Let $G$ be a compact (Lie) group, and $H \subseteq H'$ two compact (Lie) subgroups. It is clear that we have an obvious surjective map of homogeneous spaces
$$
G/H \twoheadrightarrow G/H'.
$$
Will it be true in general that this fibration gives a fibre bundle? Will the fibre be isomorphic to the quotient space $H/H'$? Will it be true that the bundle is principal if and only if $H'$ is a normal subgroup of $G$?
P.S. Do such fibre bundles have a special name?
Best Answer
I call such bundles "homogeneous bundles", but it's not a totally standard terminology.
It is true that the map $G/H\rightarrow G/H'$ is a fiber bundle map with fiber $H/H'$. One way to see this is to start with the principal bundle $H\rightarrow G \rightarrow G/H$. The group $H$ naturally acts on $H/H'$, and so we have an associated fiber bundle $H/H'\rightarrow G\times_H (H/H')\rightarrow G/H$, where the notation $G\times_H (H/H')$ refers to the quotient of $G\times (H/H')$ by the $H$-action $h\ast(g, h_1 H') = (gh^{-1}, hh_1 H')$.
The space $G\times_H (H/H')$ is diffomeorphic to $G/H'$: the map which takes $[g, hH']$ to $ghH'$ is a diffeomorphism with inverse $gH'\rightarrow [g,eH']$. With respect to this diffeomorphism, the projection $G/H'\rightarrow G/H$ is the obvious one.
Now, if $H'$ is normal in $H$ (though it doesn't have to be normal in $G$), then the map $G/H'\rightarrow G/H$ is a principal $H/H'$ bundle. The $H/H'$ action on $G/H'$ is just the $H'/H$ action on $G\times_H (H'/H)$ given by multiplication on the second coordinate: $(hH')\ast [g, h_1 H'] = [g, h_1 h H']$
Lastly, it is not necessary that $H'$ be normal in $H$ in order for the bundle to be principal. To see this, consider the chain of subgroups $$O(1)\subseteq O(2)\subseteq O(3)$$ with each $O(k)$ embedded in $O(k+1)$ as the top left $k\times k$ block. The corresponding bundle $O(2)/O(1)\rightarrow O(3)/O(1)\rightarrow O(3)/O(2)$ is the Hopf bundle, which is a principal $S^1$-bundle. However, $O(1)$ is not normal in $O(2)$ since conjugating by a matrix which swaps the standard basis of $\mathbb{R}^2$ doesn't preserve $O(1)$.