# Motivation for construction of Associated fiber bundle from a principal bundle

Given a principal $$G$$ bundle $$P(M,G)$$ and a manifold $$F$$ with an action of $$G$$ on it from left, we construct a fibre bundle over $$M$$ with fiber $$F$$ and call this the associated fiber bundle for $$P(M,G)$$.

I do not get the motivation behind the construction given in Kobayashi and Nomizu which I will write down below.

Idea is to construct a fibre bundle with fibre $$F$$ i.e., we need to construct a smooth manifold $$E$$ and a smooth map $$\pi_E:E\rightarrow M$$ that gives a fiber bundle with fiber $$F$$.

Kobayashi's proof goes as follows :

They consider the product manifold $$P\times F$$ with an action of $$G$$ as $$g.(u,\xi)=(ug,g^{-1}\xi)$$. Then they consider the quotient space $$(P\times F)/G$$ and call this $$E$$.

Consider the map projection map $$P\times F\rightarrow M$$ defined as $$(p,\xi)\mapsto \pi(p)$$.

This induces a map $$\pi_E:E=(P\times F)/G\rightarrow M$$. As $$P\rightarrow M$$ is a principal $$G$$ bundle given $$x\in M$$ there exists an open set $$U$$ containing $$x$$ and a local trivialization $$\pi^{-1}(U)\rightarrow U\times G$$. They then give a bijection $$\pi_E^{-1}(U)\rightarrow U\times F$$ and give a smooth structure on $$E$$ so that these bijections are difeomorphisms. Then, they cal $$(E,\pi_E,M,P,F)$$ the fiber bundle associated to principal $$G$$ bundle.

I am trying to understand the motivation for the above construction.

Suppose $$F=H$$, a Lie group and the action of $$G$$ on $$H$$ is given by a morphism of Lie grroup $$\phi:G\rightarrow H$$ with $$G\times H\rightarrow H$$ given by $$(g,h)\mapsto \phi(g)^{-1}h$$ do get a principal $$H$$ bundle in above construction?

Edit : I thinnk above content looks like it is asking why do we need the construction of associated fiber bundles. No, what I am asking is, suppose I have a Principal $$G$$ bundle $$P(M,G)$$ with an action of $$G$$ on a manifold $$F$$ and I want to associate some fiber bundle on $$M$$ with fiber $$F$$. Then, what suggests you to think of above construction. How does it occur naturally? Are there any other properties of Fiber bundle I should have in mind which suggest this way of construction.

• $G$ does not act on its subgroups in any reasonable way; I guess you wanted to turn that arrow around. In that case, yes. But I think the most important application of this construction is the definition of associated vector bundle to a principal bundle given a representation of $G$. – Mike Miller Sep 26 '18 at 8:21
• @MikeMiller I have edited the content... Can you say something on motivation behind such definition... As you said, given a representation $G\rightarrow Gl(V)$ which is same thing as a smooth action of $G$ on $V$ we can talk about associated fiber bundle.. Fiber in this case is vector space so we have a associated vector bundle in this case.. – Praphulla Koushik Sep 26 '18 at 8:44
• Is your question about potential applications of associated bundles as a motivation for studying them? If yes, you may want to look at gauge theories in physics. – S.Surace Sep 26 '18 at 9:53
• Have you looked closely at (what I think is) the simplest interesting case of this situation: You have a vector bundle, say over $\mathbb R$, with fibers of dimension $n$, and you want to consider bases for the vector spaces that are the fibers of your vector bundle. These bases are the points in a principal $GL(n,\mathbb R)$-bundle. If you apply the "associated bundle" construction with this principal bundle and the standard action of $GL(n,\mathbb R)$ on $\mathbb R^n$, you get back your original vector bundle (up to isomorphism). – Andreas Blass Sep 26 '18 at 14:36
• This construction exactly reverses the construction of the frame bundle from a vector bundle (e.g., the tangent bundle). The idea is that each point $f \in F_p$ in the frame bundle of a vector bundle $E$ is, by definition a basis of $E_p$. This defines a natural map of $F \times \mathbb{R}^k \rightarrow E$, where $k$ is the rank of $E$. The inverse image of any $e_p \in E$ is naturally isomorphic to $G$, so the quotient is a vector bundle isomorphic to $E$. – Deane Yang Sep 26 '18 at 14:38

