Does the Dirac belt trick work in higher dimensions? If the Dirac belt is in 4-space, is it still true that when the belt is initially given a 360 degree twist then it cannot be untwisted? 
I assume this is so because SO(n) is not simply connected, but I am still a bit fuzzy on how this all works. (For instance, even in 3-space, if the path in SO(3) is contractible, how do we know that contraction is physically realizable without self-intersection in a belt of finite width?)
 A: One can show the following: Given a noncompact connected surface $S$, any two smooth embeddings $f_0, f_1: S\to R^4$ are isotopic. (In particular, there are no twisted bands in $R^4$.) Proofs are a bit involved, but the key fact that for a generic smooth homotopy $F: S\times [0,1]\to R^4$ between $f_0, f_1$, for each $t$, the map $F(~, t): S\to R^4$ has discrete  self-intersections, i.e., each point has discrete preimage (for generic $t$  self-intersections are transversal). Since $S$ is noncompact, these self-intersections can be "pushed away to infinity'' in $S$ by modifying $F$. 
For a "Dirac belt'' one can give a more direct argument. (Mathematically speaking, a Dirac belt in $R^n$ can be interpreted as a smooth embedding $h: S^1\to R^n$ together with a vector field $X$ along $h$, normal to $h$, i.e., for each $s\in S^1$, the vector s $X(s), h'(s)$ are linearly independent. The corresponding map of the annulus $S^1\times [0,1]\to R^n$ is obtained from $(h,X)$ by the formula $f(s, t)= h(s) + ctX$, where $c$ is a sufficiently small positive number.) The proof in this setting ($n=4$) goes roughly as follows. First isotope $h$ (and regularly homotope $X$) to make it the identity embedding of a round circle $C$ into $R^4$. The vector field $X$ is deformed to a smooth vector field orthogonal to $C$ and, hence, can be identified with a smooth nowhere zero function  $C\to R^3$. But any two smooth nowhere zero maps $C\to R^3$ are smoothly homotopic to each other through maps of the same nature. (Since $R^3 - 0$ is simply connected.) 
