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

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?

• Have a look at math.stackexchange.com/questions/60522/…. Commented Feb 7, 2019 at 23:03
• @RobArthan So the Hopf Fibration is a double cover of this bundle? Commented Feb 7, 2019 at 23:11
• Yes. The double covering of the total spaces is the plate trick. Commented Feb 7, 2019 at 23:17

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.