Circle bundles inside trivial vector bundles

Question

Suppose you are given a circle bundle $p: E\rightarrow S^2$ described as a sub-bundle of the trivial vector bundle $R^2\times S^2\rightarrow S^2$ in terms of equations (meaning for each point $p\in S^2$ we have an equation that describes a smooth simple closed curve in $R^2$ depending smoothly on the point $p$). Given that the Euler class of the trivial vector bundle $R^2\times S^2\rightarrow S^2$ is zero, I would assume that also the circle bundle is trivial. Is this guess correct and does this follow from some results? More in general is it true that any circle bundle inside a trivial real 2-dimensional vector bundle is trivial? Thanks!

Answer 1

For each $p \in S^2$, you have a smooth parameterisation of a simple closed curve $\phi_p : S^1 \to \mathbb{R}^2$. Moreover, the maps $\phi_p$ vary smoothly in $p$, so the map $\Phi : S^1\times S^2 \to \mathbb{R}^2\times S^2$ given by $\Phi(t, p) = (\phi_p(t), p)$ is smooth and satisfies $\operatorname{pr}_2\circ \Phi = \operatorname{pr}_2$. Note that $\phi_p$ is a diffeomorphism onto its image for each $p$, so $\Phi$ is a diffeomorphism onto its image $Z := \Phi(S^1\times S^2)$. The total space of the circle bundle you describe is $Z$ and the projection map $Z \to S^2$ is just $\operatorname{pr}_2|_Z$. Then $\Phi^{-1} : Z \to S^1\times S^2$ is a global trivialisation of the circle bundle as $\operatorname{pr}_2\circ \Phi^{-1} = \operatorname{pr}_2|_Z$.