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Thread: SU(2) double-covers SO(3) and Plate Trick

  1. #1
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    SU(2) double-covers SO(3) and Plate Trick

    Hello,

    Has anyone heard of the plate trick ? Plate trick - Wikipedia, the free encyclopedia

    I've already proofed the double-covering of $\displaystyle SO_3(R)$ by $\displaystyle SU_2(\mathbb{R})$. I've also shown that $\displaystyle SO_3(\mathbb{R})$'s fundamental group is $\displaystyle \mathbb{Z}/2\mathbb{Z}$. But I can't find anything between these results and the fact my arm can't make a 360 turn and go back to initial position without having my soup plate fall.
    Can somebody help me modelize my arm, and explain to me the passage from theorical to practical?

    thanks!
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    There is a close path p in $\displaystyle SO_3(\mathbb{Re})$ that cannot be deformed to a point, but traversing p twice, i.e., $\displaystyle p^2$ can be deformed to a point in $\displaystyle SO_3(\mathbb{Re})$. It indicates that $\displaystyle SO_3(\mathbb{Re})$ is not simply-connected. Since $\displaystyle SO_3(\mathbb{Re})$ corresponds to the real projective space $\displaystyle \mathbb{Re}P^3$, we cannot fully draw $\displaystyle SO_3(\mathbb{Re})$ in three dimensional space. The plate trick simply shows that the first rotation of 360 cannot be deformed to a single point because the arm is twisted, but if we have one more rotation of 360 to the same direction, i.e., rotation of 720, can go back to the original state.
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  3. #3
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    Do you mean that the set of positions our hand can reach with the plate is $\displaystyle SO_3$ ? And that doing a movement with my arm is a path in $\displaystyle SO_3$ ? Why is that?
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    Quote Originally Posted by AMA View Post
    Do you mean that the set of positions our hand can reach with the plate is $\displaystyle SO_3$ ? And that doing a movement with my arm is a path in $\displaystyle SO_3$ ? Why is that?
    No, there is a correspondence, but I don't think it is an actual closed path in $\displaystyle SO_3(\mathbb{Re})$. As I said, You cannot fully draw $\displaystyle SO_3(\mathbb{Re})$ in a three dimensional space. If you already proved that $\displaystyle SU_2(\mathbb{Re})$ is the double cover of $\displaystyle SO_3(\mathbb{Re})$ and the fundamental group of $\displaystyle SO_3(\mathbb{Re})$ is $\displaystyle \mathbb{Z}/2\mathbb{Z}$, then you know that a loop $\displaystyle p$ in $\displaystyle SO_3(\mathbb{Re})$ is not contractible, but $\displaystyle p^2$ is. The plate trick is an example to show that a rotation of 360 degrees does not go back to the original state (i.e., a loop $\displaystyle p$ is not contractible) but a rotation of 720 degrees goes back to the original state (i.e., a loop $\displaystyle p^2$ is contractible).
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    Well actually I don't really see which correspondences there are. Can you also explain the two "i.e"s in your last sentence ? I think my problem comes from there.
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    Quote Originally Posted by AMA View Post
    Well actually I don't really see which correspondences there are. Can you also explain the two "i.e"s in your last sentence ? I think my problem comes from there.
    If the fundamental group of a topological space X is $\displaystyle \mathbb{Z}/2\mathbb{Z}$, then it has a presentation <x : x^2>. It indicates that there is a closed path p in X that cannot be deformed to a point, but p^2 can. The closed path p (360 degrees rotation) in SO_3(R) corresponds to a full twist in the arm, whereas p^2 in SO_3(R) restores the original state. Intuitively, this full twist keeps p from being contractible.
    Last edited by TheArtofSymmetry; Dec 25th 2010 at 06:09 PM. Reason: English grammar
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