# Math Help - SU(2) double-covers SO(3) and Plate Trick

1. ## 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 $SO_3(R)$ by $SU_2(\mathbb{R})$. I've also shown that $SO_3(\mathbb{R})$'s fundamental group is $\mathbb{Z}/2\mathbb{Z}$. 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!

2. There is a close path p in $SO_3(\mathbb{Re})$ that cannot be deformed to a point, but traversing p twice, i.e., $p^2$ can be deformed to a point in $SO_3(\mathbb{Re})$. It indicates that $SO_3(\mathbb{Re})$ is not simply-connected. Since $SO_3(\mathbb{Re})$ corresponds to the real projective space $\mathbb{Re}P^3$, we cannot fully draw $SO_3(\mathbb{Re})$ 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.

3. Do you mean that the set of positions our hand can reach with the plate is $SO_3$ ? And that doing a movement with my arm is a path in $SO_3$ ? Why is that?

4. Originally Posted by AMA
Do you mean that the set of positions our hand can reach with the plate is $SO_3$ ? And that doing a movement with my arm is a path in $SO_3$ ? Why is that?
No, there is a correspondence, but I don't think it is an actual closed path in $SO_3(\mathbb{Re})$. As I said, You cannot fully draw $SO_3(\mathbb{Re})$ in a three dimensional space. If you already proved that $SU_2(\mathbb{Re})$ is the double cover of $SO_3(\mathbb{Re})$ and the fundamental group of $SO_3(\mathbb{Re})$ is $\mathbb{Z}/2\mathbb{Z}$, then you know that a loop $p$ in $SO_3(\mathbb{Re})$ is not contractible, but $p^2$ 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 $p$ is not contractible) but a rotation of 720 degrees goes back to the original state (i.e., a loop $p^2$ is contractible).

5. 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.

6. Originally Posted by AMA
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 $\mathbb{Z}/2\mathbb{Z}$, 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.