SO(3)$ explicitly and verify that it is a version of $\gamma$. $\endgroup$
    – Harald Hanche-Olsen
    Commented Jan 27, 2011 at 10:33
  • $\begingroup$ Ah, very nice! Thanks. I also stumbled across (by complete accident) a paper in Math Magazine that uses braid groups to calculate $pi_1 SO(3)$. Looks nice: mathdl.maa.org/mathDL/46/… $\endgroup$
    – Dan Ramras
    Commented Jan 27, 2011 at 15:00
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(This is a bit long for a comment but isn't intended as an answer.)

I've broken glasses in the past demonstrating this trick so I strongly advise doing it with empty plastic mugs until you have it down pat. For those who can't quite work it out, here's a simpler way to do it: take an elastic band and two rods that have distinguishable "up" and "down". Loop the band around the two rods and keep it fairly taut (just so it doesn't fall off). Ascii picture:

   |     |
 /-|-----|-\   
/  |     |  \   
\  |     |  /   
 \---------/   
   |     |

Now turn one of the rods upside-down. By moving the rod through space but without turning it (or the system falling apart), try to untangle the elastic band.

Can't do it? Good. Can? Whoops! Either you've done something wrong or Whitehead did.

Start again from the beginning. Now turn one of the rods through a full twist (so it's right-side-up again). Now try to untangle the elastic band as before.

Can't do it? Try again! It's possible.

So in mathematical terms, what you are looking for is the position of the second rod plus it's "up-down"ness. In terms of the waiter's arm, you want the position of the hand and the expression of agony on his face: if his face is in agony, his arm is twisted; if his face is calm, his arm isn't twisted.

Mnemonic: twisting someone's arm twice gets you nowhere.

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    $\begingroup$ Isn't that nice! Markdown has interpreted my diagram as code and put the "comments" in it in a different colour. I wonder what language it thought it was ... $\endgroup$
    – Andrew Stacey
    Commented Feb 9, 2010 at 15:50
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See p. 164-167 of Bredon's Topology and Geometry. Page 166 (not viewable online) is just a sequence of nine pictures demonstrating the waiter's trick.

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    $\begingroup$ There is a vide with commentary by Lou Kauffman: try youtube.com/watch?v=CYBqIRM8GiY or search Google using: Air on the Dirac Strings This is useful but is not the complete answer so I will leave this as a comment and not an answer. $\endgroup$
    – Tim Porter
    Commented Feb 9, 2010 at 16:34

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