This construction reverses the construction of the frame bundle from a vector bundle (e.g., the tangent bundle). The idea is that each point $$f \in F_p$$ in the frame bundle of a vector bundle $$E$$ is, by definition a basis of $$E_p$$. This therefore defines a natural map of $$F \times \mathbb{R}^k \rightarrow E$$, where $$k$$ is the rank of $$E$$. The inverse image of any $$e_p \in E$$ is naturally isomorphic to $$G$$, so the quotient $$(F\times \mathbb{R}^k)/G$$ is a vector bundle isomorphic to $$E$$.

• This seem to be the thing I am looking for.. I need some time to digest... Thank you :) – Praphulla Koushik Sep 27 '18 at 6:23

To answer your specific question, you get a principal $$H$$-bundle, often denoted $$P \times^G H$$. The transition maps are clearly just given by applying $$\phi$$ to those of $$P$$.

I think you want a motivation for the general notion of relating principal and fiber bundles through associated bundles. Tensor bundles come up rather naturally, as do projectivized cotangent bundles, projectivized tangent bundles and sphere bundles (i.e. quotients of nonzero vectors by positive scaling) on the tangent and cotangent bundles. These bundles might appear to be beasts whose natural habitats are rather different parts of the forest, but they all live in the zoo of bundles associated to the one principal bundle: the frame bundle, i.e. the set of linear isomorphisms between a fixed vector space and the tangent spaces of our manifold. In this way, topological obstructions to the existence of linearly independent sections become characteristic classes on the one principal bundle. The method of the moving frame makes use of the existence of invariant differential forms on the frame bundle, and its various subbundles. G-structures are just subbundles of the frame bundle, but together they describe many of the most important geometric structures, and the theory of G-structures makes uniform construction of the local invariants of all of those structures. So the principal bundle is unifying object bringing all of those associated fiber bundles and vector bundles together, for topology and also for local differential geometry.

• I understand your first paragraph... second paragraph is completely some other foreign language for me.. :( I understand almost nothing there.. please consider using some more words in explaining... – Praphulla Koushik Sep 26 '18 at 9:50
• @PraphullaKoushik If you find the second paragraph unintelligible then you are learning the subject in the wrong order, and Kobayashi-Nomizu is not the book you should be reading. It assumes (as Ben McKay did) that you are already familiar with various examples of vector bundles on manifolds and why they are important - with that understanding, Ben's answer would tell you that many of the examples and structures that you are already familiar with are unified and clarified by the associated bundle construction. As an alternative, you might start with Marsden-Tornhave's book, for instance. – Paul Siegel Sep 26 '18 at 19:41
• I think I have conveyed wrongly what I am wanting to say... I understand that all those bundles arise naturally as associated bundles of one Principal bundle on manifold i.e., the frame bundle.. what I am not seeing is the motivation for that particular construction of fiber bundle from a Principal bundle... @PaulSiegel – Praphulla Koushik Sep 27 '18 at 1:31
• Ah sorry I just repeated what @DeaneYang has in his answer. – Michael Murray Sep 27 '18 at 5:10
• Thanks for the answer.. it did not answer my question that I asked above but it said why association of fiber bundle construction is important.. thanks again... bounty is for that .. – Praphulla Koushik Sep 29 '18 at 17:59

This is just an attempt to elaborate a bit on Ben McKay's answer beyond the confines of a mere comment.

Principal $$G$$-bundles $$P(M,G)$$ over $$M$$ can be understood as a sort of "universal generator" of transition cocycles for its associated $$G$$-bundles over $$M$$. More precisely, the transition cocycles associated to a $$G$$-bundle atlas for $$P(M,G)$$ are transition cocycles for the associated $$G$$-bundle with any given typical fiber $$F$$, up to the choice of $$G$$-action on $$F$$. The quotient $$(P\times F)/G$$ is meant to achieve precisely that. This is especially clear in the case of frame bundles and vector bundles, as depicted in Deane Yang's answer.

