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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, and is indeed the orientation covering.

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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# How does $\pi_1(SO(3))$ relate exactly to the waiters trick?

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, and is indeed the orientation covering.

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?