Timeline for How does $\pi_1(SO(3))$ relate exactly to the waiters trick?
Current License: CC BY-SA 2.5
5 events
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| Jan 27, 2011 at 15:00 | comment | added | Dan Ramras | 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/… | |
| Jan 27, 2011 at 10:33 | comment | added | Harald Hanche-Olsen |
@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 $\gamma$.
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| Jan 26, 2011 at 23:31 | comment | added | Dan Ramras | 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? | |
| Feb 9, 2010 at 17:43 | vote | accept | Andrea Ferretti | ||
| Feb 9, 2010 at 15:59 | history | answered | Harald Hanche-Olsen | CC BY-SA 2.5 |