# Fibre bundles for homogeneous spaces

Let $G$ be a Lie group and let $H$ be a closed subgroup of $G$. Then it is known that $G/H$ can be equipped with a unique differentiable structure such that $G\xrightarrow{\pi} G/H$ is a locally trivial fiber bundle. Then, the standard theory of principal bundles, connections, associated vector bundles and the associated covariant derivatives applies to $G\xrightarrow{\pi} G/H$, and I guess some simplifications appear. My question is:

The theory of principal bundles connections, associated vector bundles and the associated covariant derivatives applied to principal bundles of the form $G\xrightarrow{\pi} G/H$ goes under some special name? Could you give me a reference where this theory is developed? Is this related to the notion of "Cartan connection"?

Thanks.

• Aren't all fiber bundles locally trivial? – Robin Goodfellow Apr 21 '15 at 20:35
• @RobinGoodfellow: There "fibrations" which do not need to be locally trivial. Usually when one says "fiber bundle" in this informal discussions it is implicitly assumed that it is locally trivial. In any case your comment is irrelevant to my question. – Bilateral Apr 21 '15 at 21:03
• I think it is very relevant. We need to understand what you are asking to answer your question. You said "fiber bundle" and not "fibration," which was confusing. Also, fiber bundles are defined as locally trivial, informal discussion or not. – Robin Goodfellow Apr 21 '15 at 21:11
• @RobinGoodfellow: Your comments are completely irrelevant to the discussion. I wrote "locally trivial fiber bundle" which is perfectly fine. What is the problem? How your comments help understand what I am asking? It is not perfectly clear what I mean by "locally trivial fiber bundle"? – Bilateral Apr 21 '15 at 21:23

Vector bundles associated to the principal bundle $G\to G/H$ can be equivalently characterized as homogeneous vector bundles (i.e. as vector bundles endowed with a lift of the action of $G$ on $G/H$ to an action by vector bundle homomorphisms.) There also is a notion of homogeneous connections and so on.
There is a relation between homogeneous spaces and Cartan connection, which is that the Maurer-Cartan form on $G$ defines a Cartan connection on $G\to G/H$ and this is the so-called homogeneous model of Cartan geometries of type $(G,H)$. This is related to the fact that the tangent bundle $T(G/H)$ always is an associated bundle to $G\to G/H$. Roughly, you take the Maurer-Cartan form as defining a geometric structure on $G/H$ whose automorphisms are exactly the left actions of elements of $G$ and then look for "curved analogs" of this situation (which are Cartan geometries).
• Thanks! Then, if I understood correctly a Cartan connecion is a particular case of a connection on a principal bundle adapted to principal bundles of the form $G\to G/H$ and which can be obtained from the Maurer-Cartan form of $G$, right?. I don't understand the following: "This is related to the fact that the tangent bundle $T(G/H)$ always is an associated bundle to $G\to G/H$. " Is this related to the "soldering form"? – Bilateral Apr 22 '15 at 20:13
• No, this is a complete misunderstanding. The Maurer Cartan form on $G$ defines a Cartan connection on $G/H$, but this is a very special case. For example, a Riemannian metric on any manifold $M$ can be encoded as a Cartan connection on the orthonormal frame bundle of $M$. Morover, a general principal connection on $G\to G/H$ is not related to a Cartan connection, and generically (even on $G\to G/H$) it is not possible to obtain a principal connection from a Cartan connection. But this is an involved topic and not of the length of a comment here. – Andreas Cap Apr 23 '15 at 6:44
• @AndrasCap: What I hope it is correct is the following: Let $H\subset G$ be a closed subgroup of $G$. Then the Maurer-Cartan form $\omega\colon TG\to\mathfrak{g}$ of $G$ is a particular instance of a Cartan connection on $G\to G/H$. Moreover, if $(\mathfrak{g},\mathfrak{h})$ is reductive, then $\omega$ induces a "standard" principal $H$-bundle connection on $G\to G/H$. – Bilateral Apr 23 '15 at 11:28