I like the Hopf fibration of the 3-sphere $S^3$ enough that I found a nice way to make a physical model of it. All you need is to combine a bunch of key rings in such a way that (ii) every pair of rings is linked and (iii) every triple of rings has the same orientation (as a link in $\mathbb{R}^3$) as every other triple. The ubiquitous "split" key rings, with a tight coil winding around twice and a split running down the middle, work very well for this. The idea is to represent each coset in the homogeneous space $S^3/S^1$ with a key ring. The cosets are supposed to correspond nicely with points in $S^2$, so the model is increasingly satisfactory (and also increasingly hard to enlarge) as more rings are adjoined.
I also study Lie algebras. Recall that a Lie algebra is a vector space $\mathfrak{g}$ equipped with a bilinear product which is skew-symmetric and satisfies the Jacobi identity. These latter conditions can be stated as (ii) $[u,v]+[v,u]=0$ for all $u,v\in\mathfrak{g}$ and (iii) $[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0$ for all $u,v,w\in\mathfrak{g}$. These conditions (ii) and (iii) bear some similarity to the properties (ii) and (iii) observed in the Hopf fibration: There is a property for every pair (ii) and a property for every triple (iii), and you need both (ii) and (iii) to define the thing. What's the story?
There is likely some tie-in with braid groups. Let $B_2$ be the braid group on two strands and let $B_3$ be the braid group on three strands. Recall that $B_2$ has a presentation as $$B_2=\left<a|\emptyset\right>\cong\mathbb{Z}$$ and $B_3$ has a presentation $$B_3=\left<a,b|aba=bab\right>.$$ Notice (ii) in $B_2$, if one closes $a^2$ by identifying opposite ends of the strands, then one obtains a Hopf link and (iii) in $B_3$ if one closes $(ab)^3$ by indentifying opposite ends of the strands, then one obtains a triple of Hopf links. (Draw a picture.)
Why do these properties of the Hopf fibration of $S^3$ appear to be fundamental in the definition of a Lie algebra? I have never seen an explanation of this in print.
