Is the Hairy Ball Theorem equivalent to saying that the Hopf Fibration has no global sections?

Question

The Hairy Ball Theorem states that $S^2$ has no nonvanishing tangent vector fields. But if we did have such a field then we could normalise each vector so that it lay on the unit circle of the tangent plane at that point. These unit circles form a bundle over $S^2$ with fibre $S^1$. The Hairy Ball Theorem is therefore equivalent to saying that this bundle has no global section.

Since this is a nontrivial $S^2$-bundle with fibre $S^1$ I thought that it might be the Hopf Fibration. Is it? If not, what is the total space?

Answer 1

The total space is $SO(3) \cong \mathbb{RP}^3$. You can see this because $SO(3)$ acts freely and transitively on it, by first rotating points in $S^2$ and then by rotating tangent vectors at a given point.

It's an interesting exercise to similarly identify the total spaces of the unit tangent bundles of the other closed surfaces (fixing, say, a metric of constant curvature).