Visualizing the fundamental group of SO(3)

Recently I became interested in trying to visualize the fact that $\pi_1(\text{SO}(3)) = \mathbb{Z}/2\mathbb{Z}$. For whatever reason, the plate trick doesn't do it for me, so I've been looking for something else. I know that $\text{SO}(3)$ is homeomorphic to $P^3(\mathbb{R})$, and one concrete visualization of this space is as the solid unit ball with opposite points on the boundary identified. This model allows a straightforward visualization of the fundamental group, but what it doesn't seem to easily allow for is a corresponding visualization of the action of $\text{SO}(3)$. (In other words, it doesn't seem to be easy to visually verify that we are actually working in $\text{SO}(3)$.)

So I tried the following instead: $\text{SO}(3)$ acts simply transitively on the unit tangent bundle of $S^2$, so $UT(S^2)$ ought to be homeomorphic to $\text{SO}(3)$. This satisfies both my requirements: it is easy to visualize paths in this space, and it is also easy to see the action of $\text{SO}(3)$.

Now I would just appreciate someone confirming that I've got the details right: in this model, a representative of the nontrivial homotopy class of paths in $\pi_1(\text{SO}(3))$ is the path which parallel transports a unit vector once around a great circle, correct? How should I visualize the homotopy between the square of this path and the trivial path? (This might boil down to the plate trick or the belt trick, but at least it'll be stated in a language that I can understand.)

Here's a bit of an elaboration on my earlier comment.

You can see $SO_3$ as the quotient of a unit $3$-ball by the antipodal map on the boundary. This is realized explicitly by the exponential map for $SO_3$. The tangent space to the identity of $SO_3$ is all $3\times 3$ matrices $H$ such that $H^T=-H$. Up to conjugation by an element of $SO_3$, such a matrix looks like

$$\left(\begin{matrix} 0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right)$$

which under exponentiation is converted into rotation by $\theta$ about the $z$-axis. Every element of $SO_3$ is rotation by some angle about a fixed axis since the characteristic polynomial has $1$ as a root -- so this is a rather explicit description of how the exponential map is onto. If you choose the length of a matrix $[a_{i j}]$ to be $\sqrt{ \sum_{i j} a_{i j}^2}$, then technically $SO_3$ is the quotient of the ball of radius $\pi/\sqrt{2}$, with antipodal points on the boundary identified. But that's small potatoes.

The 3-ball with antipodal points identified, as you've seen has fundamental group $\Bbb Z_2$. In particular you can think of the 3-ball with antipodal points identified as being built up from the 1-ball with antipodal points identified, union a 2-cell (this gives a CW-decomposition of the 2-ball with antipodal points identified) then union a 3-cell. So the fundamental group all "happens" in the 2-ball with antipodal points identified. The 1-skeleton generates the fundamental group and in our model (under exponentiation) corresponds to one full rotation by $2\pi$ about any chosen axis. The relation corresponds to the 2-cell attachment and you can literally interpret it as the plate trick in a "sufficiently nice" model. Let me explain.

The trouble with the plate trick is it uses a human arm. Human arms you can think of as carrying a core underlying piece of information, a map $f : [0,1] \to SO_3$. $f(t)$ represents the orientation in space of the part of the arm distance $t$ from the shoulder, so $0$ represents the shoulder distance, and $1$ the palm of the hand. There are technical issues -- if you choose the framing corresponding to the positions of the appropriate bone in the arm, you don't get a continuous map, just a piecewise-constant map -- but if you consider the orientations of the ligaments between the bones it becomes a continuous map. So the trouble with the plate trick is the map $f$ isn't the only constraint on the arm. The arm is embedded in space. Bones are rigid, there's only so much ligaments are capable of doing. So you could be rightly concerned that the plate trick is a demonstration of a peculiarity of anatomy, rather than something about $SO_3$.

Alright, but formally what's going on with the plate trick? As mentioned, a configuration of the arm we're representing by $f$, which is an element of the path space $PSO_3$ of continuous paths in $SO_3$ with $f(0) = I \in SO_3$ fixed.

There is the path-loop fibration:

$$\Omega SO_3 \to PSO_3 \to SO_3$$

and $PSO_3$ is contractible. The plate trick "is" the induced map $\pi_1 SO_3 \to \pi_0 \Omega SO_3$ from the homotopy long exact sequence of the above fibration, which is guaranteed to be an isomorphism by the path-loop fibration. Specifically, elements of $\pi_1 SO_3$ you are thinking of as motions of your hand. The path-loop fibration is telling you that you can realize the motion as a motion of your arm (or at least, a path in $PSO_3$ but the fact that people can do the plate trick says you can actually realize it as a motion of your arm).

There is of course an element that's a bit misleading in all this. The arm isn't complicated enough to allow for you to lift the product of two generators of $\pi_1 SO_3$ and witness the null homotopy -- since the arm allows a fixed complexity, the null-homotopy happens as you lift the concatenation.

But you can see from the rather explicit exponential map above how the null homotopy actually works, as it's sitting in the image of the 2-ball with antipodal points identified.

How is that?