principal-bundles – Nontrivial Examples of Non-Trivial Principal Circle Bundles

at.algebraic-topologyfibre-bundlesprincipal-bundles

It is a well known fact that (isomorphism classes of) principal $S^1$-bundles over a base space $B$ are classified by $B$'s second integral cohomology, $H^2(B;\mathbb{Z})$.

There is always the trivial one $B\times S^1$, and for $B=S^3$ for example, these are all. The Hopf bundle is an interesting example when $B=S^2$. The principal bundles over $S^2$ correspond bijectively to $\mathbb{Z}$; much interesting information about them can be extracted using tools such as in Blair[1].

However, over the Klein bottle or $\mathbb{R}P^2$ there is only one such non-trivial bundle (since their second cohomology is $\mathbb{Z}_2$). So in order to classify all of them, one just needs to find out what is the non-trivial bundle.

Question:

  1. What are the non-trivial principal circle bundles over the Klein bottle and $\mathbb{R}P^2$ ?

  2. In particular, given a nice base-space with 2nd integral cohomology a finite group (such as $\mathbb{R}P^2$), is there a constructive way to find out what are the non-trivial principal circle bundles over it?

    [1]: D.E. Blair, Riemannian Geometry of Contact and Symplectic Manifolds.

Best Answer

A third way to think about Anton Petrunin's example is that $S^2 \to {\mathbb R}P^2$ is a ${\mathbb Z}_2$ principal bundle where the action of ${\mathbb Z}_2 = \lbrace +1, -1 \rbrace $ is the obvious action on vectors in ${\mathbb R}^3$. As ${\mathbb Z}_2$ is a subgroup of $U(1)$, the circle group of complex numbers of length one, you can use standard principal bundle constructions to extend this ${\mathbb Z}_2$ principal bundle to a $U(1)$ principal bundle. In the case at hand these constructions just give the quotient of $S^2 \times S^1$ by ${\mathbb Z}_2$ as above.

You can compute the transition functions of this $U(1)$ bundle explicitly with respect to the standard open cover of ${\mathbb R}P^2$ by just computing the same for the ${\mathbb Z}_2$ bundle $S^2 \to {\mathbb R}P^2$ and check that they give you a Cech representative for the non-zero class in $H^2({\mathbb R}P^2, {\mathbb Z}_2) = {\mathbb Z}_2$.

The fact that this $U(1)$ bundle has a reduction to ${\mathbb Z}_2$ also tells us that when we square it we will get a trivial $U(1)$ bundle. Just think of squaring the ${\mathbb Z}_2$ valued transitions functions to get transitions functions for the squared bundle. Obviously they all just take the value $1$.

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