The Unapologetic Mathematician

Mathematics for the interested outsider

The Hopf Fibration

As a nontrivial example of a foliation, I present the “Hopf fibration”. The name I won’t really explain quite yet, but we’ll see it’s a one-dimensional foliation of the three-dimensional sphere.

So, first let’s get our hands on the three-sphere S^3. This is by definition the collection of vectors of length 1 in \mathbb{R}^4, but I want to consider this definition slightly differently. Since we know that the complex plane \mathbb{C} is isomorphic to the real plane \mathbb{R}^2 as a real vector space, so we find the isomorphism \mathbb{R}^4\cong\mathbb{C}^2. Now we use the inner product on \mathbb{C} to define S^3 as the collection of vectors (z_1,z_2) with \lvert z_1\rvert^2+\lvert z_2\rvert^2=1.

Now for each \alpha\in\mathbb{R} we can define a foliation. The leaf through the point (z_1,z_2) is the curve \left(z_1e^{it},z_2e^{i\alpha t}\right). Since multiplying by e^{it} and e^{i\alpha t} doesn’t change the norm of a complex number, this whole curve is still contained within S^3. Every point in S^3 is clearly contained in some such curve, and two points being contained within the same curve is an equivalence relation: any point is in the same curve as itself; if w_1=z_1e^{it} and w_2=z_2e^{i\alpha t}, then z_1=w_1e^{i(-t)} and z_2=w_2e^{i\alpha(-t)}; and if w_1=z_1e^{is}, w_2=z_2e^{i\alpha s}, x_1=w_1e^{it} and x_2=w_2e^{i\alpha t}, then x_1=z_1e^{i(s+t)} and x_2=w_2e^{i\alpha(s+t)}. This shows that the curves do indeed partition S^3.

Now we need to show that the tangent spaces to the leaves provide a distribution on S^3. Since this will be a one-dimensional distribution, we just need to find an everywhere nonzero vector field tangent to the leaves, and the derivative of the curve through each point will do nicely. At (z_1,z_2) we get the derivative

\displaystyle\frac{d}{dt}\left(z_1e^{it},z_2e^{i\alpha t}\right)\Big\vert_0=(iz_1,i\alpha z_2)

It should be clear that this defines a smooth vector field over all of S^3, though it may not be clear from the formulas that these vectors are actually tangent to S^3. To see this we can either (messily) convert back to real coordinates or we can think geometrically and see that the tangent to a curve within a submanifold must be tangent to that submanifold.

The Hopf fibration is what results when we pick \alpha=1, but the case of irrational \alpha is very interesting. In this case we find that some leaves curve around and meet themselves, forming circles, while others never meet themselves, forming homeomorphic images of the whole real line. What this tells us is that not all the leaves of a foliation have to look like each other.

To see this, we try to solve the equations

\displaystyle\begin{aligned}z_1&=z_1e^{it}\\z_2&=z_2e^{i\alpha t}\end{aligned}

The first equation tells us that either z_1=0 or t=2\pi n. In the first case, we simply have the circle \lvert z_2\rvert=1. In the second case, the second equation tells us that either z_2=0 or 2\pi m=\alpha t=2\pi\alpha n. The case where z_2=0 is similar to the case z_1=0, but if neither coordinate is zero then we find \alpha=\frac{m}{n}. But we assumed that \alpha is irrational, so we get no nontrivial solutions for t here.

Since the curves don’t change the length of either component, we can get other examples of foliations. For instance, if we let \lvert z_1\rvert=\lvert z_2\rvert=\frac{1}{2}, then the curve will stay on the torus S^1\times S^1 where each circle has radius \frac{1}{\sqrt{2}} in its copy of \mathbb{C}. Looking at all the curves on this surface gives a foliation of the torus. If \alpha is irrational, the curve winds around and around the donut-shaped surface, never quite coming back to touch itself, but eventually coming arbitrarily close to any given point on the surface.

About these ads

July 4, 2011 - Posted by | Differential Topology, Topology

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 388 other followers

%d bloggers like this: