I am reasonably familiar with the math behind spinors, the fact that $SU(2)$ is the universal (double) cover of $SO(3)$, etc.

I've often seen the "belt trick" and the "plate trick" used to motivate these results, but I must admit that I've never understood exactly what mathematical fact these tricks are supposed to be demonstrating. Sometimes I've seen the tricks presented as giving evidence for very precise facts like (a) "$SU(2)$ double covers $SO(3)$" or (b) "$SU(2)$ is simply connected but $SO(3)$ is not" or (c) "rotating a spinor by $2\pi$ takes it to its negative". Other times, I've seen the tricks presentated as just giving very hand-wavy motivation for the broad idea that "Sometimes you need to spin things around twice in order to get back to where you started."

Obviously the belt and plate tricks don't rigorously prove any mathematical results, but which precise mathematical facts (such the ones listed above, or others) do the tricks "demonstrate"?

Part of my issue is that people often describe the belt and plate tricks as demonstrating certain results about spinors, but in practice they seem to actually employ the reverse logic. That is, instead of first (a) arguing that belts/plates are described by spinors and then (b) using the belt and plate tricks to demonstrate mathematical results about spinors, in practice people instead tend to first (a) derive mathematical results about spinors, then (b) use the belt and plate tricks to show that real-world belts and plates display similar behavior, and finally (c) hand-wavily conclude that spinors seem to describe real-world belts and plates. This is an interesting observation, but I don't see how it actually demonstrate any mathematical results.

In order to use the tricks to demonstrate mathematical results about spinors, you need to first (a) invoke some specific structure of belts and plates (but not other real-world structures) that require spinors to model without invoking the belt and plate tricks, and then (b) use the tricks to demonstrate facts about spinors. Exactly what structure of these belts and plates (and their connections to external objects like your arm) is it that requires rotations to be described by elements of $SU(2)$ instead of $SO(3)$? In order to avoid circular reasoning, this structure can't reference the belt or plate tricks.

| cite | improve this question | | | | |

1) Stretch your arm out.

2) Rotate your hand as in the plate trick. Start at time $t=0$, complete one rotation at time $t=1$, and complete the second rotation at time $t=2$.

3) Think of your arm as a copy of the unit interval, stretching from $s=0$ at your shoulder to $s=1$ at the tip of your hand.

4) Imagine a copy of the standard basis for ${\mathbb R}^3$ attached to each point of your arm at the beginning of the motion. A rotation of that basis is an element of $SO(3)$. At time $t$, the basis at point $s$ has undergone some rotation. So we have a map $f:(s,t)\mapsto SO(3)$.

5) This graph will help you visualize that map: enter image description here

6) Note that at the end of one rotation, the tip of your hand ($s=1$) has returned to its original orientation. Therefore $f(1,0)=f(1,1)=f(1,2)$. This means that the map $t\mapsto f(1,t)$ is a closed loop in $SO(3)$, the image of the upper edge of the left-square --- and also the image of the upper edge of the right square, because the second rotation is identical to the first. Call this loop $Q$. It's the loop traced out over time by the basis at the tip of your hand.

7) Note that at time $t=1$, the tip of your hand is restored to its original orientation, as is your shoulder (which in fact has maintained its orientation throughout). Therefore $f(0,1)=f(1,1)$, so that the map $s\mapsto f(s,1)$ is another closed loop. It's the loop traced out along your arm at the end of one rotation, starting from the trivial rotation at your shoulder and returning to the trivial rotation at your fingertip. Call this loop $P$.

8) Theorem: The paths $P$ and $Q$ are homotopic (that is, one can be deformed continuously into the other). Proof: Consider Square A.

9) Theorem: The path $P$ cannot be homotopic to a constant loop. Proof: Any homotopy would give a way of untwisting your arm without changing the orientation of your fingertip along the way. This is evidently impossible.

10) Theorem: The path $Q+Q$ (that is, two full rotations of your arm) is homotopic to a constant loop. Proof: Consider squares $A+B$. The trivial paths along the vertical edges tell you that this rectangle is a homotopy from Q+Q to the constant loop, keeping the endpoints fixed along the way.

11) Main theorem: $\pi_1(SO(3))$ contains an element of order $2$. Proof: By (8) and (9), $Q$ is not homotopic to a constant loop, i.e. $[Q]\neq 0$. But by (10), $2[Q]=0$.

12) In particular, $SO(3)$ is not simply connected and therefore not equal to its own universal cover (which is of course $SU(2)$).

Note: This does NOT seem to me to prove that $\pi_1(SO(3))$ has exactly two elements; only that it has at least two, and that one of those has order two. I have sometimes heard it claimed that the plate trick actually nails down the full structure of $\pi_1(SO(3))$ but I'll be very surprised (and delighted!) if someone can make that argument.

| cite | improve this answer | | | | |
  • 1
    $\begingroup$ So would you say that the only a priori logical connection between the plate trick and spinors is that the plate trick demonstrates that $SO(3)$ has a distinct universal covering group, thereby motivating the possibility that there might be physical objects that transform under the faithful representation of that new group rather than under $SO(3)$? $\endgroup$ – tparker Nov 28 '19 at 18:31
  • 1
    $\begingroup$ @tparker: Pretty much, though it does show just a little more, namely that the universal covering group has an even number of elements lying over any given element of $SO(3)$ --- because the points lying over a given point are in one-one correspondence with the elements of $\pi_1(SO(3))$. $\endgroup$ – WillO Nov 28 '19 at 21:25

Myself, I like the candle dance as an example of these, so I'll use visual images from there. I like it because there's a very obvious object involved, the candle, which continuously rotates an arbitrary amount along the up/down axis. There is also one object which is not rotating at all -- the foot of the dancer. And, because we don't like tearing our dancers into ribbons, there's going to be a smooth progression from continuously rotating to not rotating at all as one looks at parts of the body like the hand holding the candle, down through the arm, torso, and into the leg.

If the structure of this body was not simply connected, then at some point one would get stuck, unable to generate a smooth progression from candle towards foot. Indeed, this is what happens when an unskilled person tries to replicate the candle dance.

It's also what happens if someone twists your arm behind your back. If you try to solve your problems within a single rotation, you find yourself jammed up. However, if you are willing to induce a second rotation, simply walking forward with your muscles relaxed frees your muscles to undergo the second rotation needed and your arm moves free. If the person controlling your arm happens to be your dancing partner, you could then rotate it over their head (letting them do a spin), to return to the neutral position.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.