Does the Dirac belt trick work in higher dimensions? (v 2.0) The same question was asked a while back, and was correctly answered in the negative under the assumption that the "belt" was just as in 3D:  a strip of surface (perhaps infinitesimally thin as a curve with a normal vector field -- data that in 3D would, all things being oriented, determine a framing of the curve's normal bundle, but would not in higher dimensions).
I would like an answer if the "belt" is also allowed to be of higher dimension (for instance a strip of 3-manifold, or a curve with a framing of its normal bundle), ideally an answer in the affirmative for some sort of generalized higher-dimensional "belt".
In light of Pedro's reply, I should clarify:  the symmetries of a (2-)sphere in $R^3$ form $SO(3)$, but the symmetries of a sphere with a belt "to infinity" (which belt is allowed to by deformed by isotopies in the complement of the sphere and belt) form its double cover, $SU(2)$.  I'm wondering whether there is a "belt" that can be attached to an $n$-sphere in $R^{n+1}$ so that the symmetries of the sphere and "belt"-up-to-isotopy are $Spin(n)$ (for $n=4$, $SU(2)\times SU(2)$).
 A: This is a bit late, but I thought I’d offer another line of thinking from which one can conceive of a higher-dimensional belt trick.
So, to rephrase what you’re saying, what’s really going on here is that we have a nontrivial monodromy representation of $\pi_1(SO(3)) = \mathbb{Z}/2\mathbb{Z}$ that keeps track of the orientation of the belt attached to our 2-sphere. Monodromy can lead to annoying things like multi-valued functions, so we pass to the universal cover $\operatorname{Spin}(3) = SU(2)$ where non-trivial loops in $SO(3) = PSU(2)$ are untangled sensibly into paths.
Now, to be precise, the nontrivial element in $\pi_1(PSU(2))$ is the homotopy class of the loop $\alpha: S^1 \to PSU(2)$ given by
$\theta \mapsto \begin{pmatrix}
e^{\frac{1}{2}\theta i} & 0 \\
0 & e^{-\frac{1}{2}\theta i} \end{pmatrix}.$
It’s nice to visualize this by the eigenvalues (by lifting to the path in the universal cover, if you want to be pedantic), which start together at $1$ and then orbit about the unit circle in opposite directions before re-merging at $-1$. So the content of the usual belt trick is finding a based nullhomotopy of the loop $\alpha^2$, where the eigenvalues pass through each other at $-1$ and ultimately return to $1$. This is done as follows:
$t \mapsto h_t(\theta) = \left\{ \begin{array}{ll}
  \begin{pmatrix}
e^{\theta i} & 0 \\
0 & e^{-\theta i} \end{pmatrix} & 0 \leq \theta \leq \pi \\
  \begin{pmatrix}
\cos(\frac{\pi}{2}t) & -\sin(\frac{\pi}{2}t) \\
\sin(\frac{\pi}{2}t) & \cos(\frac{\pi}{2}t) \end{pmatrix} \begin{pmatrix}
e^{\theta i} & 0 \\
0 & e^{-\theta i} \end{pmatrix} \begin{pmatrix}
\cos(\frac{\pi}{2}t) & \sin(\frac{\pi}{2}t) \\
-\sin(\frac{\pi}{2}t) & \cos(\frac{\pi}{2}t) \end{pmatrix} & \pi \leq \theta \leq 2\pi \\
\end{array} \right. ,$
so that $h_0 = \alpha^2$ and $h_1 = \beta$ is the loop 
$\theta \mapsto \left\{ \begin{array}{ll}
  \begin{pmatrix}
e^{\theta i} & 0 \\
0 & e^{-\theta i} \end{pmatrix} & 0 \leq \theta \leq \pi \\
  \begin{pmatrix}
e^{-\theta i} & 0 \\
0 & e^{\theta i} \end{pmatrix} & \pi \leq \theta \leq 2\pi \\
\end{array} \right. .$
If you think about $\beta$ for a moment, you realize that the eigenvalues journey out from $1$ to $-1$ before “bouncing” off one another and marching back the way they came; this loop is clearly contractible. Of course, at any fixed $\theta$, we aren’t changing the set of eigenvalues from $\alpha(\theta)$ at all; $\beta(\theta)$ just changes how we label them halfway through their journey.
In higher dimensions, one does the same sort of work in $\operatorname{Spin}(n)$ (which is simply-connected for $n \geq 3$) to nullhomotope the square of a generator from $\pi_1( SO(n) ) \cong \mathbb{Z} / 2\mathbb{Z}$. Of course when $n=2$ we have $\pi_1( SO(2) ) \cong \mathbb{Z}$, which relates to the fact that
$ \pi_1( \operatorname{UConf}_m( \mathbb{R}^n ); \mathbb{Z} ) \cong \left\{ \begin{array}{ll}
 B_m & n = 2 \\
 S_m & n \geq 3
\end{array} \right. ,$
where $\operatorname{UConf}_m( X ) := \{ (x_1, \dots, x_m) \in X^m \; | \; x_i \not = x_j \text{ if } i \not = j \} \, / \, S_m$ is the space of unordered configurations of $m$ distinct points on $X$ and $B_m$ is the braid group on $m$ strands. This has something to do with the fact that braids are much harder to undo when you can only move side-to-side and back-to-front, but not up-and-down, and explains why exotic particle statistics (anyons!) make sense in 2-space but only fermions and bosons occur in everyday life. But I digress.
