Group structure of the local trivializations of a Lie group principal bundle

Question

Consider a Lie group $G$ and its normal Lie subgroup $H$. Then $G/H$ is a Lie group and a homogeneous space of the group $H$ with the right action. One can construct a principal $H$-bundle $$\pi: G \rightarrow G/H \,.$$ Considering now the local trivializations of this bundle $$ \psi_\alpha : \pi^{-1}(\mathcal{O}_\alpha) \rightarrow \mathcal{O}_\alpha \times H \quad , \quad \mathcal{O}_\alpha \subset G/H$$ with $ \psi_\alpha $ being diffeomorphisms, under what circumstances (conditions on the bundle in question) will the trivialization maps actually be homomorphisms (of Lie groups)? In other words, is it possible to construct the bundle and the trivialization maps in such a way that the $ \mathcal{O}_\alpha \times H $ will actually be a direct product of groups? My guess is that this would only be possible if the bundle was trivial (together perhaps with some other conditions), because the $ \mathcal{O}_\alpha $ are open and Lie subgroups need to be closed.

Answer 1

There are some problems with the question as stated. First, $G/H$ isn't a Lie group unless $H$ is a closed subgroup.

Second, before you can even ask whether $\psi_\alpha$ is a homomorphism, you have to ask whether the local trivializations can be chosen so that the sets $\mathcal O_\alpha$ are subgroups. This is almost never the case: the only open subgroups of a Lie group are unions of connected components. So at the very least, before you can even ask the question, you have to require that the bundle be trivial over the identity component at least.

Finally, your last comment, "Lie subgroups need to be closed," is not true -- a dense $1$-dimensional Lie subgroup of the torus is a counterexample. Besides, being closed does not preclude being open -- the identity component of a disconnected group is a Lie subgroup that's both closed and open.