## Spheres – 2

Down the rabbit hole I continue to go.  In the previous post on spheres, Spheres – 1, I played around with some ideas that were the result of reading a little bit about knot cobordisms.  I can’t quite recall why I started looking into fiber bundles and fibrations but I did.  Fiber bundles are something on my list as I know they’re important in field theories and I had heard of fibrations but didn’t know what they are.  They’re special cases of fiber bundles where the fibers aren’t homeomorphic to each other, just homotopic.  So every fiber bundle is a fibration but not vice versa.  What does this have to do with spheres?  So much….so much.

First a gentle introduction, as we go I’ll link references that were helpful thus far.  There’s way more to fiber bundles than I will begin to allude to right now, but as a start they seem to be generalizations of cartesian products.  Given a couple of sets A and B the cartesian product, $A\times B$ is the set of all pairs $(a,b)$ where $a\in A$ and $b\in B$.  There’s a natural map from $A\times B$ to B, the projection $\pi_B : A\times B\to B$, this project just forgets A and maps an ordered pair $(a,b)\to b$.  For any point $y\in B$ the preimage of that point $\pi_B^{-1}(y) = A\times {y}$. We call A the fiber over $y$.  We can also say that $A\times B$ fibers over B.  $A\times B$ is called the total space and B is called the base space.  The fibers are the sets A.  We can make a little diagram as well: $A\to A\times B\to B$.  The first arrow is inclusion of the fiber into $A\times B$ and the second arrow is the projection.  NOTE: $A\times B$ also fibers over A with fibers B of course.

Cartesian products are examples of fiber bundles but fiber bundles generalize the idea.  For a fiber bundle we also have a diagram: $F\to E\to B$ where F is the fiber, E is the total space and B is the base space.  The difference now is that E is not globally a cartesian product, though it will look locally like one.  This is analogous to the way manifolds generalize the notion of euclidean space.  A manifold is locally homeomorphic to $\mathbb{R}^n$ but not globally (at least not necessarily).  A fiber bundle is locally homeomorphic (maybe diffeomorphic?) to $B\times F$ but not globally.

Let’s look at some examples.  Take $[0,1]\to Cyl\to S^1$.  So in this example the fibers are unit intervals, the fiber bundle is the cylinder which fibers over the circle.  In this case the fiber bundle is globally a cartesian product.  Let’s try another one.

Take $[0,1]\to Mobius\to S^1$.  Once again the fibers are unit intervals but the Mobius band does not have the global structure of $[0,1]\times S^1$.  It does locally though, so the Mobius band is a fiber bundle over $S^1$.  One more example: $[0,1]\to HopfBand\to S^1$.  This one is even neater I think.  The Hopf Band is the surface whose boundary are two linked copies of $S^1$, or the Hopf link!  This also fibers over $S^1$ and also isn’t a cartesian product.

This is all very expository and will deserve some deeper study a bit later.  Again, what does this have to do with spheres.  There’s a very famous fibration called the Hopf fibration, $S^1\to S^3\to S^2$.  There’s a nice lecture on Youtube called “Visualizations of the Hopf Fibration.” by a fellow named Niles Johnson.  If you just want to watch the simulation it’s called “Hopf fibration — fibers and base.”  So it turns out that latex S^3$fibers over $S^2$ and that it locally looks like $S^2\times S^1$. The map from $S^3\to S^2$ is pretty cool. $S^3$ can be identified with the unit quaternions (also SU(2)) and so if we call the map $\eta$ and let $q\in S^3$ then $\eta (q)=qkq^{-1}$. Question: does this cover $S^2$ or double cover it? Niles Johnson makes a really neat statement (a couple of them actually) about this fibration. He talks about using it to understand the 3-sphere by using our knowledge of the 1-sphere (fibers) and 2-sphere (base space). That’s pretty neat. One of the students in the class asks if there are any other such fibrations of spheres and there are. Here are all of them:$latex S^0 \to S^1 \to S^1latex S^1 \to S^3 \to S^2latex S^3 \to S^7 \to S^4\$

$S^7 \to S^{15} \to S^8$

That’s it.  It’s apparently no coincidence either.  These are bound up in the fact that only $\mathbb{R}, \mathbb{C}, \mathbb{H}$, and $\mathbb{O}$ are normed division algebras!  This is pretty interesting.  I ended up looking at these fibrations because of some questions about knot cobordisms which came about because of questions about 2Cob and I end up bumping into these things!  There are some neat connections between SU(2) and SO(3) and regular polytopes of dimensions 3 & 4 that I’d like to explore further.  Moreover these Lie groups are of interest in physics as well!

I encourage you to watch the above linked lecture. It’s light and the explanations of the video demonstrating the Hopf fibrations is pretty illuminating.  Apparently the 3-sphere can be made out of tori???

There’ll be a Spheres-3 coming shortly.  The lecture also talks about using stereographic projection to visualize the 3-sphere in 3-space which makes a lot of sense.  I’ve found some nice videos demonstrating that as well.  Moreover I’ve gone down another rabbit hole to learn a bit more about the Poincare conjecture which is about spheres in higher dimensions.  So a couple more posts about Spheres and then I’ll try to get back to categories of and TQFT.

1. Oh and to boot, it turns out that Hopf’s map, $\eta$ is the generator for the infinite cyclic group $\pi_3 (S^2)$. This is really surprising. It’s easy to see that $\pi_1 (S^2)=0$ because it’s simply connected, i.e. all paths can be contracted continuously to a point. It also makes sense that $\pi_2 (S^2)\approx \mathbb{Z}$ since we can imagine mapping the 2-sphere into itself and we cannot shrink it because there’s a ‘hole’ inside the sphere. But what about $\pi_3$? Does this have anything to do with the double cover SU(2) provides? What does it mean to map the 3-sphere onto the 2-sphere in a way that cannot be contracted? so much to learn!
2. oh no…there’s more. I started looking for some information relevant to the previous post and came across this result: $\pi_4 (S^2)=\mathbb{Z}_2$