Regardless, another way to imagine generalizing the belt trick into higher dimensions is by studying loops in $PSU(n)$ for $n > 2$, which has fundamental group $\mathbb{Z}/n\mathbb{Z}$. This can be done in the universal cover $SU(n)$, by considering the (lift of) the generator $\alpha$:
$\theta \mapsto \begin{pmatrix}
e^{\frac{1}{n}\theta i} & \cdots & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots \\
0 & \cdots & e^{\frac{1}{n}\theta i} & 0 \\
0 & \cdots & 0 & e^{-\frac{n-1}{n}\theta i} \end{pmatrix}.$
The goal of the higher-dimensional belt trick here is to nullhomotope $\alpha^n$, which one does in a way analogous to before by rotating the $e^{(n-1)\theta i}$ into each diagonal entry for equal time. One then reaches the loop $\beta$ given by
$\theta \mapsto \left\{ \begin{array}{ll}
  \begin{pmatrix}
e^{\theta i} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & e^{-(n-1)\theta i} \end{pmatrix} & 0 \leq \theta \leq \frac{1}{n} 2\pi \\
\qquad \qquad \vdots & \\
  \begin{pmatrix}
e^{-(n-1)\theta i} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & e^{\theta i} \end{pmatrix} & \frac{n-1}{n} 2\pi \leq \theta \leq 2\pi \\
\end{array} \right. ,$
which is some higher-dimensional version of eigenvalues bouncing off one another as they orbit between $n$th roots of unity. The path each eigenvalue takes is contractible, so we can reel everything in just like before.
And there you have it, two different ways of generalizing the belt trick into higher dimensions (admittedly I think the first is more honest to the original spirit of the trick).
EDIT:
I think it might be worth it to add a word on $SO(n)$, since it's what you're actually asking about. In general, there is an inclusion $SO(n) \hookrightarrow S(n+1)$ given by the map
$ A \mapsto \begin{pmatrix}
A & 0 \\
0 & 1 \\
\end{pmatrix}.$
This inclusion is part of the fiber bundle $SO(n) \hookrightarrow SO(n+1) \twoheadrightarrow S^n$ induced via the smooth and transitive action of $SO(n+1)$ on the $n$-sphere. This in turn gives rise to a long exact sequence
$ \cdots \to \pi_i(SO(n)) \to \pi_i(SO(n+1)) \to \pi_i(S^n) \to \pi_{i-1}(SO(n)).$
For $n=2$, we have $\pi_1(SO(2)) \twoheadrightarrow \pi_1(SO(3)).$ This map is induced by inclusion, so the generator
$\theta \mapsto \begin{pmatrix}
\cos \theta & - \sin \theta \\
\sin \theta & \cos \theta
\end{pmatrix}$
of $\pi_1(SO(2)) \cong \mathbb{Z}$ passes to the generator of $\pi_1(SO(3)) \cong \mathbb{Z}/2\mathbb{Z}$. Thus the classic belt trick is about the nullhomotopy of the loop $\gamma: S^1 \to SO(3)$ given by
$\theta \mapsto \begin{pmatrix}
\cos (2\theta) & - \sin (2\theta) & 0 \\
\sin (2\theta) & \cos (2\theta) & 0 \\
0 & 0 & 1
\end{pmatrix}.$
We'll do that with a similar rotation as before:
$t \mapsto \left\{ \begin{array}{ll}
  \begin{pmatrix}
\cos (2\theta) & - \sin (2\theta) & 0 \\
\sin (2\theta) & \cos (2\theta) & 0 \\
0 & 0 & 1
\end{pmatrix} & 0 \leq \theta \leq \pi \\
  \begin{pmatrix}
\cos(\pi t) & 0 & -\sin(\pi t) \\
0 & 1 & 0 \\
\sin(\pi t) & 0 & \cos(\pi t)
\end{pmatrix} \begin{pmatrix}
\cos (2\theta) & - \sin (2\theta) & 0 \\
\sin (2\theta) & \cos (2\theta) & 0 \\
0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
\cos(\pi t) & 0 & \sin(\pi t) \\
0 & 1 & 0 \\
-\sin(\pi t) & 0 & \cos(\pi t)
\end{pmatrix} & \pi \leq \theta \leq 2\pi \end{array} \right. ,$
which homotopes us to the clearly contractible loop
$t \mapsto \left\{ \begin{array}{ll}
  \begin{pmatrix}
\cos (2\theta) & - \sin (2\theta) & 0 \\
\sin (2\theta) & \cos (2\theta) & 0 \\
0 & 0 & 1
\end{pmatrix} & 0 \leq \theta \leq \pi \\
  \begin{pmatrix}
\cos (2\theta) & \sin (2\theta) & 0 \\
-\sin (2\theta) & \cos (2\theta) & 0 \\
0 & 0 & 1
\end{pmatrix} & \pi \leq \theta \leq 2\pi \end{array} \right. .$
Now back to the long exact sequence. When $n \geq 3$, we have
$0 \to \pi_1(SO(n)) \stackrel{\cong}{\to} \pi_1(SO(n+1)) \to 0$.
This map is again induced by inclusion, so the generator for $\pi_1(SO(n))$ is the homotopy class of
$\theta \mapsto \begin{pmatrix}
\cos (\theta) & - \sin (\theta) & 0 \\
\sin (\theta) & \cos (\theta) & 0 \\
0 & 0 & I_{n-2}
\end{pmatrix},$
and the homotopy is done in the same way. Therefore the higher dimensional belt trick, in the sense you asked, works exactly as it does in $\mathbb{R}^3$—because the higher dimensions aren't necessary for the trick (the double rotation itself, which is done around a single axis, and then the homotopy undoing it) to be carried out.
