Erlangen-Style approach to Homogneoues Fiber Bundles In connection with a philosophy of physics project I have recently been looking at fiber bundles from an Erlangen-inspired perspective. My inspiration is the following well-known result: Every smooth homogeneous manifold, $M$, is diffeomorphic to the quotient of some two Lie groups, $M\cong H_1/H_2$, where $H_2$ is a closed subgroup of $H_1$. Therefore we can think of every homogeneous manifold , $M$, in terms of a pair of Lie groups $(H_1,H_2)$. I think I have proved an analogous result for fiber bundles (stated below). My questions are:

*

*Is there already a name for what I am calling "homogeneous fiber bundles"?

*Have the construction and characterization results given below already been proved?

*If yes to either 1 or 2 where can I find more information?

Let $(E,B,F,\pi:E\to B)$ be a fiber bundle. We will call this fiber bundle homogenous if and only if the following two conditions hold. Firstly, there must exist a finite dimensional Lie group, $T_1$, which acts smoothly and transitively on $E$ while mapping fibers to fibers. Secondly, there must exist a finite-dimensional Lie group, $T_2\subset\text{ker}(\pi)$, which acts trivially on the base space and acts smoothly and transitively on the fibers.
Let $(G_1,G_2,G_3)$ be a trio of Lie groups with $G_2$ a closed subgroup of $G_1$ and with $G_3$ a closed normal subgroup of $G_1$. One can build a homogeneous fiber bundle as follows: total space, $E_\text{G}=G_1/G_2$, base space, $B_\text{G}=G_3\backslash G_1 / G_2$, fiber type, $F_\text{G}=G_3/(G_3\cap G_2)$, and the canonical projector $\pi_\text{G}:E_\text{G}\to B_\text{G}$.
Thus at least some homogeneous fiber bundles can be built in this way, ($E_\text{G}$, $B_\text{G}$, $F_\text{G}$, $\pi_\text{G}$). Surprisingly, however, every homogeneous fiber bundle can be constructed in this way. Therefore we can think of every homogeneous fiber bundle, $(E,B,F,\pi)$, in terms of a trio of Lie groups $(G_1,G_2,G_3)$.
More generally, I am interested in analogous "homogeneity" results regarding other types of smooth topological structures, $X$. We can think of every homogeneous $X$ in terms of a n-tuple of Lie groups $(G_1,...G_n)$. Even more generally, we might allow these Lie groups to be Lie supergroups such that $X$ could be a supermanifold or a super fiber bundle.
Any comments or references would be much appreciated.
Edit: I see now that I can replace $G_3$ with the Frobenius product $G_{new} = G_2 G_3 = G_3 G_2$ since only this only way that it appears in the quotients, namely $E_G = G_1/(G_2 G_3)$ and $F_G = G_2 G_3/G_2$. This brings the fiber bundle into the form mentioned in the comments $G_{new}/G_2 \to G_1/G_2 \to G_1/G_{new}$. From here with a bit of work then drop the $T_2$ clause from the definition of "homogeneous fiber bundle" and the condition that $G_3$ has to be normal in $G_1$.
 A: *

*The name "homogeneous fiber bundle" is already a standard terminology for a related phenomenon in your answer.  Specifically, say we are given three Lie groups $H\subseteq K\subseteq G$ where each subgroup is closed.  (But no assumptions about normality)

Then the projection map $\pi:G/H\rightarrow G/K$ is a fiber bundle with fiber $K/H$.  The total space admits a transitive action by $G$.  Restricting this action to $K$ induces a transitive action of $K$ on each fiber.
As an example, the triple $\{e\}\subseteq S^1\subseteq SU(2)$ gives rise to the Hopf fibration $S^1\rightarrow S^3\rightarrow S^2$.
I don't know of a name for your specific construction, but it can be viewed as a special case of this one (at least if $G_1$ is compact and connected) as follows.  First, passing to a cover, we may assume that $G_1 = H_1\times ...\times H_s\times T^k$ is a product of simple Lie groups with a torus.  A normal subgroup is then a sub-product of these simple Lie groups, together with a subtorus.  By rearranging the factors if necessary, we can assume $G_3 = H_1\times ...\times H_r \times T^j \times \{e_{r+1}\}\times ...\times \{e_s\} \times \{e\}$ where $r\leq s$, $j\leq k$, and where $e_k$ refers to the identity in $H_k$ and $e$ is the identity in $T^{k-j}$.
Let $L$ denote the projection of $G_2$ the factors $H_{r+1}\times ...\times H_s \times T^{k-j}$.  Then we have the triple $G_2\subseteq H_1\times ...\times H_r \times T^j \times L \subseteq G_1$, and the homogeneous fiber bundle associated to this triple is the same fiber bundle you get from your construction.


*With respect to the standard notion, it's well know that that a triple gives rise to a bundle.  Conversely, given a transitive action by a Lie group with closed isotropy subgroup, the existence of intermediate closed subgroups is necessary and sufficient to create a homogeneous bundle.


*I don't know anywhere in the literature where this stuff is written down, but that's probably more from a lack of knowledge on my part.  The existence of the bundle given the triple is a simple application of the associated bunlde construction:  Starting with the principal bundle $K\rightarrow G\rightarrow G/K$, the group $K$ naturally acts on $K/H$, so we can form the associated bundle $G\times_K K/H$ and its a routine excercise to see that $G\times_K K/H$ is diffeomorphic to $G/H$.
