TITLE: Is there a higher dimensional analogue of the Dirac belt trick?
QUESTION [12 upvotes]: The Dirac belt trick produces a nice 3-dimensional geometric object with symmetry group $Spin(3) = SU(2)$:  a 2-sphere with a properly embedded framed ray (usually presented by using orientations to reduce the framing of the ray to a single normal vector field, then integrating this to give a "belt" of finite width), with the ray or "belt" regarded up to smooth isotopies fixing the sphere and trivial at infinity.
A naive generalization to higher dimensions using a (2-dimensional) "belt" does not work to produce an object with symmetry group $Spin(n)$ since the "belt" can always be untwisted.
My question:
Is there any properly embedded subspace of ${\mathbb R}^n$ with boundary on $S^{n-1}$, possibly equipped with an auxiliary structure like a framing of its normal bundle, such that when the subspace is regarded up to isotopies fixing $S^{n-1}$ and trivial at infinity, the symmetry group of the "sphere and subspace" is $Spin(n)$? 
I am particularly interested in the case of $n=4$.

REPLY [6 votes]: If I understand the question correctly, the same "belt" construction works in higher dimensions. The belt should have a full framing of its normal bundle ($n−1$ normal vectors; using the orientation of $\mathbb R^n$, you can get by with $n−2$ normal vectors; adding the orientation tangent to the belt gives $n$ vectors total).
$Spin(n)$ is the simply connected double cover of $SO(n)$, and one explicit model of this double cover is the set of pairs $(f, p)$, where $f\in SO(n)$ and $p$ is a homotopy class of path from $f$ to a base point in $SO(m)$.
The map of the sphere into $\mathbb R^n$ determines an element $f\in SO(n)$, and the framings along the belt determine a path from $f$ to a base point (that base point being the fixed framing near infinity on the belt).
I actually use a construction very similar to this when doing fermionic TQFT calculations.  I usually call the framed arc a "Dirac belt".