Setting: Let $M$ be a smooth manifold and $\{U_\alpha\}_{\alpha \in \mathcal{I}}$ a locally finite covering of open subsets. Furthermore let $G$ be a smooth Lie group. Now assume we are given a family of transition functions $$\phi_{\alpha\beta}\in \mathcal{C}^0(U_\alpha \cap U_\beta; G) \quad\text{for any}\quad \alpha,\beta \in \mathcal{I}$$ satisfying the cocyclicity conditions: $$ \phi_{\alpha\alpha} = \mathbb{1}_{U_\alpha} \\ \phi_{\alpha\beta} \phi_{\beta\gamma} = \phi_{\alpha\gamma}. $$ on the obvious domains. Now $P := \left( \coprod_{\alpha \in \mathcal{I}} U_\alpha \times G \right)\big/{\sim}$ with $$(x,g)_\alpha \sim (x, \phi_{\beta\alpha}(x)\cdot g)_\beta \quad\text{for any}\quad x \in U_\alpha \cap U_\beta$$ defines a $G$-principal fibre bundle over $M$ (in the category of topological spaces).

Question: Can we conclude that $P \rightarrow M$ is a smooth $G$-principal bundle over $M$? That is, a principal bundle in the category of smooth manifolds?

  • 2
    $\begingroup$ No, we can't. In general, there is no reason to expect continuous transition maps to yield a smooth bundle. $\endgroup$ Jan 13 '18 at 11:35
  • $\begingroup$ @AmitaiYuval I find it hard to come up with a counterexample. Do you have any in mind or do you just suspect this to hold true? When I posed the question, I was thinking about Chern-classes, which classify certain principal bundles over a manifold. They are topological invariant but over smooth base spaces we can (I think) realise every topological classe of a bundle by a smooth principal bundle. $\endgroup$ Jan 14 '18 at 21:53

Proof of continuous principal bundle is equivalent to smooth principal bundle
So the only thing left to do is to prove that your construction gives out a continuous principal bundle.
The first condition of been a continuous principal bundle is: The right $G$ action is continuous. Which in this construction, $(U_\alpha\times G)\times G\to(U_\alpha\times G)$, $(x,g)h\mapsto (x,gh)$, Because of the definition of Lie groups, it must be smooth, thus continuous.
And the second condition says that base manifold $M$ must exist local trivial cover $U_\alpha$, and that is already given in your construction.
So we conclude the $G$-principle bundle given is a continuous $G$-principal bundle. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.