Hopf Fibration – Proving Non-Triviality of the Fibre Bundle

differential-geometryfiber-bundlesgroup-cohomologyhopf-fibrationobstruction-theory

I want to show that the Hopf bundle
$$ \mathbb{S}^1 \rightarrow \mathbb{S^3} \rightarrow \mathbb{S}^2$$
is non-trivial as a principal fibre bundle.

I have seen hints of several different approaches:

  1. Hopfs original approach, using linking numbers, see Hopf fibration and $\pi_3(\mathbb{S}^2)$.
  2. Hopf invariant.
  3. Cohomology.

I want to keep it simple. My gut says cohomology is my best bet (I am slightly familiar with de Rahm Cohomology). Unfortunately, I am having trouble finding sources that treat this on my level (without a lot of general theory I think I don't need).

I have the following books:

  1. The Topology of Fibre Bundles, Steenrod.
  2. Fibre Bundles, Husemoller.
  3. Manifold and Differential Geometry, Jeffry Lee.

What is the bare minimum of theory I need to to get to this result? What route would you advise?

I am an undergraduate working on a Bsc thesis.

Best Answer

The easier way to show that is to remark that $\pi_1(S_3)=1$ and $\pi_1(S_2\times S_1)=Z$.

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