Nontrivial homogeneous Vector bundle

Question

What is an example of a Lie group $ G $ and a subgroup $ G' $, both with finitely many connected components, such that $ G\mathbin{/}G' $ is a nontrivial vector bundle? This is a follow up to

https://mathoverflow.net/questions/345905/homogeneous-manifold-deformation-retracts-onto-compact-submanifold

EDIT:

Take $ G $ to be the connected group $$ G=SE_2:= \left \{ \ \begin{bmatrix} a & b & x \\ -b & a & y \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \} $$ there is a connected group $ V $ of translations up each vertical line $$ V= \left \{ \ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix} : y \in \mathbb{R} \right \} $$ Now if we include the rotation by 180 degrees $$ \tau:=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Then $G' = \langle V, \tau \rangle$ has two connected components and $$ G\mathbin{/}G' $$ is the Moebius band (hopefully? It could still be the cylinder but I don't know how to check)

Answer 1

One example is the Möbius strip, which admits a homogeneous $E(2)$ (or $SE(2)$) structure. To see this note that the set of affine lines in $\mathbb{R}^2$ (A manifold I'll call $M$) is diffeomorphic to the Möbuis strip.

One way to see this is using the projection $M\to\mathbb{RP}^1\cong S^1$ given by mapping each line to the unique parallel line which passes through the origin. Each fiber $\pi^{-1}(l)$ of this projection can be identified with the orthogonal compliment of $l$., and the resulting bundle has typical fiber $\mathbb{R}$ and is nontrivial. (If you're familiar, this construction is equivalent to the tautological line bundle of $\mathbb{RP}^1$.)

This set of lines admits a transitive $E(2)$ action, so we have $M\cong E(2)/G$, where $G$ is the stabilizer of the $x$-axis.

Answer 2

$ SO_3(\mathbb{C})/SO_2(\mathbb{C}) $ is the tangent bundle of the sphere $ S^2 $. This vector bundle is nontrivial ($ S^2 $ is not parallelizable).