Plate trick demonstrating SO(3) not simply connected.

Question

I know there is another question about this, but I didn't find the answer adequate. How exactly does the plate trick (http://en.wikipedia.org/wiki/Plate_trick) show that $SO(3)$ isn't simply connected, but $SU(2)$ is? Which part of this is the path, and what's the homotopy?

Answer 1

Every piece of your arm has associated with it an element of $SO(3)$, which is the rotation required to move that part of your arm from the straight up position to where it is. One end of your arm doesn't move (the end attached to your shoulder). The other end (your hand) moves. If $s$ is the arclength of your arm from the shoulder, then the map $s\mapsto R(s)$, where $R(s)$ is the element of $SO(3)$ representing the rotation of that part of the arm, is a continuous map from an interval to $SO(3)$. After you have done the plate trick one time, your arm is in such an orientation that it doesn't contract to the constant map, but $R(s)$ at the ends of the intervals is the identity. So it represents a non-contractible closed curve on $SO(3)$, that is, an element of the fundamental group of $SO(3)$ which is not homotopic to the identity. But do the plate trick twice, and you are able to get your arm back to the constant map, showing that in the fundamental group of $SO(3)$ that this element squared is the identity.