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Let $G$ be a topological group and $H$ be a closed subgroup of $G$.

Consider the quotient space $G/H=\{gH:g\in G\}$ and the quotient map $q:G\rightarrow G/H$ given by $g\mapsto gH$. We have action of $H$ on $G$ from right side. This map $q:G\rightarrow G/H$ satisfies all properties to call principal bundle, except the local triviality. People called this kind of bundles to be Cartan principal bundles, which are roughly, principal bundles that does not require local triviality. So, what is the main reason for failure of local triviality in case of topological groups?

Let $G$ be a Lie group and $H$ is a closed subgroup of $G$. Then example $5.1$ of Kobayashi and Nomizu's book says that $q:G\rightarrow G/H$ is a principal bundle, local trivialization is included in this set up.

What is the main reason for failure of local triviality of $q:G\rightarrow G/H$ when $H$ is a closed subgroup of $G$ when $G$ is a topological group? What is the main reason for local triviality of $q:G\rightarrow G/H$ when $H$ is a closed subgroup of $G$ when $G$ is a Lie group?


For any one who is thinking what is a Cartan principal bundle, here is the definition.

A Cartan principal bundle is a quadrapule $P(B,G,\pi)$ where $P$ and $B$ are spaces, $G$ a topological group acting freely on $P$ to the right through $P\times G\rightarrow P$, $(u,g)\mapsto ug$, and $\pi:P\rightarrow B$ is a surjective map, subject to the following conditions:

  1. the fibres of $\pi$ equal the orbit of $G$, that is, for $u,v\in P$, the statement $\pi(u)=\pi(v)$ is equvalenet to $\exists ~ g\in G ~ : v=ug$.
  2. the division map $\delta:P\times_{\pi} P\rightarrow G, (ug,u)\mapsto g$ resulting from $(1)$, is continuous. Here $P\times_{\pi}P=\{(v,u)\in P\times P:\pi(v)=\pi(u)\}$ has the subspace topology.
  3. $\pi:P\rightarrow B$ is an identification map.

Copied from here

A simple example when $G\to G/H$ is not locally trivial can be found in the paper of Karube [On the local cross-sections in locally compact groups, J. Math. Soc. Japan 10 1958 343–347]. In the example $G$ is the product of infinitly many circles, and $H$ is the product of their order $2$ subgroups; there can be no cross-section because $G$ is locally-connected and $H$ is not, so $G$ is not even locally homeomorphic to $H\times G/H$.

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  • $\begingroup$ "What is the main reason for..." makes the question quite vague. Also it would maybe help if you provide examples without local triviality. $\endgroup$ – YCor Sep 22 at 21:23
  • $\begingroup$ @YCor sorry, I did not understand what you mean.. I understand question is quite vague.. I did not understand next sentence :O I do not know how to make this better, please suggest if you have anything in mind that might make this well posed.. $\endgroup$ – Praphulla Koushik Sep 23 at 9:54
  • $\begingroup$ Provide an example of a Hausdorff topological group $G$ and closed subgroup $H$ for which there is no local trivialization (I guess I can think of one, but I think it makes the question more telling if you have one in mind). $\endgroup$ – YCor Sep 23 at 11:27

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