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:

  1. Is there already a name for what I am calling "homogeneous fiber bundles"?
  2. Have the construction and characterization results given below already been proved?
  3. 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 $G_1$.


1 Answer 1

  1. 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.

  1. 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.

  2. 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 $G/H$.

  • $\begingroup$ By the way, it is unclear to me how to fit the Hopf bundle into the OPs proposed framework. The group $SU(2)$ acts on the total space $S^3$, mapping fibers to fibers, but where is normal subgroup $G_3$? $\endgroup$ Mar 1 at 15:16
  • $\begingroup$ Thanks for the help! I don't know about the Hopf bundle exactly, but you can do $S^1\to RP^3 \to S^2$ or equivalently $SO(2)\to SO(3)\to S^2$ as follows. Let $G_3=\{e\}\times SO(2)$ act on $E=SO(3)$ on the right. The orbits of $G_3$ are the fibers. Let $G_1=SO(3)\times SO(2)$ with the first and second factors being left and right action on $E=SO(3)$. The left action permutes the fibers and right action moves within each fiber. Let $G_2=SO(2)$ be the stabilizer subgroup of $G_1$'s action. The quotient $G_1/G_2$ gives $E=SO(3)$. Taking another quotient with $G_3$ collapses the fibers. $\endgroup$ Mar 1 at 17:06
  • $\begingroup$ Ah, so the $T_1$ is not necessarily $G_1$ in your construction. This makes sense. I think you can just change $SO(3)$ to $SU(2)$ to get the Hopf bundle then. $\endgroup$ Mar 1 at 19:22

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