Since the family of transition cocycles encode all information on the topological deviations of $$P\times_G F$$ from the trivial bundle $$M\times F$$, this means that all nontrivial topological information of $$P\times_G F$$ in this sense is already encoded in $$P(M,G)$$.

• It feels like I understand your last paragraph but not very sure... You are saying transition cocycles gives information of how far $P\times_G F$ us from $M\times F$.... So, you are saying all topological information of $P\times_G F$ is encoded in $P(M,G)$... It looks like I got it.. Need some more time to actually understand... I will respond... Thank you :) – Praphulla Koushik Sep 27 '18 at 8:01

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $$M$$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $$M$$, i. e. when I say "an object $$X$$" I will actually mean a map $$X\to M$$, and when I say "a morphism $$Y\to Z$$" I will actually mean a fibrewise map from $$Y\to M$$ to $$Z\to M$$, etc.

Now for an object $$X$$, there is another one $$\operatorname{Aut}(X)$$, with the property that morphisms from any $$Y$$ to $$\operatorname{Aut}(X)$$ are in one-to-one correspondence with fibrewise automorphisms of $$Y\times_MX\to Y$$ over $$Y$$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $$M$$) to the projection $$M\times G\to M$$ for some "usual" group $$G$$, and then one gets "your" situation.

More generally, for another object $$X'$$, there is still another one $$\operatorname{Iso}(X,X')$$ such that fibrewise maps from any $$Y$$ to $$\operatorname{Iso}(X,X')$$ are in one-to-one correspondence with fibrewise isomorphisms between $$Y\times_MX\to Y$$ and $$Y\times_MX'\to Y$$ over $$Y$$.

Now $$\operatorname{Iso}(X,X')$$ comes with a canonical right $$\operatorname{Aut}(X)$$-action and a canonical left $$\operatorname{Aut}(X')$$-action which commute, and would be a principal left $$\operatorname{Aut}(X')$$- and a principal right $$\operatorname{Aut}(X)$$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $$\operatorname{Iso}(X,X')\to M$$ may happen not to be surjective (in fact it might well happen that $$\operatorname{Iso}(X,X')$$ is empty).

However if that map is surjective, then you may try to reconstruct $$X'$$ from $$X$$ together with $$\operatorname{Iso}(X,X')$$. More precisely it is natural to hope that in this case $$X'$$ is isomorphic to the (fibrewise) quotient of $$\operatorname{Iso}(X,X')\times X$$ by $$\operatorname{Aut}(X)$$; informally speaking, you take pairs $$\left\langle\varphi,x\right\rangle$$ where $$x\in X$$ and $$\varphi:X\to X'$$, and identify $$\left\langle\varphi,\alpha(x)\right\rangle$$ with $$\left\langle\varphi\circ\alpha,x\right\rangle$$ for $$\alpha\in\operatorname{Aut}(X)$$. This is the associated bundle construction.

All this works for sure when $$X$$ is a trivial bundle, i. e. is the projection $$M\times F\to M$$ for some $$F$$; then (in good cases) $$\operatorname{Aut}(X)$$ will be the trivial group bundle $$M\times G\to M$$. Given some structure on $$F$$ (such that a vector space) you might want to further switch to the subgroup consisting of structure preserving automorphisms, and to the subbundle of $$\operatorname{Iso}(X,X')$$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $$E\to M$$ such that for a trivial one $$M\times V\to M$$ the object $$\operatorname{Iso}(M\times V,E)$$ has global support; the latter can then be identified with the frame bundle of $$E$$ once you choose a basis in $$V$$.

I cannot think of an appropriate reference right away but there surely are many, all this is well known at least from 1970ies on...

• Thanks for writing answer. I need some time to understand.. I have read this and got some idea... :) – Praphulla Koushik Sep 27 '18 at 7:58