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This is a followup of sorts to Plate trick demonstrating SO(3) not simply connected. .

Suppose I do the ``plate trick'' to demonstrate the existence of an order-2 element in $\pi_1(SO(3))$. That is, I extend my arm, pretend that I've attached the standard orthogonal basis for ${\mathbb R}^3$ to each point along my arm, and make a circle with my hand, ending up with my fingertips restored to their original position and my arm twisted.

I've now constructed two different loops in $SO(3)$ --- the loop $Q$ traced out by my fingertips over time and the loop $P$ traced out by distance along my arm after the motion is completed. I've always been a little confused about whether I'm supposed to focus on Q or on P. They are related by square A below:

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Now (writing [] for homotopy classes) we have:

Theorem 1: [P]=[Q]. Proof: Square A.

Theorem 2: [P] is non-trivial. Proof: If there were a homotopy from P to the trivial loop, I could use it to untwist my arm without changing the orientation of my fingertips. (I take it to be obvious that this is impossible).

Theorem 3: [Q]+[Q] is trivial. Proof: Square A plus Square B.

These three theorems taken together prove that [Q] (or equivalently [P]) has order 2. That is, we prove (Theorem 2) that [P] has order at least 2, we prove (Theorem 3) that [Q] has order at most 2, and then we invoke Theorem 1 to get the result.

So it appears that to get the desired result, one has to focus on both P and Q, and that it is insufficient to focus on just one of them.

My question is: Is this correct? In other words: Can we prove directly (without mentioning [P]) that [Q] has order exactly 2, and/or can we prove directly (without mentioning [Q]) that [P] has order exactly 2?

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There are other versions of this trick, see for example the well produced video Air on the Dirac Strings and my own version which relates it to the Projective Plane.

I think there is a serious point in the question. All these versions illustrate, in a way which can be made rigorous, that a certain loop $a$ is such that $a^2$ is homotopic to the identity; but they do not show that $a$ itself is not homotopic to the identity. For that it seems you do need the Seifert-van Kampen Theorem. If you are doing that, you might as well do the many pointed version, which gets the, or, strictly speaking, a, fundamental group of the circle, and much more, as well.

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